Chapter
19
Electric Forces and
Electric Fields
Chapter Outline
19.1 Historical Overview
19.2 Properties of Electric Charges
19.3 Insulators and Conductors
19.4 Coulomb’s Law
19.5 Electric Fields
19.6 Electric Field Lines
19.7 Motion of Charged Particles in
a Uniform Electric Field
Courtesy of Resonance Research Corporation
19.8 Electric Flux
19.9 Gauss’s Law
19.10 Application of Gauss’s Law to
Various Charge Distributions
19.11 Conductors in Electrostatic
Equilibrium
19.12 Context Connection: The
Atmospheric Electric Field
SUMMARY
T
his chapter is the first of three on electricity. You are probably familiar with
electrical effects, such as the static cling between articles of clothing removed from the dryer. You may also be familiar with the spark that jumps from
your finger to a doorknob after you have walked across a carpet. Much of your
daily experience involves working with devices that operate on energy transferred to
the device by means of electrical transmission and provided by the electric power
company. Even your own body is an electrochemical machine that uses electricity extensively. Nerves carry impulses as electrical signals, and electric forces are involved
in the flow of materials across cell membranes.
This chapter begins with a review of some of the basic properties of the electrostatic force that we introduced in Chapter 5 as well as some properties of the electric
field associated with stationary charged particles. Our study of electrostatics then
continues with the concept of an electric field that is associated with a continuous
charge distribution and the effect of this field on other charged particles. Once we
Mother and daughter are both enjoying
the effects of electrically charging their
bodies. Each individual hair on their heads
becomes charged and exerts a repulsive
force on the other hairs, resulting in the
“stand-up” hairdos that you see here.
619
620 CHAPTER 19 | Electric Forces and Electric Fields
understand the electric force exerted on a particle, we can include that force in the particle
under a net force model in appropriate situations.
19.1 | Historical Overview
The laws of electricity and magnetism play a central role in the operation of devices
such as cell phones, televisions, electric motors, computers, high-energy particle accelerators, and a host of electronic devices used in medicine. More fundamental,
however, is that the interatomic and intermolecular forces responsible for the formation of solids and liquids are electric in origin. Furthermore, such forces as the
pushes and pulls between objects in contact and the elastic force in a spring arise
from electric forces at the atomic level.
Chinese documents suggest that magnetism was recognized as early as about
2000 b.c. The ancient Greeks observed electric and magnetic phenomena possibly
as early as 700 b.c. They found that a piece of amber, when rubbed, attracted pieces
of straw or feathers. The existence of magnetic forces was known from observations
that pieces of a naturally occurring stone called magnetite (Fe3O4) were attracted to
iron. (The word electric comes from the Greek word for amber, elektron. The word magnetic comes from Magnesia, a city on the coast of Turkey where magnetite was found.)
In 1600, Englishman William Gilbert discovered that electrification was not limited to amber but was a general phenomenon. Scientists went on to electrify a variety
of objects, including people!
It was not until the early part of the 19th century that scientists established that
electricity and magnetism are related phenomena. In 1820, Hans Oersted discovered
that a compass needle, which is magnetic, is deflected when placed near an electric
current. In 1831, Michael Faraday in England and, almost simultaneously, Joseph
Henry in the United States showed that when a wire loop is moved near a magnet
(or, equivalently, when a magnet is moved near a wire loop) an electric current is
observed in the wire. In 1873, James Clerk Maxwell used these observations and
other experimental facts as a basis for formulating the laws of electromagnetism
as we know them today. Shortly thereafter (around 1888), Heinrich Hertz verified
Maxwell’s predictions by producing electromagnetic waves in the laboratory. This
achievement has been followed by such practical developments as radio, television,
cellular telephone systems, Bluetooth™, and Wi-Fi.
Maxwell’s contributions to the science of electromagnetism were especially significant because the laws he formulated are basic to all forms of electromagnetic
phenomena. His work is comparable in importance to Newton’s discovery of the laws
of motion and the theory of gravitation.
© Cengage Learning/Charles D. Winters
19.2 | Properties of Electric Charges
Figure 19.1 Rubbing a balloon
against your hair on a dry day causes
the balloon and your hair to become
electrically charged.
A number of simple experiments demonstrate the existence of electrostatic forces.
For example, after running a comb through your hair, you will find that the comb
attracts bits of paper. The attractive electrostatic force is often strong enough to
suspend the bits. The same effect occurs with other rubbed materials, such as glass
or rubber.
Another simple experiment is to rub an inflated balloon with wool or across your
hair (Fig. 19.1). On a dry day, the rubbed balloon will stick to the wall of a room,
often for hours. When materials behave this way, they are said to have become electrically charged. You can give your body an electric charge by walking across a wool
rug or by sliding across a car seat. You can then feel, and remove, the charge on your
body by lightly touching another person or object. Under the right conditions, a visible spark is seen when you touch and a slight tingle is felt by both parties. (Such an
experiment works best on a dry day because excessive moisture in the air can provide
a pathway for charge to leak off a charged object.)
19.2 | Properties of Electric Charges 621
Experiments also demonstrate that there are two kinds
of electric charge, given the names positive and negative
by Benjamin Franklin (1706–1790). Figure 19.2 illustrates
the interactions of the two kinds of charge. A hard rubber (or plastic) rod that has been rubbed with fur (or an
acrylic material) is suspended by a piece of string. When a
glass rod that has been rubbed with silk is brought near the
rubber rod, the rubber rod is attracted toward the glass rod
(Fig. 19.2a). If two charged rubber rods (or two charged
glass rods) are brought near each other, as in Figure 19.2b,
the force between them is repulsive. This observation demS
onstrates that the rubber and glass have different kinds
F
of charge. We use the convention suggested by Franklin;
S
S
F
the electric charge on the glass rod is called positive and
F
that on the rubber rod is called negative. On the basis of
S
F
such observations, we conclude that charges of the same
a
b
sign repel each other and charges with opposite signs
attract each other.
Figure 19.2 The electric force between (a) oppositely charged
We know that only two kinds of electric charge exist beobjects and (b) like-charged objects.
cause any unknown charge that is found experimentally to
be attracted to a positive charge is also repelled by a negative charge. No one has ever observed a charged object
that is repelled by both a positive and a negative charge or
that is attracted to both.
Attractive electric forces are responsible for the behavior of a wide variety of comElectrical attraction of
contact lenses
mercial products. For example, the plastic in many contact lenses, etafilcon, is made
up of molecules that electrically attract the protein molecules in human tears. These
protein molecules are absorbed and held by the plastic so that the lens ends up being
primarily composed of the wearer’s tears. Therefore, the lens does not behave as a
foreign object to the wearer’s eye and can be worn comfortably. Many cosmetics also
take advantage of electric forces by incorporating materials that are electrically attracted to skin or hair, causing the pigments or other chemicals to stay put once they
are applied.
Another important characteristic of electric charge is that the net charge in an
isolated system is always conserved. This represents the electric charge version of
the isolated system model. We first introduced isolated system models in Chapter 7
when we discussed conservation of energy; we now see a principle of conservation of
electric charge for an isolated system. When two initially neutral objects are charged
by being rubbed together, charge is not created in the process. The objects become
charged because electrons are transferred from one object to the other. One object
gains some amount of negative charge from the electrons transferred to it while
the other loses an equal amount of negative charge and hence is left with a positive
charge. For the isolated system of the two objects, no transfer of charge occurs across
the boundary of the system. The only change is that charge has been transferred
between two members of the system. For example, when a glass rod is rubbed with
silk, as in Figure 19.3, the silk obtains a negative charge that is equal in magnitude to
the positive charge on the glass rod as negatively charged electrons are transferred
from the glass to the silk. Likewise, when rubber is rubbed with fur, electrons are
transferred from the fur to the rubber. An uncharged object contains an enormous
Figure 19.3 When a glass rod
number of electrons (on the order of 1023). For every negative electron, however,
is rubbed with silk, electrons are
a positively charged proton is also present; hence, an uncharged object has no net
transferred from the glass to the
charge of either sign.
silk. Also, because the charges are
Another property of electric charge is that the total charge on an object is quantransferred in discrete bundles, the
charges on the two objects are 6e or
tized as integral multiples of the elementary charge e. We first saw this charge
62e or 63e, and so on.
219
e 5 1.60 3 10
C in Chapter 5. The quantization results because the charge on an
object must be due to an integral number of excess electrons or a deficiency of an
integral number of electrons.
622 CHAPTER 19 | Electric Forces and Electric Fields
a
b
c
d
QUI C K QU IZ 19.1 Three objects are brought close to one another, two at a time.
When objects A and B are brought together, they repel. When objects B and C
are brought together, they also repel. Which of the following statements are true?
(a) Objects A and C possess charges of the same sign. (b) Objects A and C possess
charges of opposite sign. (c) All three objects possess charges of the same sign.
(d) One object is neutral. (e) Additional experiments must be performed to
determine the signs of the charges.
19.3 | Insulators and Conductors
We have discussed the transfer of charge from one object to another. It is also
possible for electric charges to move from one location to another within an object;
such motion of charge is called electrical conduction. It is convenient to classify substances in terms of the ability of charges to move within the substance:
Electrical conductors are materials in which some of the electrons are free electrons1 that are not bound to atoms and can move relatively freely through the
material; electrical insulators are materials in which all electrons are bound to
atoms and cannot move freely through the material.
Materials such as glass, rubber, and dry wood are insulators. When such materials
are charged by rubbing, only the rubbed area becomes charged; the charge does
not tend to move to other regions of the material. In contrast, materials such as copper, aluminum, and silver are good conductors. When such materials are charged in
some small region, the charge readily distributes itself over the entire surface of the
material. If you hold a copper rod in your hand and rub it with wool or fur, it will not
attract a small piece of paper, which might suggest that a metal cannot be charged.
If you hold the copper rod by an insulating handle and then rub, however, the rod
remains charged and attracts the piece of paper. In the first case, the electric charges
produced by rubbing readily move from the copper through your body, which is a
conductor, and finally to the Earth. In the second case, the insulating handle prevents the flow of charge to your hand.
Semiconductors are a third class of materials, and their electrical properties are
somewhere between those of insulators and those of conductors. Charges can move
somewhat freely in a semiconductor, but far fewer charges are moving through a
semiconductor than in a conductor. Silicon and germanium are well-known examples
of semiconductors that are widely used in the fabrication of a variety of electronic devices. The electrical properties of semiconductors can be changed over many orders
of magnitude by adding controlled amounts of certain foreign atoms to the materials.
Charging by Induction
e
Figure 19.4 Charging a metallic
object by induction. (a) A neutral
metallic sphere. (b) A charged
rubber rod is placed near the sphere.
(c) The sphere is grounded. (d) The
ground connection is removed.
(e) The rod is removed.
When a conductor is connected to the Earth by means of a conducting wire or pipe,
it is said to be grounded. For present purposes, the Earth can be modeled as an infinite reservoir for electrons, which means that it can accept or supply an unlimited
number of electrons. In this context, the Earth serves a purpose similar to our energy reservoirs introduced in Chapter 17. With that in mind, we can understand how
to charge a conductor by a process known as charging by induction.
To understand how to charge a conductor by induction, consider a neutral
(uncharged) metallic sphere insulated from the ground as shown in Figure 19.4a.
There are an equal number of electrons and protons in the sphere if the charge
on the sphere is exactly zero. When a negatively charged rubber rod is brought near
1A metal atom contains one or more outer electrons, which are weakly bound to the nucleus. When many atoms combine
to form a metal, the free electrons are these outer electrons, which are not bound to any one atom. These electrons move
about the metal in a manner similar to that of gas molecules moving in a container.
19.3 | Insulators and Conductors 623
the sphere, electrons in the region nearest the rod experience a repulsive force
and migrate to the opposite side of the sphere. This migration leaves the side of
the sphere near the rod with an effective positive charge because of the diminished
number of electrons as in Figure 19.4b. (The left side of the sphere in Figure 19.4b
is positively charged as if positive charges moved into this region, but in a metal it
is only electrons that are free to move.) This migration occurs even if the rod never
actually touches the sphere. If the same experiment is performed with a conducting wire connected from the sphere to the Earth (Fig. 19.4c), some of the electrons
in the conductor are so strongly repelled by the presence of the negative charge in
the rod that they move out of the sphere through the wire and into the Earth. The
symbol
at the end of the wire in Figure 19.4c indicates that the wire is connected
to ground, which means a reservoir such as the Earth. If the wire to ground is then
removed (Fig. 19.4d), the conducting sphere contains an excess of induced positive
charge because it has fewer electrons than it needs to cancel out the positive charge
of the protons. When the rubber rod is removed from the vicinity of the sphere
(Fig. 19.4e), this induced positive charge remains on the ungrounded sphere. Note
that the rubber rod loses none of its negative charge during this process.
Charging an object by induction requires no contact with the object inducing the
charge. This behavior is in contrast to charging an object by rubbing, which does
require contact between the two objects.
A process similar to the first step in charging by induction in conductors takes
place in insulators. In most neutral atoms and molecules, the average position of
the positive charge coincides with the average position of the negative charge. In the
presence of a charged object, however, these positions may shift slightly because of
the attractive and repulsive forces from the charged object, resulting in more positive charge on one side of the molecule than on the other. This effect is known as
polarization. The polarization of individual molecules produces a layer of charge on
the surface of the insulator as shown in Figure 19.5a, in which a charged balloon on
the left is placed against a wall on the right. In the figure, the negative charge layer
in the wall is closer to the positively charged balloon than the positive charges at the
other ends of the molecules. Therefore, the attractive force between the positive and
negative charges is larger than the repulsive force between the positive charges. The
result is a net attractive force between the charged balloon and the neutral insulator.
Your knowledge of induction in insulators should help you explain why a charged
rod attracts bits of electrically neutral paper (Fig. 19.5b) or why a balloon that has
been rubbed against your hair can stick to a neutral wall.
a
© Cengage Learning/Charles D. Winters
b
Figure 19.5 (a) A charged balloon is brought near an insulating wall. (b) A
charged rod is brought close to bits of paper.
624 CHAPTER 19 | Electric Forces and Electric Fields
QUI C K QU IZ 19.2 Three objects are brought close to one another, two at a time.
When objects A and B are brought together, they attract. When objects B and C are
brought together, they repel. Which of the following are necessarily true? (a) Objects
A and C possess charges of the same sign. (b) Objects A and C possess charges of
opposite sign. (c) All three of the objects possess charges of the same sign. (d) One
object is neutral. (e) Additional experiments must be performed to determine information about the charges on the objects.
19.4 | Coulomb’s Law
Electric forces between charged objects were measured quantitatively by Charles Coulomb using the torsion balance, which he invented (Fig. 19.6). Coulomb confirmed
that the electric force between two small charged spheres is proportional to the inverse
square of their separation distance r, that is, Fe ~ 1/r 2. The operating principle of the
torsion balance is the same as that of the apparatus used by Sir Henry Cavendish to
measure the density of the Earth (Section 11.1), with the electrically neutral spheres
replaced by charged ones. The electric force between charged spheres A and B in Figure 19.6 causes the spheres to either attract or repel each other, and the resulting motion causes the suspended fiber to twist. Because the restoring torque of the twisted
fiber is proportional to the angle through which it rotates, a measurement of this angle
provides a quantitative measure of the electric force of attraction or repulsion. Once
the spheres are charged by rubbing, the electric force between them is very large compared with the gravitational attraction, and so the gravitational force can be ignored.
In Chapter 5, we introduced Coulomb’s law, which describes the magnitude of
the electrostatic force between two charged particles with charges q 1 and q 2 and
separated by a distance r :
u q 1 uu q 2 u
Fe 5 k e
19.1b
r2
Figure 19.6 Coulomb’s torsion
balance, which was used to establish
the inverse-square law for the
electrostatic force between two
charges.
where ke (5 8.987 6 3 109 N ? m2/C2) is the Coulomb constant and the force is in
newtons if the charges are in coulombs and if the separation distance is in meters.
The constant ke is also written as
1
ke 5
40
where the constant 0 (Greek letter epsilon), known as the permittivity of free space,
has the value
© INTERFOTO/Alamy
0 5 8.854 2 3 10212 C2/N ? m2
Charles Coulomb
French Physicist (1736–1806)
Coulomb’s major contributions to science
were in the areas of electrostatics and
magnetism. During his lifetime, he also
investigated the strengths of materials and
determined the forces that affect objects
on beams, thereby contributing to the
field of structural mechanics. In the field
of ergonomics, his research provided a
fundamental understanding of the ways
in which people and animals can best do
work.
Note that Equation 19.1 gives only the magnitude of the force. The direction of the
force on a given particle must be found by considering where the particles are located
with respect to one another and the sign of each charge. Therefore, a pictorial representation of a problem in electrostatics is very important in analyzing the problem.
The charge of an electron is q 5 2e 5 21.60 3 10219 C, and the proton has a charge
of q 5 1e 5 1.60 3 10219 C; therefore, 1 C of charge is equal to the magnitude of the
charge of (1.60 3 10219)21 5 6.25 3 1018 electrons. (The elementary charge e was introduced in Section 5.5.) Note that 1 C is a substantial amount of charge. In typical electrostatic experiments, where a rubber or glass rod is charged by friction, a net charge
on the order of 1026 C (5 1 C) is obtained. In other words, only a very small fraction
of the total available electrons (on the order of 1023 in a 1-cm3 sample) are transferred
between the rod and the rubbing material. The experimentally measured values of the
charges and masses of the electron, proton, and neutron are given in Table 19.1.
TABLE 19.1 | Charge and Mass of the Electron, Proton, and Neutron
Particle
Charge (C)
Mass (kg)
Electron (e)
21.602 176 5 3 10219
9.109 4 3 10231
Proton (p)
11.602 176 5 3 10219
1.672 62 3 10227
Neutron (n)
0
1.674 93 3 10227
19.4 | Coulomb’s Law 625
When dealing with Coulomb’s law, remember that
force is a vector quantity and must be treated accordingly. Furthermore, Coulomb’s law applies exactly
only to particles.2 The electrostatic force exerted by q 1
:
on q 2, written F 12, can be expressed in vector form as3
:
F 12 5 ke
q 1q 2
r2
r̂12
S
F
S
F
19.2b
S
F
r
where r̂ 12 is a unit vector directed from q 1 toward q 2
S
as in Active Figure 19.7a. Equation 19.2 can be used
F
to find the direction of the force in space, although a
a
b
carefully drawn pictorial representation is needed to
clearly identify the direction of r̂ 12. From Newton’s third Active Figure 19.7 Two point charges separated by a distance r exert a
:
law, we see that the electric force exerted by q 2 on q 1 is force on each other given by Coulomb’s law. Note that the force F 21 exerted
:
equal in magnitude to the force exerted by q 1 on q 2 and by q 2 on q 1 is equal in magnitude and opposite in direction to the force F 12
:
:
exerted by q 1 on q 2.
in the opposite direction; that is, F 21 5 2F 12. From
Equation 19.2, we see that if q 1 and q 2 have the same sign, the product q 1q 2 is positive
and the force is repulsive as in Active Figure 19.7a. The force on q 2 is in the same direction as r̂ 12 and is directed away from q 1. If q 1 and q 2 are of opposite sign as in Active
Figure 19.7b, the product q 1q 2 is negative and the force is attractive. In this case, the
force on q 2 is in the direction opposite to r̂ 12, directed toward q 1.
When more than two charged particles are present, the force between any pair
is given by Equation 19.2. Therefore, the resultant force on any one particle equals
the vector sum of the individual forces due to all other particles. This principle of
superposition as applied to electrostatic forces is an experimentally observed fact
and simply represents the traditional vector sum of forces introduced in Chapter 4.
As an example, if four charged particles are present, the resultant force on particle
1 due to particles 2, 3, and 4 is given by the vector sum
:
:
:
:
F 1 5 F 21 1 F 31 1 F 41
Q U I CK QUI Z 19.3 Object A has a charge of 12 C, and object B has a charge
of 16 C. Which statement is true about the electric forces on the objects?
:
:
:
:
:
:
:
:
(a) F AB 5 23F BA (b) F AB 5 2F BA (c) 3F AB 5 2F BA (d) F AB 5 3F BA
:
:
:
:
(e) F AB 5 F BA (f) 3F AB 5 F BA
Exampl e 19.1 | Where Is the Net Force Zero?
Three point charges lie along the x axis as shown
in Figure 19.8. The positive charge q 1 5 15.0 C is
at x 5 2.00 m, the positive charge q 2 5 6.00 C is
at the origin, and the net force acting on q 3 is zero.
What is the x coordinate of q 3?
SOLUTION
Figure 19.8 (Example 19.1)
Three point charges are placed
along the x axis. If the resultant
force acting on q 3 is zero, the force
:
F 13 exerted by q 1 on q 3 must be
equal in magnitude and opposite in
:
direction to the force F 23 exerted by
q 2 on q 3.
S
F
S
F
Conceptualize Because q 3 is near two other charges, it experiences two electric forces. The forces lie along the same line as
:
:
indicated in Figure 19.8. Because q 3 is negative and q 1 and q 2 are positive, the forces F 13 and F 23 are both attractive.
Categorize Because the net force on q 3 is zero, we model the point charge as a particle in equilibrium.
Analyze Write an expression for the net force on charge
:
:
:
F 3 5 F 23 1 F 13 5 2ke
q3 when it is in equilibrium:
2Coulomb’s law can also be used for larger objects to which the particle model can be applied.
3Notice that we use “q ” as shorthand notation for “the particle with charge q .” This usage is common when discussing
2
2
charged particles, similar to the use in mechanics of “m 2” for “the particle with mass m 2.” The context of the sentence
will tell you whether the symbol represents an amount of charge or a particle with that charge.
u q2 uu q3 u ˆ
u q1 uu q3 u ˆ
i 1 ke
i50
2
x
(2.00 2 x)2
continued
626 CHAPTER 19 | Electric Forces and Electric Fields
19.1 cont.
ke
Move the second term to the right side of the equation
and set the coefficients of the unit vector î equal:
Eliminate ke and uq3u and rearrange the equation:
u q2 u uq3 u
x2
5 ke
u q1 uu q3 u
(2.00 2 x)2
(2.00 2 x)2uq2u 5 x2uq1u
(4.00 2 4.00x 1 x2)(6.00 3 1026 C) 5 x2(15.0 3 1026 C)
Reduce the quadratic equation to a simpler form:
3.00x 2 1 8.00x 2 8.00 5 0
Solve the quadratic equation for the positive root:
x 5 0.775 m
Finalize The second root to the quadratic equation is x 5 23.44 m. That is another location where the magnitudes of the
forces on q3 are equal, but both forces are in the same direction, so they do not cancel.
Exampl e 19.2 | The Hydrogen Atom
The electron and proton of a hydrogen atom are separated (on the average) by a distance of approximately
5.3 3 10211 m. Find the magnitudes of the electric force and the gravitational force between the two particles.
SOLUTION
Conceptualize Think about the two particles separated by the very small distance given in the problem statement. In Chapter 5, we mentioned that the gravitational force between an electron and a proton is very small compared to the electric
force between them, so we expect this to be the case with the results of this example.
Categorize The electric and gravitational forces will be evaluated from universal force laws, so we categorize this example
as a substitution problem.
u e uu 2e u
(1.60 3 10219 C)2
9 N ? m2/C2)
5
(8.99
3
10
Fe 5 ke
Use Coulomb’s law to find the magnitude of
r2
(5.3 3 10211 m)2
the electric force:
28
5 8.2 3 10 N
Use Newton’s law of universal gravitation and
Table 19.1 (for the particle masses) to find
the magnitude of the gravitational force:
Fg 5 G
m em p
r2
5 (6.67 3 10211 N ? m2/kg2)
(9.11 3 10231 kg)(1.67 3 10227 kg)
(5.3 3 10211 m)2
5 3.6 3 10247 N
The ratio Fe /Fg < 2 3 1039. Therefore, the gravitational force between charged atomic particles is negligible when compared
with the electric force. Notice the similar forms of Newton’s law of universal gravitation and Coulomb’s law of electric forces.
Other than the magnitude of the forces between elementary particles, what is a fundamental difference between the two forces?
Exampl e 19.3 | Find the Charge on the Spheres
Two identical small charged spheres, each having a mass of
3.00 3 1022 kg, hang in equilibrium as shown in Figure 19.9a.
The length L of each string is 0.150 m, and the angle is 5.008.
Find the magnitude of the charge on each sphere.
S
T
u
u u
u
u
S
F
SOLUTION
u
⫹
Conceptualize Figure 19.9a helps us conceptualize this exam-
ple. The two spheres exert repulsive forces on each other. If they
are held close to each other and released, they move outward
from the center and settle into the configuration in Figure 19.9a
after the oscillations have vanished due to air resistance.
Categorize The key phrase “in equilibrium” helps us model
each sphere as a particle in equilibrium. This example is similar
⫹
⫹
g
a
b
Figure 19.9 (Example 19.3) (a) Two identical spheres, each
carrying the same charge q, suspended in equilibrium. (b) Diagram
of the forces acting on the sphere on the left part of (a).
19.5 | Electric Fields 627
19.3 cont.
to the particle in equilibrium problems in Chapter 4 with the added feature that one of the forces on a sphere is an electric
force.
Analyze The force diagram for the left-hand sphere is shown in Figure 19.9b. The sphere is in equilibrium under the ap:
:
plication of the force T from the string, the electric force F e from the other sphere, and the gravitational force m:
g.
o F 5 T sin 2 F 5 0 : T sin 5 F
(2) o F 5 T cos 2 mg 5 0 : T cos 5 mg
(1)
Write Newton’s second law for the left-hand sphere in
component form:
x
e
e
y
Divide Equation (1) by Equation (2) to find Fe :
tan
5
Fe
mg
Use the geometry of the right triangle in Figure 19.9a to
find a relationship between a, L, and :
sin
5
a
L
Solve Coulomb’s law (Eq. 19.1) for the charge uq u on each
sphere:
uq u 5
uq u 5
Substitute numerical values:
:
:
Fer 2
B ke
5
Fe 5 mg tan
a 5 L sin
Fe(2a)2
B
ke
5
mg tan (2L sin )2
B
ke
(3.00 3 1022 kg)(9.80 m/s2) tan (5.008)[2(0.150 m) sin (5.008)]2
B
8.99 3 109 N ? m2/C2
5 4.42 3 1028 C
Finalize If the sign of the charges were not given in Figure 19.9, we could not determine them. In fact, the sign of the
charge is not important. The situation is the same whether both spheres are positively charged or negatively charged.
19.5 | Electric Fields
In Section 4.1, we discussed the differences between contact forces and field forces.
Two field forces—the gravitational force in Chapter 11 and the electric force here—
have been introduced into our discussions so far. As pointed out earlier, field forces can
act through space, producing an effect even when no physical contact occurs between
interacting objects. The gravitational field :
g at a point in space due to a source particle
:
was defined in Section 11.1 to be equal to the gravitational force F g acting on a test par:
:
ticle of mass m divided by that mass: g ; F g /m. The concept of a field was developed
by Michael Faraday (1791–1867) in the context of electric forces and is of such practical
value that we shall devote much attention to it in the next several chapters. In this approach, an electric field is said to exist in the region of space around a charged object,
the source charge. When another charged object—the test charge—enters this electric
field, an electric force acts on it. As an example, consider Figure 19.10, which shows a
small positive test charge q 0 placed near a second object carrying a much greater positive charge Q. We define the electric field due to the source charge at the location of
the test charge to be the electric force on the test charge per unit charge, or, to be more
:
:
specific, the electric field vector E at a point in space is defined as the electric force F e
4
acting on a positive test charge q0 placed at that point divided by the test charge:
:
E;
:
:
Fe
q0
19.3b
:
The vector E has the SI units of newtons per coulomb (N/C). The direction of E
as shown in Figure 19.10 is the direction of the force a positive test charge experi:
ences when placed in the field. Note that E is the field produced by some charge or
charge distribution separate from the test charge; it is not the field produced by the
4When using Equation 19.3, we must assume the test charge q is small enough that it does not disturb the charge distri0
bution responsible for the electric field. If the test charge is great enough, the charge on the metallic sphere is redistributed and the electric field it sets up is different from the field it sets up in the presence of the much smaller test charge.
S
E
Figure 19.10 A small positive test
charge q0 placed at point P near an
object carrying a much larger positive
charge Q experiences an electric
:
field E at point P established by
the source charge Q. We will always
assume that the test charge is so small
that the field of the source charge is
unaffected by its presence.
c Definition of electric field
628 CHAPTER 19 | Electric Forces and Electric Fields
Active Figure 19.11 (a), (c) When
a test charge q0 is placed near a
source charge q, the test charge experiences a force. (b), (d) At a point
P near a source charge q, there exists
an electric field.
S
F
S
r
F
a
r
c
S
E
S
E
r
b
Pitfall Prevention | 19.1
Particles Only
Equation 19.4 is valid only for a
particle of charge q, that is, an object
of zero size. For a charged object of
finite size in an electric field, the field
may vary in magnitude and direction
over the size of the object, so the
corresponding force equation may be
more complicated.
r
d
test charge itself. Also note that the existence of an electric field is a property of its
source; the presence of the test charge is not necessary for the field to exist. The test
charge serves as a detector of the electric field: an electric field exists at a point if a test
charge at that point experiences an electric force.
Once the electric field is known at some point, the force on any particle with charge q
placed at that point can be calculated from a rearrangement of Equation 19.3:
:
:
Fe 5 q E
19.4b
Once the electric force on a particle is evaluated, its motion can be determined from
the particle under a net force model or the particle in equilibrium model (the electric
force may have to be combined with other forces acting on the particle), and the techniques of earlier chapters can be used to find the motion of the particle.
Consider a point charge5 q located a distance r from a test particle with charge q0.
According to Coulomb’s law, the force exerted on the test particle by q is
qq 0
:
F e 5 ke 2 r̂
r
where r̂ is a unit vector directed from q toward q0. This force in Active Figure 19.11a is
directed away from the source charge q. Because the electric field at P, the position of
:
:
the test charge, is defined by E 5 Fe /q0, we find that at P, the electric field created by q is
c Electric field due to a point
charge
q
:
E 5 ke
r2
19.5b
r̂
If the source charge q is positive, Active Figure 19.11b shows the situation with the
test charge removed; the source charge sets up an electric field at point P, directed
away from q. If q is negative as in Active Figure 19.11c, the force on the test charge
is toward the source charge, so the electric field at P is directed toward the source
charge as in Active Figure 19.11d.
To calculate the electric field at a point P due to a group of point charges, we first
calculate the electric field vectors at P individually using Equation 19.5 and then add
them vectorially. In other words, the total electric field at a point in space due to a
group of charged particles equals the vector sum of the electric fields at that point
due to all the particles. This superposition principle applied to fields follows directly
from the vector addition property of forces. Therefore, the electric field at point P
of a group of source charges can be expressed as
c Electric field due to a finite
number of point charges
:
E 5 ke o
qi
i ri
2
r̂i
19.6b
5We have used the phrase “charged particle” so far. The phrase “point charge” is somewhat misleading because charge
is a property of a particle, not a physical entity. It is similar to misleading phrasing in mechanics such as “a mass m is
placed . . .” (which we have avoided) rather than “a particle with mass m is placed. . . .” This phrase is so ingrained in
physics usage, however, that we will use it and hope that this footnote suffices to clarify its use.
19.5 | Electric Fields 629
where ri is the distance from the i th charge qi to the point P (the location at which
the field is to be evaluated) and r̂ i is a unit vector directed from qi toward P.
Q U I CK QUI Z 19.4 A test charge of 13 C is at a point P where an external electric
field is directed to the right and has a magnitude of 4 3 106 N/C. If the test charge
is replaced with another charge of 23 C, what happens to the external electric field
at P ? (a) It is unaffected. (b) It reverses direction. (c) It changes in a way that cannot
be determined.
Exampl e 19.4 | Electric Field of a Dipole
An electric dipole consists of a point charge q and a point charge 2q
separated by a distance of 2a as in Figure 19.12. As we shall see in later
chapters, neutral atoms and molecules behave as dipoles when placed
in an external electric field. Furthermore, many molecules, such as
HCl, are permanent dipoles. (HCl can be effectively modeled as an H1
ion combined with a Cl2 ion.) The effect of such dipoles on the behavior of materials subjected to electric fields is discussed in Chapter 20.
:
(A) Find the electric field E due to the dipole along the y axis at
the point P, which is a distance y from the origin.
SOLUTION
Conceptualize In Example 19.1, we added vector forces to find the
net force on a particle. Here, we add electric field vectors to find the
net electric field at a point in space.
Categorize We have two source charges and wish to find the total elec-
tric field, so we categorize this example as one in which we can use the
superposition principle represented by Equation 19.6.
:
S
E
u
S
E
u
Figure 19.12 (Example
19.4) The total electric
:
field E at P due to
two charges of equal
magnitude and opposite
sign (an electric dipole)
equals the vector sum
:
:
:
E 1 1 E 2 . The field E 1 is
due to the positive charge
:
q, and E 2 is the field due
to the negative charge 2q.
S
E
u
u
:
Analyze At P, the fields E 1 and E 2 due to the two particles are equal in magnitude because P is equidistant from the two
:
:
:
charges. The total field at P is E 5 E 1 1 E 2.
q
q
5 ke 2
r2
y 1 a2
:
:
The y components of E 1 and E 2 are equal in magnitude and opposite in sign, so they cancel. The x components are equal
:
and add because they have the same sign. The total field E is therefore parallel to the x axis.
q
Find an expression for the magnitude of the electric field
(1) E 5 2ke 2
cos y
1
a2
at P :
q
a
E 5 2ke 2
From the geometry in Figure 19.12 we see that
(y 1 a 2) (y 2 1 a 2)1/2
cos 5 a/r 5 a/(y2 1 a2)1/2. Substitute this result
2qa
into Equation (1):
(2) E 5 ke 2
(y 1 a 2)3/2
Find the magnitudes of the fields at P :
E 1 5 E 2 5 ke
(B) Find the electric field for points y .. a far from the dipole.
SOLUTION
Equation (2) gives the value of the electric field on the y
axis at all values of y. For points far from the dipole, for
which y .. a, neglect a2 in the denominator and write
the expression for E in this case:
(3) E < ke
2qa
y3
Finalize Therefore, we see that along the y axis the field of a dipole at a distant point varies as 1/r 3, whereas the more slowly
varying field of a point charge varies as 1/r 2. (Note: In the geometry of this example, r 5 y.) At distant points, the fields of
the two charges in the dipole almost cancel each other. The 1/r 3 variation in E for the dipole is also obtained for a distant
point along the x axis (see Problem 20) and for a general distant point.
630 CHAPTER 19 | Electric Forces and Electric Fields
Electric Field Due to Continuous Charge Distributions
r̂
r̂
r̂
S
S
E
S
E
E
Figure 19.13 The electric field
:
E at P due to a continuous charge
distribution is the vector sum of the
:
fields DE due to all the elements Dq i
of the charge distribution. Three
sample elements are shown.
In most practical situations (e.g., an object charged by rubbing), the average separation between source charges is small compared with their distances from the point
at which the field is to be evaluated. In such cases, the system of source charges can
be modeled as continuous. That is, we imagine that the system of closely spaced discrete charges is equivalent to a total charge that is continuously distributed through
some volume or over some surface.
To evaluate the electric field of a continuous charge distribution, the following
procedure is used. First, we divide the charge distribution into small elements, each
of which contains a small amount of charge Dq as in Figure 19.13. Next, modeling
the element as a point charge, we use Equation 19.5 to calculate the electric field
:
DE at a point P due to one of these elements. Finally, we evaluate the total field at P
due to the charge distribution by performing a vector sum of the contributions of all
the charge elements (i.e., by applying the superposition principle).
The electric field at P in Figure 19.13 due to one element of charge Dq i is given
by
:
DE i 5 ke
Dqi
r̂
ri2 i
where the index i refers to the ith element in the distribution, ri is the distance from
the element to point P, and r̂ i is a unit vector directed from the element toward
:
P. The total electric field E at P due to all elements in the charge distribution is
approximately
:
E < ke o
i
Dqi
ri2
r̂ i
Now, we apply the model in which the charge distribution is continuous, and we let
the elements of charge become infinitesimally small. With this model, the total field
at P in the limit Dqi : 0 becomes
:
E 5 lim ke o
Dqi : 0
i
Dqi
ri2
r̂ i 5 ke
E
dq
r2
r̂
19.7b
where dq is an infinitesimal amount of charge and the integration is over all the
charge creating the electric field. The integration is a vector operation and must
be treated with caution. It can be evaluated in terms of individual components, or
perhaps symmetry arguments can be used to reduce it to a scalar integral. We shall
illustrate this type of calculation with several examples in which we assume that the
charge is uniformly distributed on a line or a surface or throughout some volume.
When performing such calculations, it is convenient to use the concept of a charge
density along with the following notations:
c Volume charge density
c Surface charge density
c Linear charge density
• If a total charge Q is uniformly distributed throughout a volume V, the volume
charge density is defined by
Q
;
19.8b
V
where has units of coulombs per cubic meter.
• If Q is uniformly distributed on a surface of area A, the surface charge density is defined by
Q
;
19.9b
A
where has units of coulombs per square meter.
• If Q is uniformly distributed along a line of length /, the linear charge density is defined by
Q
;
19.10b
,
where has units of coulombs per meter.
19.5 | Electric Fields 631
PRO B LE M -SO LVI N G ST RATEGY: Calculating the Electric Field
The following procedure is recommended for solving
problems that involve the determination of an electric
field due to individual charges or a charge distribution.
1. Conceptualize. Establish a mental representation
of the problem: think carefully about the individual
charges or the charge distribution and imagine what
type of electric field it would create. Appeal to any
symmetry in the arrangement of charges to help you
visualize the electric field.
2. Categorize. Are you analyzing a group of individual
charges or a continuous charge distribution? The
answer to this question tells you how to proceed in
the Analyze step.
3. Analyze.
(a) If you are analyzing a group of individual
charges, use the superposition principle: when
several point charges are present, the resultant
field at a point in space is the vector sum of the
individual fields due to the individual charges
(Eq. 19.6). Be very careful in the manipulation
of vector quantities. It may be useful to review
the material on vector addition in Chapter 1. Example 19.4 demonstrated this procedure.
(b) If you are analyzing a continuous charge distribution, replace the vector sums for evaluating
the total electric field from individual charges
by vector integrals. The charge distribution is
divided into infinitesimal pieces, and the vector
sum is carried out by integrating over the entire
charge distribution (Eq. 19.7). Examples 19.5
and 19.6 demonstrate such procedures.
Consider symmetry when dealing with either
a distribution of point charges or a continuous
charge distribution. Take advantage of any symmetry in the system you observed in the Conceptualize
step to simplify your calculations. The cancellation
of field components perpendicular to the axis in
Example 19.6 is an example of the application of
symmetry.
4. Finalize. Check to see if your electric field expres-
sion is consistent with the mental representation and
if it reflects any symmetry that you noted previously.
Imagine varying parameters such as the distance of
the observation point from the charges or the radius
of any circular objects to see if the mathematical result changes in a reasonable way.
Exampl e 19.5 | The Electric Field Due to a Charged Rod
A rod of length , has a uniform positive charge per unit length and a total charge Q. Calculate the electric field at a point P that
is located along the long axis of the rod and a distance a from one
end (Fig. 19.14).
S
E
ᐉ
SOLUTION
:
Conceptualize The field d E at P due to each segment of charge on
Figure 19.14 (Example 19.5) The electric field at P due
the rod is in the negative x direction because every segment carries a
positive charge.
to a uniformly charged rod lying along the x axis.
Categorize Because the rod is continuous, we are evaluating the field due to a continuous charge distribution rather than
a group of individual charges. Because every segment of the rod produces an electric field in the negative x direction, the
sum of their contributions can be handled without the need to add vectors.
Analyze Let’s assume the rod is lying along the x axis, dx is the length of one small segment, and dq is the charge on that
segment. Because the rod has a charge per unit length , the charge dq on the small segment is dq 5 dx.
dq
dx
Find the magnitude of the electric field at P due to one
dE 5 ke 2 5 ke 2
x
x
segment of the rod having a charge dq :
Find the total field at P using6 Equation 19.7:
E5
E
,1a
a
ke dx
x2
continued
6To carry out integrations such as this one, first express the charge element dq in terms of the other variables in the
integral. (In this example, there is one variable, x, so we made the change dq 5 dx.) The integral must be over scalar
quantities; therefore, express the electric field in terms of components, if necessary. (In this example, the field has only
an x component, so this detail is of no concern.) Then, reduce your expression to an integral over a single variable (or
to multiple integrals, each over a single variable). In examples that have spherical or cylindrical symmetry, the single
variable is a radial coordinate.
632 CHAPTER 19 | Electric Forces and Electric Fields
19.5 cont.
Noting that ke and 5 Q /, are constants and can be removed from the integral, evaluate the integral:
E
E 5 ke ,1a dx
x
a
(1) E 5 ke
1 ,1a
x a
3 4
5 ke 2
2
Q 1
keQ
1
2
5
, a
,1a
a(, 1 a)
1
2
Finalize If a : 0, which corresponds to sliding the bar to the left until its left end is at the origin, then E : `. That repre-
sents the condition in which the observation point P is at zero distance from the charge at the end of the rod, so the field
becomes infinite. We explore large values of a below.
What If? Suppose point P is very far away from the rod. What is the nature of the electric field at such a point?
Answer If P is far from the rod (a .. ,), then , in the denominator of Equation (1) can be neglected and E < keQ /a2. That
is exactly the form you would expect for a point charge. Therefore, at large values of a/,, the charge distribution appears to
be a point charge of magnitude Q ; the point P is so far away from the rod we cannot distinguish that it has a size. The use
of the limiting technique (a/, : `) is often a good method for checking a mathematical expression.
Example 19.6 | The Electric Field of a Uniform Ring of Charge
A ring of radius a carries a uniformly distributed positive total charge Q. Calculate
the electric field due to the ring at a point P
lying a distance x from its center along the
central axis perpendicular to the plane of
the ring (Fig. 19.15a).
S
E
u
u
›
S
S
E
E
SOLUTION
Conceptualize Figure 19.15a shows the elec-
a
b
:
tric field contribution d E at P due to a single
Figure 19.15 (Example 19.6) A uniformly charged ring of radius a. (a) The field
segment of charge at the top of the ring. This
at P on the x axis due to an element of charge dq. (b) The total electric field at P is
along the x axis. The perpendicular component of the field at P due to segment 1 is
field vector can be resolved into components
canceled by the perpendicular component due to segment 2.
dEx parallel to the axis of the ring and dE'
perpendicular to the axis. Figure 19.15b
shows the electric field contributions from two segments on opposite sides of the ring. Because of the symmetry of the situation, the perpendicular components of the field cancel. That is true for all pairs of segments around the ring, so we can
ignore the perpendicular component of the field and focus solely on the parallel components, which simply add.
Categorize Because the ring is continuous, we are evaluating the field due to a continuous charge distribution rather than
a group of individual charges.
Analyze Evaluate the parallel component of an electric
(1) dEx 5 ke
field contribution from a segment of charge dq on the ring:
From the geometry in Figure 19.15a, evaluate cos :
(2) cos
Substitute Equation (2) into Equation (1):
dEx 5 ke
All segments of the ring make the same contribution to
the field at P because they are all equidistant from this
point. Integrate to obtain the total field at P :
Ex 5
E
5
dq
r2
cos
5 ke
dq
a2 1 x 2
cos
x
x
5 2
r
(a 1 x 2)1/2
dq
ke x
x
5 2
dq
a 2 1 x 2 (a 2 1 x 2)1/2
(a 1 x 2)3/2
ke x
ke x
dq 5 2
2
2
3/2
(a 1 x )
(a 1 x 2)3/2
(3) E 5
ke x
Q
2
(a 1 x 2)3/2
E
dq
19.6 | Electric Field Lines 633
19.6 cont.
Finalize This result shows that the field is zero at x 5 0. Is that consistent with the symmetry in the problem? Furthermore,
notice that Equation (3) reduces to keQ /x 2 if x .. a, so the ring acts like a point charge for locations far away from the ring.
What If? Suppose a negative charge is placed at the center of the ring in Figure 19.15 and displaced slightly by a distance
x ,, a along the x axis. When the charge is released, what type of motion does it exhibit?
Answer In the expression for the field due to a ring of charge, let x ,, a, which results in
Ex 5
keQ
a3
x
Therefore, from Equation 19.4, the force on a charge 2q placed near the center of the ring is
Fx 5 2
ke qQ
a3
x
Because this force has the form of Hooke’s law (Eq. 12.1), the motion of the negative charge is simple harmonic!
19.6 | Electric Field Lines
A convenient specialized pictorial representation for visualizing electric field patterns is created by drawing lines showing the direction of the electric field vector at
any point. These lines, called electric field lines, are related to the electric field in
any region of space in the following manner:
:
• The electric field vector E is tangent to the electric field line at each point.
• The number of electric field lines per unit area through a surface that is perpendicular to the lines is proportional to the magnitude of the electric field in that
region. Therefore, E is large where the field lines are close together and small
where they are far apart.
These properties are illustrated in Figure 19.16. The density of lines through surFigure 19.16 Electric field lines
face A is greater than the density of lines through surface B. Therefore, the magnipenetrating two surfaces.
tude of the electric field on surface A is larger than on surface B. Furthermore, the
field drawn in Figure 19.16 is nonuniform because the lines at different locations
point in different directions.
Some representative electric field lines for a single positive point charge are
shown in Figure 19.17a. Note that in this two-dimensional drawing we show only
the field lines that lie in the plane of the page.
The lines are actually directed radially outward
in all directions from the charge, somewhat like
the needles of a porcupine. Because a positively
charged test particle placed in this field would
be repelled by the charge q, the lines are directed
radially away from q. Similarly, the electric field lines
for a single negative point charge are directed toward
the charge (Fig. 19.17b). In either case, the lines are
radial and extend to infinity. Note that the lines are
closer together as they come nearer to the charge,
indicating that the magnitude of the field is increasing. The electric field lines end in Figure 19.17a and
begin in Figure 19.17b on hypothetical charges we assume to be located infinitely far away.
Is this visualization of the electric field in terms of
a
b
field lines consistent with Equation 19.5? To answer
this question, consider an imaginary spherical sur- Figure 19.17 The electric field lines for a point charge. Notice that the
face of radius r, concentric with the charge. From figures show only those field lines that lie in the plane of the page.
634 CHAPTER 19 | Electric Forces and Electric Fields
Pitfall Prevention | 19.2
Electric Field Lines Are Not Paths
of Particles
Electric field lines represent the field
at various locations. Except in very
special cases, they do not represent the
path of a charged particle released in
an electric field.
symmetry, we see that the magnitude of the electric field is the same everywhere
on the surface of the sphere. The number of lines N emerging from the charge
is equal to the number penetrating the spherical surface. Hence, the number of
lines per unit area on the sphere is N/4r 2 (where the surface area of the sphere
is 4r 2). Because E is proportional to the number of lines per unit area, we see that
E varies as 1/r 2. This result is consistent with that obtained from Equation 19.5;
that is, E 5 keq/r 2.
The rules for drawing electric field lines for any charge distribution are as follows:
• The lines must begin on a positive charge and terminate on a negative charge.
In the case of an excess of one type of charge, some lines will begin or end infinitely far away.
• The number of lines drawn leaving a positive charge or approaching a negative
charge is proportional to the magnitude of the charge.
• No two field lines can cross.
Figure 19.18 The electric field
lines for two charges of equal
magnitude and opposite sign (an
electric dipole).
Figure 19.19 The electric field
lines for two positive point charges.
(The locations A, B, and C are
discussed in Quick Quiz 19.5.)
Because charge is quantized, the number of lines leaving any positively charged
object must be 0, ae, 2ae, . . . , where a is an arbitrary (but fixed) proportionality
constant chosen by the person drawing the lines. Once a is chosen, the number of
lines is no longer arbitrary. For example, if object 1 has charge Q 1 and object 2 has
charge Q 2, the ratio of the number of lines connected to object 2 to those connected
to object 1 is N2/N1 5 Q 2/Q 1.
The electric field lines for two point charges of equal magnitude but opposite
signs (the electric dipole) are shown in Figure 19.18. In this case, the number of
lines that begin at the positive charge must equal the number that terminate at the
negative charge. At points very near the charges, the lines are nearly radial. The high
density of lines between the charges indicates a region of strong electric field. The
attractive nature of the force between the particles is also suggested by Figure 19.18,
with the lines from one particle ending on the other particle.
Figure 19.19 shows the electric field lines in the vicinity of two equal positive point
charges. Again, close to either charge the lines are nearly radial. The same number
of lines emerges from each particle, because the charges are equal in magnitude,
and end on hypothetical charges infinitely far away. At great distances from the particles, the field is approximately equal to that of a single point charge of magnitude
2q. The repulsive nature of the electric force between particles of like charge is suggested in the figure in that no lines connect the particles and that the lines bend
away from the region between the charges.
Finally, we sketch the electric field lines associated with a positive point charge
12q and a negative point charge 2q in Active Figure 19.20. In this case, the number
of lines leaving 12q is twice the number terminating on 2q. Hence, only half the
lines that leave the positive charge end at the negative charge. The remaining half
terminate on hypothetical negative charges infinitely far away. At large distances
from the particles (large compared with the particle separation), the electric field
lines are equivalent to those of a single point charge 1q.
QUI C K QU IZ 19.5 Rank the magnitudes of the electric field at points A, B, and C in
Figure 19.19 (greatest magnitude first).
19.7 | Motion of Charged Particles
in a Uniform Electric Field
:
Active Figure 19.20 The electric
field lines for a point charge 12q and
a second point charge 2q.
When a particle of charge q and mass m is placed in an electric field E , the electric
:
:
force exerted on the charge is given by Equation 19.4, F e 5 qE . If this force is the
only force exerted on the particle, it is the net force. If other forces also act on
the particle, the electric force is simply added to the other forces vectorially to
determine the net force. According to the particle under a net force model from
19.7 | Motion of Charged Particles in a Uniform Electric Field 635
Chapter 4, the net force causes the particle to accelerate. If the electric force is the
only force on the particle, Newton’s second law applied to the particle gives
:
:
a
F e 5 qE 5 m:
The acceleration of the particle is therefore
:
:
a 5
qE
m
19.11b
:
If E is uniform (i.e., constant in magnitude and direction), the acceleration is constant and the particle under constant acceleration analysis model can be used to
describe the motion of the particle. If the particle has a positive charge, its acceleration is in the direction of the electric field. If the particle has a negative charge, its
acceleration is in the direction opposite the electric field.
Exampl e 19.7 | An Accelerating Positive Charge: Two Models
S
:
E
A uniform electric field E is directed along the x axis between parallel
plates of charge separated by a distance d as shown in Figure 19.21. A
positive point charge q of mass m is released from rest at a point 𝖠 next to
the positive plate and accelerates to a point 𝖡 next to the negative plate.
(A) Find the speed of the particle at 𝖡 by modeling it as a particle
under constant acceleration.
S
v
훽
S
v
훾
SOLUTION
Conceptualize When the positive charge is placed at 𝖠, it experiences
an electric force toward the right in Figure 19.21 due to the electric field
directed toward the right.
Categorize Because the electric field is uniform, a constant electric
force acts on the charge. Therefore, as suggested in the problem statement, the point charge can be modeled as a charged particle under constant acceleration.
Analyze Use Equation 2.14 to express the velocity of the
Figure 19.21 (Example
19.7) A positive point
charge q in a uniform
:
electric field E undergoes
constant acceleration in
the direction of the field.
vf 2 5 vi2 1 2a(xf 2 xi) 5 0 1 2a(d 2 0) 5 2ad
particle as a function of position:
Solve for vf and substitute for the magnitude of the acceleration from Equation 19.11:
vf 5 "2ad 5
qE
2qEd
2
d5
B m
B m
1 2
(B) Find the speed of the particle at 𝖡 by modeling it as a nonisolated system.
SOLUTION
Categorize The problem statement tells us that the charge is a nonisolated system. Energy is transferred to this charge by
work done by the electric force exerted on the charge. The initial configuration of the system is when the particle is at 𝖠,
and the final configuration is when it is at 𝖡.
Analyze Write the appropriate reduction of the conservation of energy equation, Equation 7.2, for the system of
the charged particle:
W 5 DK
Replace the work and kinetic energies with values appropriate for this situation:
Fe Dx 5 K 𝖡 2 K 𝖠 5 12mvf 2 2 0 : vf 5
Substitute for the electric force Fe and the displacement Dx:
vf 5
2(qE)(d)
2qEd
5
m
B
B m
Finalize The answer to part (B) is the same as that for part (A), as we expect.
2Fe Dx
B m
636 CHAPTER 19 | Electric Forces and Electric Fields
Example 19.8 | An Accelerated Electron
An electron enters the region of a uniform electric field as shown
in Active Figure 19.22, with vi 5 3.00 3 106 m/s and E 5 200 N/C.
The horizontal length of the plates is , 5 0.100 m.
S
E
(A) Find the acceleration of the electron while it is in the electric field.
ᐉ
SOLUTION
i
Conceptualize This example differs from the preceding one because
the velocity of the charged particle is initially perpendicular to the
electric field lines. (In Example 19.7, the velocity of the charged particle is always parallel to the electric field lines.) As a result, the electron
in this example follows a curved path as shown in Active Figure 19.22.
S
E
Categorize Because the electric field is uniform, a constant electric
force is exerted on the electron. To find the acceleration of the electron, we can model it as a particle under a net force.
S
v
Active Figure 19.22 (Example 19.8) An electron is
projected horizontally into a uniform electric field produced by two charged plates.
Analyze The direction of the electron’s acceleration is downward in
Active Figure 19.22, opposite the direction of the electric field lines.
The particle under a net force model was used to develop
Equation 19.11 in the case in which the electric force on
a particle is the only force. Use this equation to evaluate
the y component of the acceleration of the electron:
ay 5 2
eE
me
Substitute numerical values:
ay 5 2
(1.60 3 10219 C)(200 N/C)
5 23.51 3 1013 m/s2
9.11 3 10231 kg
(B) Assuming the electron enters the field at time t 5 0, find the time at which it leaves the field.
SOLUTION
Categorize Because the electric force acts only in the vertical direction in Active Figure 19.22, the motion of the particle
in the horizontal direction can be analyzed by modeling it as a particle under constant velocity.
Analyze Solve Equation 2.5 for the time at which the elec-
xf 5 xi 1 vx t : t 5
tron arrives at the right edges of the plates:
Substitute numerical values:
t5
xf 2 xi
vx
,20
0.100 m
5
5 3.33 3 1028 s
vx
3.00 3 106 m/s
Finalize We have neglected the gravitational force acting on the electron, which represents a good approximation when
dealing with atomic particles. For an electric field of 200 N/C, the ratio of the magnitude of the electric force eE to the
magnitude of the gravitational force mg is on the order of 1012 for an electron and on the order of 109 for a proton.
19.8 | Electric Flux
S
E
Figure 19.23 Field lines of a
uniform electric field penetrating a
plane of area A perpendicular to the
field. The electric flux FE through
this area is equal to EA.
Now that we have described the concept of electric field lines qualitatively, let us use
a new concept, electric flux, to approach electric field lines on a quantitative basis.
Electric flux is a quantity proportional to the number of electric field lines penetrating some surface. (We can define only a proportionality because the number of lines
we choose to draw is arbitrary.)
First consider an electric field that is uniform in both magnitude and direction
as in Figure 19.23. The field lines penetrate a plane rectangular surface of area A,
which is perpendicular to the field. Recall that the number of lines per unit area is
proportional to the magnitude of the electric field. The number of lines penetrating
the surface of area A is therefore proportional to the product EA. The product of the
19.8 | Electric Flux 637
electric field magnitude E and a surface area A perpendicular to the field is called
the electric flux FE :
›
19.12b
FE 5 EA
From the SI units of E and A, we see that electric flux has the units N ? m2/C.
If the surface under consideration is not perpendicular to the field, the number
of lines through it must be less than that given by Equation 19.12. This concept
can be understood by considering Figure 19.24, where the normal to the surface of
area A is at an angle of to the uniform electric field. Note that the number of lines
that cross this area is equal to the number that cross the projected area A', which is
perpendicular to the field. From Figure 19.24, we see that the two areas are related
by A' 5 A cos . Because the flux through area A equals the flux through A', we
conclude that the desired flux is
FE 5 EA cos 19.13b
From this result, we see that the flux through a surface of fixed area has the maximum value EA when the angle between the normal to the surface and the electric
field is zero. This situation occurs when the normal is parallel to the field and the
surface is perpendicular to the field. The flux is zero when the surface is parallel to
the field because the angle in Equation 19.13 is then 908.
In more general situations, the electric field may vary in both magnitude and
direction over the surface in question. Unless the field is uniform, our definition
of flux given by Equation 19.13 therefore has meaning only over a small element of
area. Consider a general surface divided up into a large number of small elements,
each of area DA. The variation in the electric field over the element can be ignored
:
if the element is small enough. It is convenient to define a vector DAi whose magnitude represents the area of the i th element and whose direction is defined to be
perpendicular to the surface as in Figure 19.25. The electric flux DFE through this
small element is
:
u
u
›
S
E
u
Figure 19.24 Field lines
representing a uniform electric field
penetrating an area A that is at an
angle to the field.
u
S
A
S
E
u
S
A
:
DFE 5 Ei DAi cos i 5 E i ? DAi
where we have used the definition of the scalar product of two vectors
: :
(A ? B 5 AB cos ). By summing the contributions of all elements, we obtain the
total flux through the surface. If we let the area of each element approach zero, the
number of elements approaches infinity and the sum is replaced by an integral. The
general definition of electric flux is therefore
:
:
FE ; lim o E i ? DAi 5
DAi : 0
E
:
:
E ? dA
19.14b
surface
Equation 19.14 is a surface integral, which must be evaluated over the surface in
question. In general, the value of FE depends both on the field pattern and on the
specified surface.
We shall often be interested in evaluating electric flux through a closed surface.
A closed surface is defined as one that completely divides space into an inside region and an outside region so that movement cannot take place from one region
to the other without penetrating the surface. This definition is similar to that of
the system boundary in system models, in which the boundary divides space into
a region inside the system and the outer region, the environment. The surface
of a sphere is an example of a closed surface, whereas a drinking glass is an open
surface.
Consider the closed surface in Active Figure 19.26 (page 638). Note that the vec:
tors DAi point in different directions for the various surface elements. At each point,
these vectors are perpendicular to the surface and, by convention, always point outward
:
from the inside region. At the element labeled ➀, E is outward and i , 908; hence,
:
:
the flux DFE 5 E ? DAi through this element is positive. For element ➁, the field
:
lines graze the surface (perpendicular to the vector DAi ); therefore, i 5 908 and the
Figure 19.25 A small element of a
surface of area DAi .
c Electric flux
638 CHAPTER 19 | Electric Forces and Electric Fields
flux is zero. For elements such as ➂, where the field lines are
crossing the surface from the outside to the inside, 1808 . i .
908 and the flux is negative because cos i is negative.
The net flux through the surface is proportional to the net
number of lines penetrating the surface, where the net number
means the number leaving the volume surrounded by the surface
minus the number entering the volume. If more lines are leaving
the surface than entering, the net flux is positive. If more lines
enter than leave the surface, the net flux is negative. Using the
symbol r to represent an integral over a closed surface, we can
write the net flux FE through a closed surface as
쩸
쩺
쩹
FE 5
S
R
:
:
E ? dA 5
R
19.15b
En dA
A
S
A
S
u
E
S
E
쩹
쩺
u
S
E
쩸
S
A
where En represents the component of the electric field normal
to the surface.
Evaluating the net flux through a closed surface can be very
cumbersome. If the field is perpendicular or parallel to the surface at each point and constant in magnitude, however, the calculation is straightforward. The following example illustrates this
point.
Active Figure 19.26 A closed surface in an electric field. The area vectors
are, by convention, normal to the surface and point outward.
Example 19.9 | Flux Through a Cube
:
Consider a uniform electric field E oriented in the x direction in
empty space. A cube of edge length , is placed in the field, oriented as
shown in Figure 19.27. Find the net electric flux through the surface of
the cube.
S
A
쩺
ᐉ
S
S
E
A
SOLUTION
S
ᐉ
Conceptualize Examine Figure 19.27 carefully. Notice that the electric
field lines pass through two faces perpendicularly and are parallel to four
other faces of the cube.
A
쩸
ᐉ
Categorize We evaluate the flux from its definition, so we categorize this
쩻
example as a substitution problem.
The flux through four of the faces (➂, ➃, and the unnumbered faces)
:
is zero because E is parallel to the four faces and therefore perpendicular
:
to d A on these faces.
Write the integrals for the net flux through faces ➀ and ➁:
:
:
For face ➀, E is constant and directed inward but d A1 is directed outward ( 5 1808). Find the flux through this face:
:
For face ➁, E is constant and outward and in the same
:
direction as d A2 ( 5 08). Find the flux through this face:
Find the net flux by adding the flux over all six faces:
FE 5
E
E
:
E
:
:
E ? dA 5
1
2
in the shape of a cube in a uniform electric field
oriented parallel to the x axis. Side ➃ is the bottom
of the cube, and side ➀ is opposite side ➁.
:
E
E
E
:
:
E ? dA
2
E (cos 1808) dA 5 2E
1
:
:
E ? dA 5
쩹
Figure 19.27 (Example 19.9) A closed surface
E ? dA 1
1
S
A
2
E
dA 5 2EA 5 2E,2
1
E (cos 08) dA 5 E
E
2
dA 5 1EA 5 E ,2
FE 5 2E ,2 1 E ,2 1 0 1 0 1 0 1 0 5 0
19.9 | Gauss’s Law 639
19.9 | Gauss’s Law
In this section, we describe a general relation between the net electric flux through a
closed surface and the charge enclosed by the surface. This relation, known as Gauss’s
law, is of fundamental importance in the study of electrostatic fields.
First, let us consider a positive point charge q located at the center of a spherical
surface of radius r as in Figure 19.28. The field lines radiate outward and hence are
perpendicular (or normal) to the surface at each point. That is, at each point on the
:
:
surface, E is parallel to the vector DAi representing the local element of area DAi.
Therefore, at all points on the surface,
:
S
E
S
A
:
E ? DAi 5 En DAi 5 E DAi
and, from Equation 19.15, we find that the net flux through the surface is
FE 5
R
En dA 5
R
E dA 5 E
R
dA 5 EA
because E is constant over the surface. From Equation 19.5, we know that the magnitude of the electric field everywhere on the surface of the sphere is E 5 keq/r 2.
Furthermore, for a spherical surface, A 5 4r 2 (the surface area of a sphere). Hence,
the net flux through the surface is
FE 5 EA 5
ke q
Figure 19.28 A spherical surface
of radius r surrounding a point
charge q.
1 r 2 (4r ) 5 4k q
2
2
e
Recalling that ke 5 1/40, we can write this expression in the form
q
FE 5
19.16b
0
This result, which is independent of r, says that the net flux through a spherical
surface is proportional to the charge q at the center inside the surface. This result
mathematically represents that (1) the net flux is proportional to the number of
field lines, (2) the number of field lines is proportional to the charge inside the
surface, and (3) every field line from the charge must pass through the surface.
That the net flux is independent of the radius is a consequence of the inverse-square
dependence of the electric field given by Equation 19.5. That is, E varies as 1/r 2,
but the area of the sphere varies as r 2. Their combined effect produces a flux that is
independent of r.
Now consider several closed surfaces surrounding a charge q as in Figure 19.29.
Surface S1 is spherical, whereas surfaces S 2 and S 3 are nonspherical. The flux that
passes through surface S1 has the value q /0. As we discussed in Section 19.8, the flux
is proportional to the number of electric field lines passing through that surface. The
construction in Figure 19.29 shows that the number of electric field lines through
the spherical surface S1 is equal to the number of electric field lines through the
nonspherical surfaces S2 and S3. It is therefore reasonable to conclude that the net
flux through any closed surface is independent of the shape of that surface. (One
can prove that conclusion using E ~ 1/r 2.) In fact,
Figure 19.29 Closed surfaces of
various shapes surrounding a positive
charge.
the net flux through any closed surface surrounding the point charge q is given
by q/0 and is independent of the position of the charge within the surface.
Now consider a point charge located outside a closed surface of arbitrary shape as
in Figure 19.30. As you can see from this construction, electric field lines enter the
surface and then leave it. Therefore, the number of electric field lines entering
the surface equals the number leaving the surface. Consequently, we conclude
that the net electric flux through a closed surface that surrounds no net charge is
zero. If we apply this result to Example 19.9, we see that the net flux through the
cube is zero because there was no charge inside the cube. If there were charge in
the cube, the electric field could not be uniform throughout the cube as specified
in the example.
Figure 19.30 A point charge
located outside a closed surface.
640 CHAPTER 19 | Electric Forces and Electric Fields
Let us extend these arguments to the generalized case of many point charges. We
shall again make use of the superposition principle. That is, we can express the net
flux through any closed surface as
R
:
E ? dA 5
R
:
:
:
(E 1 1 E 2 1 ? ? ?) ? dA
:
:
Active Figure 19.31 The net electric flux through any closed surface
depends only on the charge inside
that surface. The net flux through
surface S is q l/0, the net flux through
surface S9 is (q 2 1 q 3)/0, and the net
flux through surface S 0 is zero.
Pitfall Prevention | 19.3
Zero Flux Is Not Zero Field
In two situations, there is zero flux
through a closed surface: either
(1) there are no charged particles
enclosed by the surface or (2) there
are charged particles enclosed, but
the net charge inside the surface is
zero. For either situation, it is incorrect
to conclude that the electric field on
the surface is zero. Gauss’s law states
that the electric flux is proportional
to the enclosed charge, not the
electric field.
:
:
where E is the total electric field at any point on the surface and E 1, E 2, . . . are
the fields produced by the individual charges at that point. Consider the system of
charges shown in Active Figure 19.31. The surface S surrounds only one charge, q1;
hence, the net flux through S is q1/0. The flux through S due to the charges outside
it is zero because each electric field line from these charges that enters S at one point
leaves it at another. The surface S9 surrounds charges q2 and q3; hence, the net flux
through S9 is (q2 1 q3)/0. Finally, the net flux through surface S 0 is zero because no
charge exists inside this surface. That is, all electric field lines that enter S 0 at one
point leave S 0 at another. Notice that charge q4 does not contribute to the net flux
through any of the surfaces because it is outside all the surfaces.
Gauss’s law, which is a generalization of the foregoing discussion, states that the
net flux through any closed surface is
FE 5
R
:
:
E ? dA 5
q in
0
19.17b
:
where q in represents the net charge inside the surface and E represents the electric field
at any point on the surface. In words, Gauss’s law states that the net electric flux
through any closed surface is equal to the net charge inside the surface divided by
0. The closed surface used in Gauss’s law is called a gaussian surface.
Gauss’s law is valid for the electric field of any system of charges or continuous
distribution of charge. In practice, however, the technique is useful for calculating
the electric field only in situations where the degree of symmetry is high. As we shall
see in the next section, Gauss’s law can be used to evaluate the electric field for
charge distributions that have spherical, cylindrical, or plane symmetry. We do so by
:
choosing an appropriate gaussian surface that allows E to be removed from the integral in Gauss’s law and performing the integration. Note that a gaussian surface is a
mathematical surface and need not coincide with any real physical surface.
QUI C K QU IZ 19.6 If the net flux through a gaussian surface is zero, the following
four statements could be true. Which of the statements must be true? (a) There are
no charges inside the surface. (b) The net charge inside the surface is zero. (c) The
electric field is zero everywhere on the surface. (d) The number of electric field
lines entering the surface equals the number leaving the surface.
QUI C K QU IZ 19.7 Consider the charge distribution shown in Active Figure 19.31.
(i) What are the charges contributing to the total electric flux through surface S9?
(a) q 1 only (b) q 4 only (c) q 2 and q 3 (d) all four charges (e) none of the charges
(ii) What are the charges contributing to the total electric field at a chosen point on the
surface S9? (a) q 1 only (b) q 4 only (c) q 2 and q 3 (d) all four charges (e) none
of the charges
THINKING PHYSICS 19.1
A spherical gaussian surface surrounds a point charge q.
Describe what happens to the net flux through the surface
if (a) the charge is tripled, (b) the volume of the sphere is
doubled, (c) the surface is changed to a cube, and (d) the
charge is moved to another location inside the surface.
Reasoning (a) If the charge is tripled, the flux through the
surface is also tripled because the net flux is proportional
to the charge inside the surface. (b) The net flux remains
constant when the volume changes because the surface
surrounds the same amount of charge, regardless of its volume. (c) The net flux does not change when the shape of
the closed surface changes. (d) The net flux through the
closed surface remains unchanged as the charge inside the
surface is moved to another location as long as the new
location remains inside the surface. b
19.10 | Application of Gauss’s Law to Various Charge Distributions 641
19.10 | Application of Gauss’s Law to Various
Charge Distributions
As mentioned earlier, Gauss’s law is useful in determining electric fields when the
charge distribution has a high degree of symmetry. The following examples show
ways of choosing the gaussian surface over which the surface integral given by Equation 19.17 can be simplified and the electric field determined. The surface should
always be chosen to take advantage of the symmetry of the charge distribution so
that we can remove E from the integral and solve for it. The crucial step in applying Gauss’s law is to determine a useful gaussian surface. Such a surface should be
a closed surface for which each portion of the surface satisfies one or more of the
following conditions:
1. The value of the electric field can be argued by symmetry to be constant over
the portion of the surface.
2. The dot product in Equation 19.17 can be expressed as a simple algebraic
:
:
product E dA because E and dA are parallel.
:
:
3. The dot product in Equation 19.17 is zero because E and dA are
perpendicular.
4. The electric field is zero over the portion of the surface.
Note that different portions of the gaussian surface can satisfy different conditions as
long as every portion satisfies at least one condition. We will see all four of these conditions used in the examples and discussions in the remainder of this chapter. If the
charge distribution does not have sufficient symmetry such that a gaussian surface
that satisfies these conditions can be found, Gauss’s law is not useful for determining
the electric field for that charge distribution.
Exampl e 19.10 | A Spherically Symmetric Charge Distribution
An insulating solid sphere of radius a has a uniform
volume charge density and carries a total positive charge
Q (Fig. 19.32).
(A) Calculate the magnitude of the electric field at a
point outside the sphere.
SOLUTION
Conceptualize Notice how this problem differs from
our previous discussion of Gauss’s law. The electric field
due to point charges was discussed in Section 19.9. Now
we are considering the electric field due to a distribution
of charge. We found the field for various distributions of
charge in Section 19.5 by integrating over the distribution.
This example demonstrates a difference from our discussions in Section 19.5. In this example, we find the electric
field using Gauss’s law.
Categorize Because the charge is distributed uniformly
a
b
Figure 19.32 (Example 19.10) A uniformly charged insulating
sphere of radius a and total charge Q. In diagrams such as this one,
the dotted line represents the intersection of the gaussian surface
with the plane of the page.
throughout the sphere, the charge distribution has spherical symmetry and we can apply Gauss’s law to find the electric field.
Analyze To reflect the spherical symmetry, let’s choose a spherical gaussian surface of radius r, concentric with the sphere,
:
:
as shown in Figure 19.32a. For this choice, condition (2) is satisfied everywhere on the surface and E ? dA 5 E dA.
Q
:
:
:
:
Replace E ? d A in Gauss’s law with E dA:
FE 5 E ? d A 5 E dA 5
0
continued
R
R
642 CHAPTER 19 | Electric Forces and Electric Fields
19.10 cont.
By symmetry, E is constant everywhere on the surface,
which satisfies condition (1), so we can remove E from
the integral:
Solve for E:
R
E dA 5 E
(1) E 5
R
dA 5 E (4r 2) 5
Q
40r 2
5 ke
Q
Q
0
(for r
r2
a)
Finalize This field is identical to that for a point charge. Therefore, the electric field due to a uniformly charged sphere in
the region external to the sphere is equivalent to that of a point charge located at the center of the sphere.
(B) Find the magnitude of the electric field at a point inside the sphere.
SOLUTION
Analyze In this case, let’s choose a spherical gaussian surface having radius r , a, concentric with the insulating sphere
(Fig. 19.32b). Let V 9 be the volume of this smaller sphere. To apply Gauss’s law in this situation, recognize that the charge
q in within the gaussian surface of volume V 9 is less than Q.
Calculate q in by using q in5 V 9:
Notice that conditions (1) and (2) are satisfied everywhere on the gaussian surface in Figure 19.32b. Apply
Gauss’s law in the region r , a:
q in 5 V 9 5 143r 32
R
E dA 5 E
q in
Solve for E and substitute for q in:
E5
Substitute 5 Q /43a3 and 0 5 1/4ke :
(2) E 5
40r
R
5
2
dA 5 E (4r 2) 5
143r 32
40r 2
Q /43a3
3(1/4ke)
5
r 5 ke
q in
0
r
30
Q
a3
r (for r
a)
Finalize This result for E differs from the one obtained in part (A). It shows that E : 0 as r : 0. Therefore, the result eliminates the problem that would exist at r 5 0 if E varied as 1/r 2 inside the sphere as it does outside the sphere. That is, if E ~
1/r 2 for r , a, the field would be infinite at r 5 0, which is physically impossible. Notice also that Equations (1) and (2) both
give the same value of the field at the surface of the sphere (r 5 a), showing that the field is continuous.
Example 19.11 | A Cylindrically Symmetric Charge Distribution
Find the electric field a distance r from a line of positive
charge of infinite length and constant charge per unit
length (Fig. 19.33a).
SOLUTION
S
E
Conceptualize The line of charge is infinitely long. There-
fore, the field is the same at all points equidistant from
the line, regardless of the vertical position of the point in
Figure 19.33a.
Analyze The symmetry of the charge distribution requires
:
that E be perpendicular to the line charge and directed
outward as shown in Figure 19.33b. To reflect the symmetry
E
A
Categorize Because the charge is distributed uniformly
along the line, the charge distribution has cylindrical
symmetry and we can apply Gauss’s law to find the electric
field.
S
S
ᐉ
a
b
Figure 19.33 (Example 19.11) (a) An infinite line of charge
surrounded by a cylindrical gaussian surface concentric with the line.
(b) An end view shows that the electric field at the cylindrical surface
is constant in magnitude and perpendicular to the surface.
19.10 | Application of Gauss’s Law to Various Charge Distributions 643
19.11 cont.
of the charge distribution, let’s choose a cylindrical gaussian surface of radius r and length , that is coaxial with the line
:
charge. For the curved part of this surface, E is constant in magnitude and perpendicular to the surface at each point, sat:
isfying conditions (1) and (2). Furthermore, the flux through the ends of the gaussian cylinder is zero because E is parallel
to these surfaces. That is the first application we have seen of condition (3).
:
:
We must take the surface integral in Gauss’s law over the entire gaussian surface. Because E ? d A is zero for the flat ends
of the cylinder, however, we restrict our attention to only the curved surface of the cylinder.
R
:
:
E ? dA 5 E
Apply Gauss’s law and conditions (1) and (2) for the
curved surface, noting that the total charge inside our
gaussian surface is ,:
FE 5
Substitute the area A 5 2r , of the curved surface:
E (2r ,) 5
Solve for the magnitude of the electric field:
E5
R
dA 5 EA 5
q in
,
5
0
0
,
0
5 2ke
r
20r
19.18b
Finalize This result shows that the electric field due to a cylindrically symmetric charge distribution varies as 1/r, whereas
the field external to a spherically symmetric charge distribution varies as 1/r 2. Equation 19.18 can also be derived by direct
integration over the charge distribution. (See Problem 18.)
What If? What if the line segment in this example were not infinitely long?
Answer If the line charge in this example were of finite
length, the electric field would not be given by Equation 19.18. A finite line charge does not possess sufficient symmetry to make use of Gauss’s law because the
magnitude of the electric field is no longer constant over
the surface of the gaussian cylinder: the field near the
ends of the line would be different from that far from
the ends. Therefore, condition (1) would not be satisfied
:
in this situation. Furthermore, E is not perpendicular
to the cylindrical surface at all points: the field vectors
near the ends would have a component parallel to the
line. Therefore, condition (2) would not be satisfied. For
points close to a finite line charge and far from the ends,
Equation 19.18 gives a good approximation of the value
of the field.
It is left for you to show (see Problem 48) that the electric field inside a uniformly charged rod of finite radius and
infinite length is proportional to r.
Exampl e 19.12 | A Plane of Charge
Find the electric field due to an infinite plane of positive charge with uniform surface
charge density .
SOLUTION
S
E
Conceptualize Notice that the plane of charge is infinitely large. Therefore, the electric
field should be the same at all points equidistant from the plane.
Categorize Because the charge is distributed uniformly on the plane, the charge distri-
bution is symmetric; hence, we can use Gauss’s law to find the electric field.
:
S
E
Analyze By symmetry, E must be perpendicular to the plane at all points. The direc:
:
tion of E is away from positive charges, indicating that the direction of E on one side
Figure 19.34 (Example 19.12)
of the plane must be opposite its direction on the other side as shown in Figure 19.34.
A cylindrical gaussian surface
A gaussian surface that reflects the symmetry is a small cylinder whose axis is perpenpenetrating an infinite plane of
dicular to the plane and whose ends each have an area A and are equidistant from
charge. The flux is EA through each
:
end of the gaussian surface and zero
the plane. Because E is parallel to the curved surface of the cylinder—and therefore
:
through its curved surface.
perpendicular to d A at all points on this surface—condition (3) is satisfied and there is
no contribution to the surface integral from this surface. For the flat ends of the cylinder, conditions (1) and (2) are satisfied. The flux through each end of the cylinder is EA; hence, the total flux through the entire gaussian surface is just that
through the ends, FE 5 2EA.
continued
644 CHAPTER 19 | Electric Forces and Electric Fields
19.12 cont.
Write Gauss’s law for this surface, noting that the enclosed charge is q in 5 A:
FE 5 2EA 5
Solve for E :
E5
q in
A
5
0
0
20
19.19b
Finalize Because the distance from each flat end of the cylinder to the plane does not appear in Equation 19.19, we conclude that E 5 /20 at any distance from the plane. That is, the field is uniform everywhere.
What If? Suppose two infinite planes of charge are parallel to each other, one positively charged and the other negatively
charged. Both planes have the same surface charge density. What does the electric field look like in this situation?
Answer The electric fields due to the two planes add in the region between the planes, resulting in a uniform field of mag-
nitude /0, and cancel elsewhere to give a field of zero. This method is a practical way to achieve uniform electric fields
with finite-sized planes placed close to each other.
19.11 | Conductors in Electrostatic Equilibrium
S
E
S
E
Figure 19.35 A conducting slab in
:
an external electric field E . The
charges induced on the two surfaces of
the slab produce an electric field that
opposes the external field, giving
a resultant field of zero inside the slab.
Figure 19.36 A conductor of
arbitrary shape. The broken line
represents a gaussian surface that can
be just inside the conductor’s surface.
A good electrical conductor, such as copper, contains charges (electrons) that are
not bound to any atom and are free to move about within the material. When no motion of charge occurs within the conductor (other than thermal motion), the conductor is in electrostatic equilibrium. As we shall see, an isolated conductor (one that
is insulated from ground) in electrostatic equilibrium has the following properties:
1. The electric field is zero everywhere inside the conductor, whether the conductor is solid or hollow.
2. If the conductor is isolated and carries a charge, the charge resides on its
surface.
3. The electric field at a point just outside a charged conductor is perpendicular
to the surface of the conductor and has a magnitude /0, where is the surface charge density at that point.
4. On an irregularly shaped conductor, the surface charge density is greatest at
locations where the radius of curvature of the surface is smallest.
We will verify the first three properties in the following discussion. The fourth
property is presented here so that we have a complete list of properties for conductors in electrostatic equilibrium. The verification of it, however, requires concepts
from Chapter 20, so we will postpone its verification until then.
The first property can be understood by considering a conducting slab placed in
:
an external field E (Fig. 19.35). The electric field inside the conductor must be zero
under the assumption that we have electrostatic equilibrium. If the field were not
zero, free charges in the conductor would accelerate under the action of the electric force. This motion of electrons, however, would mean that the conductor is not
in electrostatic equilibrium. Therefore, the existence of electrostatic equilibrium is
consistent only with a zero field in the conductor.
Let us investigate how this zero field is accomplished. Before the external field is
applied, free electrons are uniformly distributed throughout the conductor. When
the external field is applied, the free electrons accelerate to the left in Figure 19.35,
causing a plane of negative charge to be present on the left surface. The movement
of electrons to the left results in a plane of positive charge on the right surface. These
planes of charge create an additional electric field inside the conductor that opposes
the external field. As the electrons move, the surface charge density increases until
the magnitude of the internal field equals that of the external field, giving a net field
of zero inside the conductor.
We can use Gauss’s law to verify the second property of a conductor in electrostatic
equilibrium. Figure 19.36 shows an arbitrarily shaped conductor. A gaussian surface is
drawn just inside the conductor and can be as close to the surface as we wish. As we have
19.12 | Context Connection: The Atmospheric Electric Field 645
just shown, the electric field everywhere inside a conductor in electrostatic equilibrium
is zero. Therefore, the electric field must be zero at every point on the gaussian surface
(condition 4 in Section 19.10). From this result and Gauss’s law, we conclude that the
net charge inside the gaussian surface is zero. Because there can be no net charge inside the gaussian surface (which is arbitrarily close to the conductor’s surface), any net
charge on the conductor must reside on its surface. Gauss’s law does not tell us how
this excess charge is distributed on the surface, only that it must reside on the surface.
Conceptually, we can understand the location of the charges on the surface by
imagining placing many charges at the center of the conductor. The mutual repulsion of the charges causes them to move apart. They will move as far as they can,
which is to various points on the surface.
To verify the third property, we can also use Gauss’s law. We draw a gaussian surface
in the shape of a small cylinder having its end faces parallel to the surface (Fig. 19.37).
Part of the cylinder is just outside the conductor and part is inside. The field is nor:
mal to the surface because the conductor is in electrostatic equilibrium. If E had a
component parallel to the surface, an electric force would be exerted on the charges
parallel to the surface, free charges would move along the surface, and so the conductor would not be in equilibrium. Therefore, we satisfy condition 3 in Section 19.10 for
the curved part of the cylinder in that no flux exists through this part of the gaussian
:
surface because E is parallel to this part of the surface. No flux exists through the flat
:
face of the cylinder inside the conductor because E 5 0 (condition 4). Hence, the net
flux through the gaussian surface is the flux through the flat face outside the conductor
where the field is perpendicular to the surface. Using conditions 1 and 2 for this face,
the flux is EA, where E is the electric field just outside the conductor and A is the area
of the cylinder’s face. Applying Gauss’s law to this surface gives
FE 5
R
E dA 5 EA 5
S
E
Figure 19.37 A gaussian surface
in the shape of a small cylinder is
used to calculate the electric field just
outside a charged conductor.
q in
A
5
0
0
where we have used that q in 5 A. Solving for E gives
E5
0
19.20b
THINKING PHYSICS 19.2
Suppose a point charge 1Q is in empty space. We surround the charge with a spherical, uncharged conducting
shell so that the charge is at the center of the shell. What
effect does that have on the field lines from the charge?
Reasoning When the spherical shell is placed around
the charge, the free charges in the shell adjust so as to satisfy the rules for a conductor in equilibrium and Gauss’s
law. A net charge of 2Q moves to the interior surface of
the conductor, so the electric field within the conductor
is zero (a spherical gaussian surface totally within the
shell encloses no net charge). A net charge of 1Q resides
on the outer surface, so a gaussian surface outside the
sphere encloses a net charge of 1Q , just as if the shell
were not there. Therefore, the only change in the field
lines from the initial situation is the absence of field lines
over the thickness of the conducting shell. b
19.12 | Context Connection: The Atmospheric
Electric Field
In this chapter, we discussed the electric field due to various charge distributions. On
the surface of the Earth and in the atmosphere, a number of processes create charge
distributions, resulting in an electric field in the atmosphere. These processes include cosmic rays entering the atmosphere, radioactive decay at the Earth’s surface,
and lightning, the focus of our study in this Context.
The result of these processes is an average negative charge distributed over
the surface of the Earth of about 5 3 105 C, which is a tremendous amount of
charge. (The Earth is neutral overall; the positive charges corresponding to this
negative surface charge are spread through the atmosphere, as we shall discuss in
646 CHAPTER 19 | Electric Forces and Electric Fields
Figure 19.38 A typical
tripolar charge distribution in a
thundercloud. The dots indicate
the average position of each charge
distribution.
Chapter 20.) We can calculate the average surface charge density over the surface
of the Earth:
avg 5
Q
A
5
Q
4r
5
2
5 3 105 C
, 1029 C/m2
4(6.37 3 106 m)2
Throughout this Context, we will be adopting a number of simplification models.
Consequently, we will consider our calculations to be order-of-magnitude estimates
of the actual values, as suggested by the , sign above.
The Earth is a good conductor. Therefore, we can use the third property of conductors in Section 19.11 to find the average magnitude of the electric field at the
surface of the Earth:
Eavg 5
avg
0
5
1029 C/m2
, 102 N/C
8.85 3 10212 C2/N ? m2
which is a typical value of the fair-weather electric field that exists in the absence
of a thunderstorm. The direction of the field is downward because the charge on
the Earth’s surface is negative. During a thunderstorm, the electric field under the
thundercloud is significantly higher than the fair-weather electric field, because
of the charge distribution in the thundercloud.
Figure 19.38 shows a typical charge distribution in a thundercloud. The charge distribution can be modeled as a tripole, although the positive charge at the bottom of the
cloud tends to be smaller than the other two charges. The mechanism of charging in
thunderclouds is not well understood and continues to be an active area of research.
It is this high concentration of charge in the thundercloud that is responsible for
the very strong electric fields that cause lightning discharge between the cloud and
the ground. Typical electric fields during a thunderstorm are as high as 25 000 N/C.
The distribution of negative charges in the center of the cloud in Figure 19.38 is the
source of negative charge that moves downward in a lightning strike.
Transient Luminous Events
Normal lightning is related to atmospheric electric fields in the troposphere between a thundercloud and the ground. Let us consider the effects of electric fields
above thunderclouds as shown in Figure 19.39. We find a number of visual effects
| Summary 647
Figure 19.39 A representation of
several types of transient luminous
events in the atmosphere above
thunderclouds.
associated with storms and lightning occurring in this region of the atmosphere. In
general, these phenomena are called transient luminous events. One type of event is
called a sprite, which occurs above thunderstorm clouds, with the light from the event
originating between 90 and 100 km above the earth’s surface. A sprite is triggered
by normal tropospheric lightning from the thundercloud below it and appears as
a luminous red flash, possibly with vertical tendrils hanging below. These displays
last less than a second and are not easily seen with the naked eye. A sprite was first
photographed by accident in 1989. Since then, additional photographic evidence
has been made available and several astronauts at the International Space Station
have reported seeing sprites while they were above a violent storm. Scientists believe
that the electric fields at these altitudes are strong enough to ionize air molecules.
Red light is released when the electrons recombine with molecular nitrogen ions, in
a manner similar to the source of light in a fluorescent lamp.
Other lightning-induced transient luminous events are called elves. The light associated with these events lasts for less than 1 ms and has been observed using high speed
photometers and CCD cameras. These events, which precede the onset of sprites, appear as expanding halos of light in the ionosphere at altitudes between 75 and 105 km.
The expansion of the halos is faster than the speed of light, but there is no violation of
the principles of relativity that we studied in Chapter 9 because no particles travel that
fast. Current theories relate to an expanding spherical electromagnetic pulse from a
lightning strike that interacts with the ionosphere to create the luminous display.
Also seen in Figure 19.39 is an optical event in the stratosphere called a blue jet. These
displays occur as an upward-propagating ejection from the top of a thundercloud, disappearing at about 40–50 km from the ground. Blue jets are associated with storm
clouds but do not appear to be directly triggered by lightning flashes as are the sprites.
Other types of luminous events include blue starters, trolls, gnomes, and pixies. Research on the origin of transient luminous events is ongoing.
SUMMARY |
Electric charges have the following important properties:
1. Two kinds of charges exist in nature, positive and negative, with the property that charges of opposite sign attract
each other and charges of the same sign repel each other.
2. The force between charged particles varies as the inverse
square of their separation distance.
3. Charge is conserved.
4. Charge is quantized.
Conductors are materials in which charges move relatively
freely. Insulators are materials in which charges do not move
freely.
Coulomb’s law states that the electrostatic force between
two stationary, charged particles separated by a distance r has
the magnitude
u q 1 uu q 2 u
Fe 5 ke
19.1b
r2
648 CHAPTER 19 | Electric Forces and Electric Fields
where the Coulomb constant ke 5 8.99 3 109 N ? m2/C2. The
vector form of Coulomb’s law is
:
q1q2
r̂
19.2b
r2 12
An electric field exists at a point in space if a positive test
charge q0 placed at that point experiences an electric force.
The electric field is defined as
F 12 5 ke
:
E ;
:
Fe
19.3b
q0
The force on a particle with charge q placed in an electric
:
field E is
:
:
Fe 5 qE
19.4b
The electric field due to the point charge q at a distance r
from the charge is
q
E 5 ke 2 r̂
r
:
19.5b
where r̂ is a unit vector directed from the charge toward the
point in question. The electric field is directed radially outward
from a positive charge and is directed toward a negative charge.
The electric field due to a group of charges can be obtained using the superposition principle. That is, the total
electric field equals the vector sum of the electric fields of all
the charges at some point:
:
E 5 ke o
qi
r̂
2 i
19.6b
i ri
Similarly, the electric field of a continuous charge distribution at some point is
:
E 5 ke
E
dq
r2
r̂
19.7b
where dq is the charge on one element of the charge distribution and r is the distance from the element to the point in
question.
OBJECTIVE QUESTIONS |
1. A point charge of 24.00 nC is located at (0, 1.00) m. What
is the x component of the electric field due to the point
charge at (4.00, 22.00) m? (a) 1.15 N/C (b) 20.864 N/C
(c) 1.44 N/C (d) 21.15 N/C (e) 0.864 N/C
2. Charges of 3.00 nC, 22.00 nC, 27.00 nC, and 1.00 nC are
contained inside a rectangular box with length 1.00 m, width
2.00 m, and height 2.50 m. Outside the box are charges
of 1.00 nC and 4.00 nC. What is the electric flux through
the surface of the box? (a) 0 (b) 25.64 3 102 N ? m2/C
(c) 21.47 3 103 N ? m2/C (d) 1.47 3 103 N ? m2/C
(e) 5.64 3 102 N ? m2/C
3. An object with negative charge is placed in a region of space
where the electric field is directed vertically upward. What
is the direction of the electric force exerted on this charge?
(a) It is up. (b) It is down. (c) There is no force. (d) The
force can be in any direction.
Electric field lines are useful for describing the electric
:
field in any region of space. The electric field vector E is
always tangent to the electric field lines at every point. Furthermore, the number of lines per unit area through a
surface perpendicular to the lines is proportional to the
:
magnitude of E in that region.
Electric flux is proportional to the number of electric field
lines that penetrate a surface. If the electric field is uniform
and makes an angle of with the normal to the surface, the
electric flux through the surface is
FE 5 EA cos 19.13b
In general, the electric flux through a surface is defined by
the expression
FE ;
E
:
:
E ? dA
19.14b
surface
Gauss’s law says that the net electric flux FE through any
closed gaussian surface is equal to the net charge inside the
surface divided by 0:
q in
:
:
19.17b
FE 5 E ? dA 5
0
R
Using Gauss’s law, one can calculate the electric field due to
various symmetric charge distributions.
A conductor in electrostatic equilibrium has the following
properties:
1. The electric field is zero everywhere inside the conductor,
whether the conductor is solid or hollow.
2. If the conductor is isolated and carries a charge, the
charge resides on its surface.
3. The electric field at a point just outside a charged conductor is perpendicular to the surface of the conductor
and has a magnitude /0, where is the surface charge
density at that point.
4. On an irregularly shaped conductor, the surface charge
density is greatest at locations where the radius of curvature
of the surface is smallest.
denotes answer available in Student
Solutions Manual/Study Guide
4. A particle with charge q is located inside a cubical gaussian
surface. No other charges are nearby. (i) If the particle is at
the center of the cube, what is the flux through each one
of the faces of the cube? (a) 0 (b) q/20 (c) q/60 (d) q/80
(e) depends on the size of the cube (ii) If the particle can
be moved to any point within the cube, what maximum
value can the flux through one face approach? Choose from
the same possibilities as in part (i).
5. The magnitude of the electric force between two protons
is 2.30 3 10226 N. How far apart are they? (a) 0.100 m
(b) 0.022 0 m (c) 3.10 m (d) 0.005 70 m (e) 0.480 m
6. Estimate the magnitude of the electric field due to the proton in a hydrogen atom at a distance of 5.29 3 10211 m, the
expected position of the electron in the atom. (a) 10211 N/C
(b) 108 N/C (c) 1014 N/C (d) 106 N/C (e) 1012 N/C
| Conceptual Questions 649
7. Rank the electric fluxes through each gaussian surface
shown in Figure OQ19.7 from largest to smallest. Display
any cases of equality in your ranking.
plane (c) inside a uniformly charged ball (d) outside a
uniformly charged sphere (e) Gauss’s law can be readily applied to find the electric field in all these contexts.
13. Two point charges attract each other with an electric force of
magnitude F. If the charge on one of the particles is reduced
to one-third its original value and the distance between the
particles is doubled, what is the resulting magnitude of the elec1
tric force between them? (a) 12
F (b) 13 F (c) 16 F (d) 34 F (e) 32 F
a
bb
c
d
Figure OQ19.7
8. A circular ring of charge with radius b has total charge q
uniformly distributed around it. What is the magnitude of
the electric field at the center of the ring? (a) 0 (b) keq/b 2
(c) keq 2/b 2 (d) keq 2/b (e) none of those answers
9. Two solid spheres, both of radius 5 cm, carry identical total
charges of 2 C. Sphere A is a good conductor. Sphere B is
an insulator, and its charge is distributed uniformly throughout its volume. (i) How do the magnitudes of the electric
fields they separately create at a radial distance of 6 cm
compare? (a) EA . E B 5 0 (b) EA . E B . 0 (c) EA 5 E B . 0
(d) 0 , EA , E B (e) 0 5 EA , E B (ii) How do the magnitudes
of the electric fields they separately create at radius 4 cm
compare? Choose from the same possibilities as in part (i).
10. An electron with a speed of 3.00 3 106 m/s moves into a uniform electric field of magnitude 1.00 3 103 N/C. The field
lines are parallel to the electron’s velocity and pointing in
the same direction as the velocity. How far does the electron
travel before it is brought to rest? (a) 2.56 cm (b) 5.12 cm
(c) 11.2 cm (d) 3.34 m (e) 4.24 m
11. A very small ball has a mass of 5.00 3 1023 kg and a charge
of 4.00 C. What magnitude electric field directed upward
will balance the weight of the ball so that the ball is suspended motionless above the ground?
(a) 8.21 3 102 N/C (b) 1.22 3 104 N/C (c) 2.00 3 1022 N/C
(d) 5.11 3 106 N/C (e) 3.72 3 103 N/C
12. In which of the following contexts can Gauss’s law not be
readily applied to find the electric field? (a) near a long, uniformly charged wire (b) above a large, uniformly charged
CONCEPTUAL QUESTIONS |
14. Three charged particles are arranged on corners of a square as shown in Figure OQ19.14,
with charge 2Q on both the
particle at the upper left corner and the particle at the
lower right corner and with
charge 12Q on the particle at
the lower left corner. (i) What
Figure OQ19.14
is the direction of the electric
field at the upper right corner, which is a point in empty
space? (a) It is upward and to the right. (b) It is straight to
the right. (c) It is straight downward. (d) It is downward and
to the left. (e) It is perpendicular to the plane of the picture
and outward. (ii) Suppose the 12Q charge at the lower left
corner is removed. Then does the magnitude of the field
at the upper right corner (a) become larger, (b) become
smaller, (c) stay the same, or (d) change unpredictably?
15. Assume the charged ob
jects in Figure OQ19.15 are
fixed. Notice that there is no
sight line from the location
Figure OQ19.15
of q2 to the location of q1. If
you were at q 1, you would be unable to see q 2 because it is behind q 3. How would you calculate the electric force exerted on
the object with charge q 1? (a) Find only the force exerted by q 2
on charge q 1. (b) Find only the force exerted by q 3 on charge q 1.
(c) Add the force that q 2 would exert by itself on charge q 1 to the
force that q 3 would exert by itself on charge q 1. (d) Add the force
that q 3 would exert by itself to a certain fraction of the force that
q 2 would exert by itself. (e) There is no definite way to find the
force on charge q 1.
denotes answer available in Student
Solutions Manual/Study Guide
1. A uniform electric field exists in a region of space containing no charges. What can you conclude about the net electric flux through a gaussian surface placed in this region of
space?
5. Would life be different if the electron were positively
charged and the proton was negatively charged? (b) Does
the choice of signs have any bearing on physical and chemical interactions? Explain your answers.
2. A glass object receives a positive charge by rubbing it with a
silk cloth. In the rubbing process, have protons been added
to the object or have electrons been removed from it?
6. A student who grew up in a tropical country and is studying in the United States may have no experience with static
electricity sparks and shocks until his or her first American
winter. Explain.
3. If more electric field lines leave a gaussian surface than
enter it, what can you conclude about the net charge enclosed by that surface?
4. Why must hospital personnel wear special conducting shoes
while working around oxygen in an operating room? What
might happen if the personnel wore shoes with rubber
soles?
7. If a suspended object A is attracted to a charged object B,
can we conclude that A is charged? Explain.
8. A cubical surface surrounds a point charge q. Describe what
happens to the total flux through the surface if (a) the
charge is doubled, (b) the volume of the cube is doubled,
(c) the surface is changed to a sphere, (d) the charge is
650 CHAPTER 19 | Electric Forces and Electric Fields
moved to another location inside the surface, and (e) the
charge is moved outside the surface.
9. A person is placed in a large, hollow, metallic sphere that is
insulated from ground. (a) If a large charge is placed on the
sphere, will the person be harmed upon touching the inside
of the sphere? (b) Explain what will happen if the person also
has an initial charge whose sign is opposite that of the charge
on the sphere.
10. Consider point A in Figure CQ19.10 located an arbitrary distance from two positive
point charges in otherwise
empty space. (a) Is it possible
for an electric field to exist at
point A in empty space? Explain. (b) Does charge exist at
this point? Explain. (c) Does
a force exist at this point?
Explain.
Figure CQ19.10
11. Consider two identical conducting spheres whose surfaces
are separated by a small distance. One sphere is given a
large net positive charge, and the other is given a small
net positive charge. It is found that the force between
the spheres is attractive even though they both have net
charges of the same sign. Explain how this attraction is
possible.
12. If the total charge inside a closed surface is known but
the distribution of the charge is unspecified, can you use
Gauss’s law to find the electric field? Explain.
13. Consider an electric field that is uniform in direction
throughout a certain volume. Can it be uniform in magnitude? Must it be uniform in magnitude? Answer these
questions (a) assuming the volume is filled with an insulating material carrying charge described by a volume charge
density and (b) assuming the volume is empty space. State
reasoning to prove your answers.
14. On the basis of the repulsive nature of the force between
like charges and the freedom of motion of charge within a
conductor, explain why excess charge on an isolated conductor must reside on its surface.
15. A common demonstration involves charging a rubber balloon, which is an insulator, by rubbing it on your hair and
then touching the balloon to a ceiling or wall, which is also
an insulator. Because of the electrical attraction between
the charged balloon and the neutral wall, the balloon sticks
to the wall. Imagine now that we have two infinitely large,
flat sheets of insulating material. One is charged, and the
other is neutral. If these sheets are brought into contact,
does an attractive force exist between them as there was for
the balloon and the wall?
PROBLEMS |
denotes Master It tutorial available in Enhanced
WebAssign
denotes asking for quantitative and conceptual reasoning
denotes symbolic reasoning problem
denotes “paired problems” that develop reasoning with
symbols and numerical values
denotes Watch It video solution available in Enhanced
WebAssign
The problems found in this chapter may be assigned
online in Enhanced WebAssign.
1. denotes straightforward problem; 2. denotes intermediate
problem; 3. denotes challenging problem
1. denotes full solution available in the Student Solutions Manual/
Study Guide
1. denotes problems most often assigned in Enhanced
WebAssign.
denotes biomedical problem
denotes guided problem
Section 19.2 Properties of Electric Charges
Section 19.4 Coulomb’s Law
1. Find to three significant digits the charge and the mass of
the following particles. Suggestion: Begin by looking up the
mass of a neutral atom on the periodic table of the elements
in Appendix C. (a) an ionized hydrogen atom, represented
as H1 (b) a singly ionized sodium atom, Na1 (c) a chloride
ion Cl2 (d) a doubly ionized calcium atom, Ca11 5 Ca21
(e) the center of an ammonia molecule, modeled as an
N32 ion (f) quadruply ionized nitrogen atoms, N41, found
in plasma in a hot star (g) the nucleus of a nitrogen atom
(h) the molecular ion H2O2
3. Nobel laureate Richard Feynman (1918–1988) once said
that if two persons stood at arm’s length from each other
and each person had 1% more electrons than protons, the
force of repulsion between them would be enough to lift a
“weight” equal to that of the entire Earth. Carry out an order-of-magnitude calculation to substantiate this assertion.
2.
(a) Calculate the number of electrons in a small, electrically neutral silver pin that has a mass of 10.0 g. Silver has
47 electrons per atom, and its molar mass is 107.87 g/mol.
(b) Imagine adding electrons to the pin until the negative
charge has the very large value 1.00 mC. How many electrons are added for every 109 electrons already present?
4. Two protons in an atomic nucleus are typically separated by
a distance of 2 3 10215 m. The electric repulsion force between the protons is huge, but the attractive nuclear force
is even stronger and keeps the nucleus from bursting apart.
What is the magnitude of the electric force between two
protons separated by 2.00 3 10215 m?
5.
Two identical conducting small spheres are placed with
their centers 0.300 m apart. One is given a charge of 12.0 nC
and the other a charge of 218.0 nC. (a) Find the electric force
exerted by one sphere on the other. (b) What If ? The spheres
| Problems 651
are connected by a conducting wire. Find the electric force
each exerts on the other after they have come to equilibrium.
6. Why is the following situation impossible? Two identical dust particles of mass 1.00 g are floating in empty space, far from any
external sources of large gravitational or electric fields, and
at rest with respect to each other. Both particles carry electric
charges that are identical in magnitude and sign. The gravitational and electric forces between the particles happen to have
the same magnitude, so each particle experiences zero net
force and the distance between the particles remains constant.
7.
8.
9.
10.
Two small
beads having positive
charges q1 5 3q and
q2 5 q are fixed at the
opposite ends of a horizontal insulating rod Figure P19.7 Problems 7 and 8.
of length d 5 1.50 m.
The bead with charge q1 is at the origin. As shown in Figure P19.7, a third small, charged bead is free to slide on the rod.
(a) At what position x is the third bead in equilibrium? (b) Can
the equilibrium be stable?
Two small beads having charges q1 and q2 of the
same sign are fixed at the opposite ends of a horizontal insulating rod of length d. The bead with charge q1 is at the origin. As shown in Figure P19.7, a third small, charged bead
is free to slide on the rod. (a) At what position x is the third
bead in equilibrium? (b) Can the equilibrium be stable?
Three charged particles
are located at the corners of an
equilateral triangle as shown
in Figure P19.9. Calculate
the total electric force on the
7.00-C charge.
Particle A of charge 3.00 3
1024 C is at the origin, particle
13.
Section 19.5 Electric Fields
14.
Two charged particles are located on the x axis.
The first is a charge 1Q at x 5 2a. The second is an unknown charge located at x 5 13a. The net electric field
these charges produce at the origin has a magnitude
of 2keQ /a 2. Explain how many values are possible for the
unknown charge and find the possible values.
15.
A uniformly charged ring of radius 10.0 cm has a total
charge of 75.0 C. Find the electric field on the axis of
the ring at (a) 1.00 cm, (b) 5.00 cm, (c) 30.0 cm, and
(d) 100 cm from the center of the ring.
16. What are the magnitude and direction of the electric field
that will balance the weight of (a) an electron and (b) a proton? (You may use the data in Table 19.1.)
17.
18.
m
m
m
B of charge 26.00 3 1024 C is
at (4.00 m, 0), and particle C of
Figure P19.9
charge 1.00 3 1024 C is at (0,
3.00 m). We wish to find the net electric force on C. (a) What
is the x component of the electric force exerted by A on C?
(b) What is the y component of the force exerted by A on
C? (c) Find the magnitude of the force exerted by B on C.
(d) Calculate the x component of the force exerted by B
on C. (e) Calculate the y component of the force exerted by
B on C. (f) Sum the two x components from parts (a) and
(d) to obtain the resultant x component of the electric force
acting on C. (g) Similarly, find the y component of the resultant force vector acting on C. (h) Find the magnitude and direction of the resultant electric force acting on C.
11. Review. In the Bohr theory of the hydrogen atom, an electron moves in a circular orbit about a proton, where the radius of the orbit is 5.29 3 10211 m. (a) Find the magnitude
of the electric force exerted on each particle. (b) If this
force causes the centripetal acceleration of the electron,
what is the speed of the electron?
12. A charged particle A exerts a force of 2.62 N to the right on
charged particle B when the particles are 13.7 mm apart. Particle B moves straight away from A to make the distance between
them 17.7 mm. What vector force does it then exert on A?
Review. A molecule of DNA (deoxyribonucleic acid)
is 2.17 m long. The ends of the molecule become singly
ionized: negative on one end, positive on the other. The helical molecule acts like a spring and compresses 1.00% upon
becoming charged. Determine the effective spring constant
of the molecule.
In Figure P19.17, determine the point (other than
infinity) at which the elec
m
tric field is zero.
A thin rod of length /
Figure P19.17
and uniform charge per
unit length lies along
the x axis as shown in Figure P19.18. (a) Show that
the electric field at P, a distance y from the rod along
its perpendicular bisector,
has no x component and is
given by E 5 2ke sin 0 /y.
(b) Using your result to part
(a), show that the field of
a rod of infinite length is
E 5 2ke /y. (Suggestion:
ᐉ
First, calculate the field at P
due to an element of length
Figure P19.18
dx, which has a charge dx.
Then, change variables from x to , using the relationships
x 5 y tan and dx 5 y sec2 d , and integrate over .)
19. Three point charges are
arranged as shown in Figure P19.19. (a) Find the
vector electric field that
the 6.00-nC and 23.00-nC
charges together create at
the origin. (b) Find the
vector force on the 5.00-nC
charge.
20.
m
Consider the electric dipole shown in Figure P19.20. Show that
the electric field at a distant point on the 1x axis
is Ex < 4keqa/x3.
Figure P19.19
⫺
⫹
Figure P19.20
652 CHAPTER 19 | Electric Forces and Electric Fields
21. A uniformly charged insulating rod of length
14.0 cm is bent into the shape of a semicircle
as shown in Figure P19.21. The rod has a total
charge of 27.50 C. Find (a) the magnitude
and (b) the direction of the electric field at O,
the center of the semicircle.
Section 19.7 Motion of Charged Particles in a Uniform
Electric Field
22. Two 2.00-C point charges are located on the
x axis. One is at x 5 1.00 m, and the other is at Figure P19.21
x 5 21.00 m. (a) Determine the electric field
on the y axis at y 5 0.500 m. (b) Calculate the electric force on
a 23.00-C charge placed on the y axis at y 5 0.500 m.
23.
A rod 14.0 cm long is uniformly charged and has a total
charge of 222.0 C. Determine (a) the magnitude and
(b) the direction of the electric field along the axis of the
rod at a point 36.0 cm from its center.
24.
A continuous line of charge lies along the x axis,
extending from x 5 1x0 to positive infinity. The line carries positive charge with a uniform linear charge density 0.
What are (a) the magnitude and (b) the direction of the
electric field at the origin?
31.
A proton is projected in the positive x direction into a
:
region of a uniform electric field E 5 (26.00 3 105) î N/C at
t 5 0. The proton travels 7.00 cm as it comes to rest. Determine (a) the acceleration of the proton, (b) its initial speed,
and (c) the time interval over which the proton comes to rest.
32.
Protons are projected with an initial speed vi 5
9.55 km/s from a field-free region through a plane and into
:
a region where a uniform electric field E 5 2720 ĵ N/C is
present above the plane as shown in Figure P19.32. The initial velocity vector of the protons makes an angle with the
plane. The protons are to hit a target that lies at a horizontal
distance of R 5 1.27 mm from the point where the protons cross the plane and enter the electric field. We wish to
find the angle at which the protons must pass through the
plane to strike the target. (a) What analysis model describes
the horizontal motion of the protons above the plane?
(b) What analysis model describes the vertical motion of
the protons above the plane? (c) Argue that Equation 3.16
would be applicable to the protons in this situation. (d) Use
Equation 3.16 to write an expression for R in terms of vi ,
E, the charge and mass of the proton, and the angle .
(e) Find the two possible values of the angle . (f) Find the
time interval during which the proton is above the plane in
Figure P19.32 for each of the two possible values of .
25. Three solid plastic cylinders all have radius 2.50 cm and length
6.00 cm. Find the charge of each cylinder given the following additional information about each one. Cylinder (a) carries charge with uniform density 15.0 nC/m2 everywhere on
its surface. Cylinder (b) carries charge with uniform density
15.0 nC/m2 on its curved lateral surface only. Cylinder (c) carries
charge with uniform density 500 nC/m3 throughout the plastic.
26.
27.
Four charged particles are
at the corners of a square of side a
as shown in Figure P19.27. Determine (a) the electric field at the location of charge q and (b) the total
electric force exerted on q.
Section 19.6 Electric Field Lines
28.
S
E
Show that the maximum magnitude Emax of the electric field along the axis of a uniformly charged ring occurs at
x 5a/"2 (see Fig. 19.15) and has the value Q /(6"30a2).
Three equal positive charges q are
at the corners of an equilateral triangle
of side a as shown in Figure P19.28.
Assume the three charges together
create an electric field. (a) Sketch the
field lines in the plane of the charges.
(b) Find the location of one point
(other than `) where the electric field
is zero. What are (c) the magnitude and
(d) the direction of the electric field at
P due to the two charges at the base?
S
v
Figure P19.30 shows the electric
field lines for two charged particles
separated by a small distance. (a) Determine the ratio q 1/q 2. (b) What are
the signs of q 1 and q 2?
u
ⴛ
S
E
Figure P19.32
Figure P19.27
33.
A proton accelerates from rest in a uniform electric field
of 640 N/C. At one later moment, its speed is 1.20 Mm/s
(nonrelativistic because v is much less than the speed of
light). (a) Find the acceleration of the proton. (b) Over
what time interval does the proton reach this speed?
(c) How far does it move in this time interval? (d) What is
its kinetic energy at the end of this interval?
34.
The electrons in a particle beam each have a kinetic energy
K. What are (a) the magnitude and (b) the direction of the
electric field that will stop these electrons in a distance d ?
35.
A proton moves at 4.50 3 105 m/s in the horizontal
direction. It enters a uniform vertical electric field with a
magnitude of 9.60 3 103 N/C. Ignoring any gravitational
effects, find (a) the time interval required for the proton
to travel 5.00 cm horizontally, (b) its vertical displacement
during the time interval in which it travels 5.00 cm horizontally, and (c) the horizontal and vertical components of its
velocity after it has traveled 5.00 cm horizontally.
Figure P19.28
29. A negatively charged rod of finite
length carries charge with a uniform
charge per unit length. Sketch the
electric field lines in a plane containing the rod.
30.
j
Section 19.8 Electric Flux
36.
Figure P19.30
A vertical electric field of magnitude 2.00 3 104 N/C exists above the Earth’s surface on a day when a thunderstorm
| Problems 653
is brewing. A car with a rectangular size of 6.00 m by 3.00 m is
traveling along a dry gravel roadway sloping downward at 10.08.
Determine the electric flux through the bottom of the car.
37.
A 40.0-cm-diameter circular loop is rotated in a uniform electric field until the position of maximum electric flux
is found. The flux in this position is measured to be 5.20 3
105 N ? m2/C. What is the magnitude of the electric field?
Section 19.10 Application of Gauss’s Law to Various
Charge Distributions
45.
A solid sphere of radius 40.0 cm has a total positive charge
of 26.0 C uniformly distributed throughout its volume. Calculate the magnitude of the electric field (a) 0 cm, (b) 10.0 cm,
(c) 40.0 cm, and (d) 60.0 cm from the center of the sphere.
46.
A nonconducting wall carries charge with a uniform density of 8.60 C/cm2. (a) What is the electric field
7.00 cm in front of the wall if 7.00 cm is small compared with
the dimensions of the wall? (b) Does your result change as
the distance from the wall varies? Explain.
Section 19.9 Gauss’s Law
38.
39.
40.
A particle with charge Q
is located a small distance
immediately above the
center of the flat face of a
hemisphere of radius R as
shown in Figure P19.38.
What is the electric flux
(a) through the curved surface and (b) through the
flat face as : 0?
d
48.
Figure P19.38
The following charges
are located inside a submarine: 5.00 C, 29.00 C, 27.0 C,
and 284.0 C. (a) Calculate the net electric flux through
the hull of the submarine. (b) Is the number of electric
field lines leaving the submarine greater than, equal to, or
less than the number entering it?
50.
An insulating solid sphere of radius a has a uniform volume charge density and carries a total positive charge Q. A
spherical gaussian surface of radius r, which shares a common center with the insulating sphere, is inflated starting
from r 5 0. (a) Find an expression for the electric flux passing through the surface of the gaussian sphere as a function
of r for r , a. (b) Find an expression for the electric flux for
r . a. (c) Plot the flux versus r.
51.
A large, flat, horizontal sheet of charge has a charge per
unit area of 9.00 C/m2. Find the electric field just above
the middle of the sheet.
52.
A cylindrical shell of radius 7.00 cm and length 2.40 m
has its charge uniformly distributed on its curved surface.
The magnitude of the electric field at a point 19.0 cm radially outward from its axis (measured from the midpoint of
the shell) is 36.0 kN/C. Find (a) the net charge on the shell
and (b) the electric field at a point 4.00 cm from the axis,
measured radially outward from the midpoint of the shell.
53.
Consider a thin, spherical shell of radius 14.0 cm with a
total charge of 32.0 C distributed uniformly on its surface.
Find the electric field (a) 10.0 cm and (b) 20.0 cm from the
center of the charge distribution.
54.
A uniformly charged, straight filament 7.00 m in length
has a total positive charge of 2.00 C. An uncharged cardboard cylinder 2.00 cm in length and 10.0 cm in radius
surrounds the filament at its center, with the filament as
the axis of the cylinder. Using reasonable approximations,
find (a) the electric field at the surface of the cylinder and
(b) the total electric flux through the cylinder.
Figure P19.41 Problems 41
A particle with charge Q and 42.
is located at the center of a
cube of edge L. In addition, six other identical charged particles
q are positioned symmetrically around Q as shown in Figure
P19.41. For each of these particles, q is a negative number. Determine the electric flux through one face of the cube.
43.
The electric field everywhere on the surface of a
thin, spherical shell of radius 0.750 m is of magnitude
890 N/C and points radially toward the center of the
sphere. (a) What is the net charge within the sphere’s surface? (b) What is the distribution of the charge inside the
spherical shell?
44.
A charge of 170 C is at the center of a cube of edge
80.0 cm. No other charges are nearby. (a) Find the flux
through each face of the cube. (b) Find the flux through
the whole surface of the cube. (c) What If? Would your
answers to either part (a) or part (b) change if the charge
were not at the center? Explain.
Consider a long, cylindrical charge distribution of radius R with a uniform charge density . Find the electric
field at distance r from the axis, where r , R.
49. A 10.0-g piece of Styrofoam carries a net charge of 20.700 C
and is suspended in equilibrium above the center of a large,
horizontal sheet of plastic that has a uniform charge density on
its surface. What is the charge per unit area on the plastic sheet?
A particle with charge of 12.0 C is placed at the
center of a spherical shell of radius 22.0 cm. What is the
total electric flux through (a) the surface of the shell and
(b) any hemispherical surface of the shell? (c) Do the results depend on the radius? Explain.
41. A particle with charge
Q 5 5.00 C is located at
the center of a cube of edge
L 5 0.100 m. In addition, six
other identical charged particles having q 5 21.00 C
are positioned symmetrically around Q as shown in
Figure P19.41. Determine
the electric flux through
one face of the cube.
42.
47. In nuclear fission, a nucleus of uranium-238, which contains
92 protons, can divide into two smaller spheres, each having 46
protons and a radius of 5.90 3 10215 m. What is the magnitude
of the repulsive electric force pushing the two spheres apart?
Section 19.11 Conductors in Electrostatic Equilibrium
55.
A long, straight metal rod has a radius of 5.00 cm and a
charge per unit length of 30.0 nC/m. Find the electric field
(a) 3.00 cm, (b) 10.0 cm, and (c) 100 cm from the axis of
the rod, where distances are measured perpendicular to the
rod’s axis.
654 CHAPTER 19 | Electric Forces and Electric Fields
56. Why is the following situation
impossible? A solid copper
sphere of radius 15.0 cm is
in electrostatic equilibrium
and carries a charge of
40.0 nC. Figure P19.56 shows
the magnitude of the electric
field as a function of radial
position r measured from
the center of the sphere.
Figure P19.64. Determine the
total electric flux through the
surface of a sphere of radius R
centered at O resulting from
this line charge. Consider
both cases, where (a) R , d
and (b) R . d.
Figure P19.56
57. A solid conducting sphere of radius 2.00 cm has a charge
8.00 C. A conducting spherical shell of inner radius 4.00 cm
and outer radius 5.00 cm is concentric with the solid sphere
and has a total charge 24.00 C. Find the electric field at
(a) r 5 1.00 cm, (b) r 5 3.00 cm, (c) r 5 4.50 cm, and (d) r 5
7.00 cm from the center of this charge configuration.
58.
A very large, thin, flat plate of aluminum of area A has a
total charge Q uniformly distributed over its surfaces. Assuming the same charge is spread uniformly over the upper surface of an otherwise identical glass plate, compare the electric
fields just above the center of the upper surface of each plate.
59.
A thin, square, conducting plate 50.0 cm on a side lies in the
xy plane. A total charge of 4.00 3 1028 C is placed on the plate.
Find (a) the charge density on each face of the plate, (b) the
electric field just above the plate, and (c) the electric field just
below the plate. You may assume the charge density is uniform.
60.
A long, straight wire is surrounded by a hollow metal
cylinder whose axis coincides with that of the wire. The wire
has a charge per unit length of , and the cylinder has a net
charge per unit length of 2. From this information, use
Gauss’s law to find (a) the charge per unit length on the
inner surface of the cylinder, (b) the charge per unit length
on the outer surface of the cylinder, and (c) the electric
field outside the cylinder a distance r from the axis.
61. A square plate of copper with 50.0-cm sides has no net
charge and is placed in a region of uniform electric field
of 80.0 kN/C directed perpendicularly to the plate. Find
(a) the charge density of each face of the plate and (b) the
total charge on each face.
Section 19.12 Context Connection: The Atmospheric
Electric Field
62.
Review. In fair weather, the electric field in the air at a particular location immediately above the Earth’s surface is 120 N/C
directed downward. (a) What is the surface charge density on
the ground? Is it positive or negative? (b) Imagine the surface
charge density is uniform over the planet. What then is the
charge of the whole surface of the Earth? (c) Imagine the Moon
had a charge 27.3% as large as that on the surface of the Earth,
with the same sign. Find the electric force the Earth would then
exert on the Moon. (d ) State how the answer to part (e) compares with the gravitational force the Earth exerts on the Moon.
63. In the air over a particular region at an altitude of 500 m above
the ground, the electric field is 120 N/C directed downward.
At 600 m above the ground, the electric field is 100 N/C downward. What is the average volume charge density in the layer of
air between these two elevations? Is it positive or negative?
Additional Problems
64.
An infinitely long line charge having a uniform charge
per unit length lies a distance d from point O as shown in
65. Four identical charged particles
(q 5 110.0 C) are located on
the corners of a rectangle as
shown in Figure P19.65. The
dimensions of the rectangle are
L 5 60.0 cm and W 5 15.0 cm.
Calculate (a) the magnitude
and (b) the direction of the
total electric force exerted on
the charge at the lower left corner by the other three charges.
l
Figure P19.64
Figure P19.65
66. Why is the following situation impossible? An electron enters a
region of uniform electric field between two parallel plates.
The plates are used in a cathode-ray tube to adjust the position of an electron beam on a distant fluorescent screen.
The magnitude of the electric field between the plates is
200 N/C. The plates are 0.200 m in length and are separated
by 1.50 cm. The electron enters the region at a speed of
3.00 3 106 m/s, traveling parallel to the plane of the plates
in the direction of their length. It leaves the plates heading
toward its correct location on the fluorescent screen.
67. A small, 2.00-g plastic ball
is suspended by a 20.0-cmlong string in a uniform
electric field as shown in
Figure P19.67. If the ball
is in equilibrium when the
string makes a 15.08 angle
with the vertical, what is
the net charge on the ball?
68.
⫻
Two point charges
Figure P19.67
q A5 212.0 C and q B 5
45.0 C and a third particle with unknown charge q C are located on the x axis. The particle q A is at the origin, and q B
is at x 5 15.0 cm. The third particle is to be placed so that
each particle is in equilibrium under the action of the electric
forces exerted by the other two particles. (a) Is this situation
possible? If so, is it possible in more than one way? Explain.
Find (b) the required location and (c) the magnitude and the
sign of the charge of the third particle.
69. Review. Two identical blocks resting
on a frictionless,
horizontal surface
are connected by
a
a light spring having a spring constant k 5 100 N/m
and an unstretched length Li 5
b
0.400 m as shown
in Figure P19.69a. Figure P19.69 Problems 69 and 70.
A charge Q is
slowly placed on each block, causing the spring to stretch to an
| Problems 655
equilibrium length L 5 0.500 m as shown in Figure P19.69b.
Determine the value of Q , modeling the blocks as charged
particles.
70.
Review. Two identical blocks resting on a frictionless,
horizontal surface are connected by a light spring having a
spring constant k and an unstretched length Li as shown in
Figure P19.69a. A charge Q is slowly placed on each block,
causing the spring to stretch to an equilibrium length L as
shown in Figure P19.69b. Determine the value of Q , modeling the blocks as charged particles.
71. A line of charge with uniform density 35.0 nC/m lies along
the line y 5 215.0 cm, between the points with coordinates
x 5 0 and x 5 40.0 cm. Find the electric field it creates at the
origin.
72.
74.
76.
Two small spheres of mass m are suspended from
strings of length ℓ that are connected at a common point. One
sphere has charge Q and the other charge 2Q. The strings
make angles 1 and 2 with the vertical. (a) Explain how 1
and 2 are related. (b) Assume 1 and 2 are small. Show that
the distance r between the spheres is approximately
r<
73.
wish to understand completely the charges and electric fields
at all locations. (a) Find the charge contained within a sphere
of radius r , a. (b) From this value, find the magnitude of the
electric field for r , a. (c) What charge is contained within a
sphere of radius r when a , r , b ? (d) From this value, find the
magnitude of the electric field for r when a , r , b. (e) Now
consider r when b , r , c. What is the magnitude of the electric field for this range of values of r ? (f) From this value, what
must be the charge on the inner surface of the hollow sphere?
(g) From part (f ), what must be the charge on the outer surface of the hollow sphere? (h) Consider the three spherical
surfaces of radii a, b, and c. Which of these surfaces has the
largest magnitude of surface charge density?
4keQ 2, 1/3
1 mg 2
Two infinite, nonconducting sheets of
charge are parallel to each other as shown
in Figure P19.73. The sheet on the left has
a uniform surface charge density , and
the one on the right has a uniform charge
density 2. Calculate the electric field at
points (a) to the left of, (b) in between, s
and (c) to the right of the two sheets.
s
(d) What If? Find the electric fields in all
Figure P19.73
three regions if both sheets have positive
uniform surface charge densities of value .
Consider the charge distribution shown in Figure P19.74.
(a) Show that the magnitude of the electric field at the center of any face of the cube has a value of 2.18keq/s2. (b) What
is the direction of the electric field at the center of the top
face of the cube?
f5
Figure P19.74
A solid, insulating sphere of radius a has
a uniform charge density
throughout its volume
and a total charge Q. Concentric with this sphere is
an uncharged, conducting,
hollow sphere whose inner
and outer radii are b and c as
shown in Figure P19.75. We
Figure P19.75
1 keqQ 1/2
2 ma3
1 2
77. Inez is putting up decorations for her sister’s
quinceañera
(fifteenth
birthday party). She ties
three light silk ribbons together to the top of a gateway and hangs a rubber
balloon from each ribbon
(Fig. P19.77). To include
the effects of the gravitational and buoyant forces
on it, each balloon can be
modeled as a particle of
Figure P19.77
mass 2.00 g, with its center
50.0 cm from the point of support. Inez rubs the whole surface
of each balloon with her woolen scarf, making the balloons
hang separately with gaps between them. Looking directly upward from below the balloons, Inez notices that the centers of
the hanging balloons form a horizontal equilateral triangle with
sides 30.0 cm long. What is the common charge each balloon
carries?
78.
75.
Review. A negatively
charged particle 2q is
placed at the center of a
uniformly charged ring,
where the ring has a total
positive charge Q as shown
in Figure P19.76. The particle, confined to move
along the x axis, is moved
a small distance x along the
Figure P19.76
axis (where x ,, a) and released. Show that the particle oscillates in simple harmonic
motion with a frequency given by
A sphere of radius 2a is made of
a nonconducting material that has
a uniform volume charge density .
(Assume the material does not affect
the electric field.) A spherical cavity
of radius a is now removed from the
sphere as shown in Figure P19.78.
Show that the electric field within
Figure P19.78
the cavity is uniform and is given by
Ex 5 0 and Ey 5 a/30 (Suggestion:
The field within the cavity is the superposition of the field due
to the original uncut sphere plus the field due to a sphere the
size of the cavity with a uniform negative charge density 2.)
Chapter
20
Electric Potential and
Capacitance
Chapter Outline
20.1 Electric Potential and Potential Difference
20.2 Potential Difference in a Uniform Electric
Field
20.3 Electric Potential and Potential Energy Due
to Point Charges
20.4 Obtaining the Value of the Electric Field
from the Electric Potential
© Cengage Learning/George Semple
20.5 Electric Potential Due to Continuous
Charge Distributions
20.6 Electric Potential Due to a Charged Conductor
20.7 Capacitance
20.8 Combinations of Capacitors
20.9 Energy Stored in a Charged Capacitor
20.10 Capacitors with Dielectrics
20.11 Context Connection: The Atmosphere as
a Capacitor
SUMMARY
T
he concept of potential energy was introduced in Chapter 6 in connection with
such conservative forces as gravity and the elastic force of a spring. By using
the principle of conservation of mechanical energy in an isolated system, we are
often able to avoid working directly with forces when solving mechanical problems. In
this chapter, we shall use the energy concept in our study of electricity. Because the electrostatic force (given by Coulomb’s law) is conservative, electrostatic phenomena can
conveniently be described in terms of an electric potential energy function. This concept
enables us to define a quantity called electric potential, which is a scalar quantity and
which therefore leads to a simpler means of describing some electrostatic phenomena
than the electric field method. As we shall see in subsequent chapters, the concept of
electric potential is of great practical value in many applications.
This chapter also addresses the properties of capacitors, devices that store
charge. The ability of a capacitor to store charge is measured by its capacitance. Capacitors are used in common applications such as frequency tuners in radio receivers, filters in power supplies, and energy-storing devices in electronic flash units.
656
This device is a variable capacitor, used
to tune radios to a selected station.
When one set of metal plates is rotated
so as to lie between a fixed set of plates,
the capacitance of the device changes.
Capacitance is a parameter that depends
on electric potential, the primary topic of
this chapter.
20.1 | Electric Potential and Potential Difference
657
20.1 | Electric Potential and Potential Difference
:
When a test charge q 0 is placed in an electric field E created by some source charge
:
:
:
distribution, the electric force acting on the test charge is q 0 E. The force F e 5 q 0 E is
conservative because the force between charges described by Coulomb’s law is conservative. When the test charge is moved in the field at constant velocity by some
external agent, the work done by the field on the charge is equal to the negative
of the work done by the external agent causing the displacement. This situation is
analogous to that of lifting an object with mass in a gravitational field: the work done
by the external agent is mgh, and the work done by the gravitational force is 2mgh.
When analyzing electric and magnetic fields, it is common practice to use the
notation d :
s to represent an infinitesimal displacement vector that is oriented tangent to a path through space. This path may be straight or curved, and an integral
performed along this path is called either a path integral or a line integral (the two
terms are synonymous).
For an infinitesimal displacement d :
s of a point charge q 0 immersed in an electric
field, the work done within the charge–field system by the electric field on the charge
:
:
is Wint 5 Fe ? d :
s 5 q 0E ? d :
s . As this amount of work is done by the field, the potential
:
energy of the charge–field system is changed by an amount dU 5 2Wint 5 2q 0 E ? d :
s.
For a finite displacement of the charge from point 𝖠 to point 𝖡, the change in potential energy of the system DU 5 U𝖡 2 U𝖠 is
DU 5 2q 0
E
𝖡
:
E ? d:
s
20.1b
The integration is performed along the path that q0 follows as it moves from 𝖠 to 𝖡.
:
Because the force q0 E is conservative, this line integral does not depend on the path
taken from 𝖠 to 𝖡.
For a given position of the test charge in the field, the charge–field system has
a potential energy U relative to the configuration of the system that is defined as
U 5 0. Dividing the potential energy by the test charge gives a physical quantity that
depends only on the source charge distribution and has a value at every point in an
electric field. This quantity is called the electric potential (or simply the potential)V :
V5
U
q0
20.2b
Because potential energy is a scalar quantity, electric potential also is a scalar quantity.
As described by Equation 20.1, if the test charge is moved between two positions
𝖠 and 𝖡 in an electric field, the charge–field system experiences a change in potential energy. The potential difference DV 5 V𝖡 2 V𝖠 between two points 𝖠 and 𝖡 in
an electric field is defined as the change in potential energy of the system when a test
charge q 0 is moved between the points divided by the test charge:
DV ;
E
𝖡
DU
:
52
E ? d:
s
q0
𝖠
c Change in electric potential
energy of a charge–field system
𝖠
20.3b
In this definition, the infinitesimal displacement d :
s is interpreted as the displacement between two points in space rather than the displacement of a point charge as
in Equation 20.1.
Just as with potential energy, only differences in electric potential are meaningful.
We often take the value of the electric potential to be zero at some convenient point
in an electric field.
Potential difference should not be confused with difference in potential energy.
The potential difference between 𝖠 and 𝖡 exists solely because of a source charge
and depends on the source charge distribution (consider points 𝖠 and 𝖡 without
the presence of the test charge). For a potential energy to exist, we must have a system of two or more charges. The potential energy belongs to the system and changes
only if a charge is moved relative to the rest of the system.
Pitfall Prevention | 20.1
Potential and Potential Energy
The potential is characteristic of the field
only, independent of a charged test
particle that may be placed in the
field. Potential energy is characteristic
of the charge–field system due to an
interaction between the field and a
charged particle placed in the field.
c Potential difference between
two points
658
CHAPTER 20 | Electric Potential and Capacitance
Pitfall Prevention | 20.2
Voltage
A variety of phrases are used to
describe the potential difference
between two points, the most
common being voltage, arising
from the unit for potential. A
voltage applied to a device, such as
a television, or across a device is the
same as the potential difference
across the device. Despite popular
language, voltage is not something
that moves through a device.
Pitfall Prevention | 20.3
The Electron Volt
The electron volt is a unit of energy,
NOT of potential. The energy of
any system may be expressed in eV,
but this unit is most convenient
for describing the emission and
absorption of visible light from
atoms. Energies of nuclear processes
are often expressed in MeV.
S
훽
If an external agent moves a test charge from 𝖠 to 𝖡 without changing the kinetic energy of the test charge, the agent performs work that changes the potential
energy of the system: W 5 DU. Imagine an arbitrary charge q located in an electric
field. From Equation 20.3, the work done by an external agent in moving a charge q
through an electric field at constant velocity is
20.4b
W 5 q DV
Because electric potential is a measure of potential energy per unit charge, the SI
unit of both electric potential and potential difference is joules per coulomb, which
is defined as a volt (V):
1 V ; 1 J/C
That is, 1 J of work must be done to move a 1-C charge through a potential difference of 1 V.
Equation 20.3 shows that potential difference also has units of electric field times
distance. It follows that the SI unit of electric field (N/C) can also be expressed in
volts per meter:
1 N/C 5 1 V/m
Therefore, we can interpret the electric field as a measure of the rate of change of
the electric potential with respect to position.
As discussed in Section 9.7, a unit of energy commonly used in atomic and nuclear
physics is the electron volt (eV), which is defined as the energy a charge–field system
gains or loses when a charge of magnitude e (that is, an electron or a proton) is moved
through a potential difference of 1 V. Because 1 V 5 1 J/C and the fundamental
charge is equal to 1.602 3 10219 C, the electron volt is related to the joule as follows:
E
1 eV 5 1.602 3 10219 C ? V 5 1.602 3 10219 J
훾
For instance, an electron in the beam of a typical dental x-ray machine may have a
speed of 1.4 3 108 m/s. This speed corresponds to a kinetic energy 1.1 3 10214 J
(using relativistic calculations as discussed in Chapter 9), which is equivalent to
6.7 3 104 eV. Such an electron has to be accelerated from rest through a potential
difference of 67 kV to reach this speed.
Figure 20.1 (Quick Quiz 20.1) Two
points in an electric field.
20.5b
QUICK QUIZ 20.1 In Figure 20.1, two points 𝖠 and 𝖡 are located within a region in
which there is an electric field. (i) How would you describe the potential difference
DV 5 V𝖡 2 V𝖠? (a) It is positive. (b) It is negative. (c) It is zero. (ii) A negative charge
is placed at 𝖠 and then moved to 𝖡. How would you describe the change in potential
energy of the charge–field system for this process? Choose from the same possibilities.
20.2 | Potential Difference in a Uniform
Electric Field
Equations 20.1 and 20.3 hold in all electric fields, whether uniform or varying, but they
can be simplified for a uniform field. First, consider a uniform electric field directed
along the negative y axis as shown in Active Figure 20.2a. Let’s calculate the potential
difference between two points 𝖠 and 𝖡 separated by a distance d, where the displacement :
s points from 𝖠 toward 𝖡 and is parallel to the field lines. Equation 20.3 gives
E
V𝖡 2 V 𝖠 5 DV 5 2
𝖡
E
:
E ? d:
s 52
𝖠
𝖡
E
𝖡
E ds (cos 08) 5 2
𝖠
E ds
𝖠
Because E is constant, it can be removed from the integral sign, which gives
c Potential difference between
two points in a uniform electric
field
DV 5 2E
E
𝖡
𝖠
ds 5 2Ed
20.6b
20.2 | Potential Difference in a Uniform Electric Field
The negative sign indicates that the electric potential at point 𝖡 is lower than at point 𝖠; that is,
V𝖡 , V𝖠. Electric field lines always point in the
direction of decreasing electric potential as shown
in Active Figure 20.2a.
Now suppose a test charge q 0 moves from 𝖠 to
𝖡. We can calculate the change in the potential energy of the charge–field system from Equations 20.3
and 20.6:
DU 5 q 0 DV 5 2q 0Ed
659
When a positive test charge moves
from point 훽 to point 훾, the
electric potential energy of the
charge–field system decreases.
When an object with mass moves
from point 훽 to point 훾, the
gravitational potential energy of
the object–field system decreases.
훽
훽
d
d
20.7b
q0
m
This result shows that if q 0 is positive, then DU is
훾
훾
negative. Therefore, in a system consisting of a
S
S
E
g
positive charge and an electric field, the electric
potential energy of the system decreases when the
charge moves in the direction of the field. Equiva
b
alently, an electric field does work on a positive
:
Active Figure 20.2 (a) When the electric field E is directed downward,
charge when the charge moves in the direction
point 𝖡 is at a lower electric potential than point 𝖠. (b) An object of mass m
of the electric field. That is analogous to the work
moving downward in a gravitational field :
g.
done by the gravitational field on a falling object
as shown in Active Figure 20.2b. If a positive test charge is released from rest in
:
:
this electric field, it experiences an electric force q 0 E in the direction of E (downPoint 훾 is at a lower electric
ward in Active Fig. 20.2a). Therefore, it accelerates downward, gaining kinetic
potential than point 훽.
energy. As the charged particle gains kinetic energy, the potential energy of the
charge–field system decreases by an equal amount. This equivalence should not
be surprising; it is simply conservation of mechanical energy in an isolated system
as introduced in Chapter 7.
The comparison between a system of an electric field with a positive test
charge and a gravitational field with a test mass in Active Figure 20.2 is useful
for conceptualizing electrical behavior. The electrical situation, however, has
one feature that the gravitational situation does not: the test charge can be
negative. If q 0 is negative, then DU in Equation 20.7 is positive and the situation
is reversed. A system consisting of a negative charge and an electric field gains
electric potential energy when the charge moves in the direction of the field.
If a negative charge is released from rest in an electric field, it accelerates in a
direction opposite the direction of the field. For the negative charge to move in
the direction of the field, an external agent must apply a force and do positive
work on the charge.
Now consider the more general case of a charged particle that moves between 𝖠
and 𝖡 in a uniform electric field such that the vector :
s is not parallel to the field
lines as shown in Figure 20.3. In this case, Equation 20.3 gives
E
𝖡
DV 5 2
𝖠
:
:
E ? d:
s 5 2E ?
E
𝖡
:
d:
s 5 2E ? :
s
20.8b
:
where again E was removed from the integral because it is constant. The change in
potential energy of the charge–field system is
:
E
훾
S
s
훽
u
d
훿
Points 훾 and 훿 are at the
same electric potential.
Figure 20.3 A uniform electric
field directed along the positive x
axis.
c Change in potential between
two points in a uniform electric
field
𝖠
DU 5 q 0 DV 5 2q 0 E ? :
s
S
20.9b
Finally, we conclude from Equation 20.8 that all points in a plane perpendicular
to a uniform electric field are at the same electric potential. We can see that in Figure 20.3, where the potential difference V𝖡 2 V𝖠 is equal to the potential difference
:
V𝖢 2 V𝖠. (Prove this fact to yourself by working out two dot products for E ? :
s : one
:
for :
s 𝖠:𝖡, where the angle between E and :
s is arbitrary as shown in Figure 20.3,
and one for :
s 𝖠:𝖢, where 5 0.) Therefore, V𝖡 5 V𝖢. The name equipotential surface is given to any surface consisting of a continuous distribution of points having
the same electric potential.
660
CHAPTER 20 | Electric Potential and Capacitance
훾
9V
8V
7V
훽

훿
6V

Figure 20.4 (Quick Quiz 20.2)
Four equipotential surfaces.
The equipotential surfaces associated with a uniform electric field consist of
a family of parallel planes that are all perpendicular to the field. Equipotential
surfaces associated with fields having other symmetries are described in later
sections.
QUI C K QU IZ 20.2 The labeled points in Figure 20.4 are on a series of equipotential
surfaces associated with an electric field. Rank (from greatest to least) the work done
by the electric field on a positively charged particle that moves from 𝖠 to 𝖡, from
𝖡 to 𝖢, from 𝖢 to 𝖣, and from 𝖣 to 𝖤.
Example 20.1 | The Electric Field Between Two Parallel Plates of Opposite Charge
A battery has a specified potential difference DV between its terminals and establishes that potential difference between conductors attached to the terminals. A 12-V
battery is connected between two parallel plates as shown in Figure 20.5. The separation between the plates is d 5 0.30 cm, and we assume the electric field between the
plates to be uniform. (This assumption is reasonable if the plate separation is small
relative to the plate dimensions and we do not consider locations near the plate
edges.) Find the magnitude of the electric field between the plates.
SOLUTION
Conceptualize In Chapter 19 we investigated the uniform electric field between parallel plates. The new feature to this problem is that the electric field is related to the new
concept of electric potential.
A
B
V = 12 V
d
Figure 20.5 (Example 20.1)
A 12-V battery connected to two
parallel plates. The electric field
between the plates has a magnitude
given by the potential difference DV
divided by the plate separation d.
Categorize The electric field is evaluated from a relationship between field and potential given in this section, so we categorize this example as a substitution problem.
Use Equation 20.6 to evaluate the magnitude of the
electric field between the plates:
E5
u VB 2 VA u
d
5
12 V
5 4.0 3 103 V/m
0.30 3 1022 m
The configuration of plates in Figure 20.5 is called a parallel-plate capacitor and is examined in greater detail in Section 20.7.
Example 20.2 | Motion of a Proton in a Uniform Electric Field
A proton is released from rest at point 𝖠 in a uniform
electric field that has a magnitude of 8.0 3 104 V/m
(Fig. 20.6). The proton undergoes a displacement of mag:
nitude d 5 0.50 m to point 𝖡 in the direction of E. Find
the speed of the proton after completing the displacement.
SOLUTION
Conceptualize Visualize the proton in Figure 20.6 moving
downward through the potential difference. The situation is
analogous to an object falling through a gravitational field.
훽
S
v훽 0
S
d
Figure 20.6 (Example
20.2) A proton accelerates
from 𝖠 to 𝖡 in the direction
of the electric field.
훾
E
S
v훾
Categorize The system of the proton and the two plates in Figure 20.6 does not interact with the environment, so we
model it as an isolated system.
Analyze Use Equation 20.6 to find the potential differ-
DV 5 2Ed 5 2(8.0 3 104 V/m)(0.50 m) 5 24.0 3 104 V
ence between points 𝖠 and 𝖡:
Write the appropriate reduction of Equation 7.2, the
conservation of energy equation, for the isolated system
of the charge and the electric field:
DK 1 DU 5 0
661
20.3 | Electric Potential and Potential Energy Due to Point Charges
20.2 cont.
Substitute the changes in energy for both terms:
(12mv 2 2 0) 1 e DV 5 0
Solve for the final speed of the proton:
v5
Substitute numerical values:
v5
B
22e DV
m
22(1.6 3 10219 C)(24.0 3 104 V)
B
1.67 3 10227 kg
5 2.8 3 106 m/s
Finalize Because DV is negative for the field, DU is also
negative for the proton–field system. The negative value of
DU means the potential energy of the system decreases as
the proton moves in the direction of the electric field. As
the proton accelerates in the direction of the field, it gains
kinetic energy while the electric potential energy of the
system decreases at the same time.
Figure 20.6 is oriented so that the proton moves
downward. The proton’s motion is analogous to that of
an object falling in a gravitational field. Although the
gravitational field is always downward at the surface of
the Earth, an electric field can be in any direction, depending on the orientation of the plates creating the
field. Therefore, Figure 20.6 could be rotated 908 or 1808
and the proton could move horizontally or upward in the
electric field!
20.3 | Electric Potential and Potential
Energy Due to Point Charges
As discussed in Section 19.6, an isolated positive point charge q produces an electric
field directed radially outward from the charge. To find the electric potential at a
point located a distance r from the charge, let’s begin with the general expression
for potential difference,
V 𝖡 2 V𝖠 5 2
E
𝖡
:
E ? d:
s
𝖠
where 𝖠 and 𝖡 are the two arbitrary points shown in Figure 20.7. At any point in
:
space, the electric field due to the point charge is E 5 (keq /r 2)r̂ (Eq. 19.5), where
:
r̂ is a unit vector directed radially outward from the charge. The quantity E ? d :
s
can be expressed as
q
:
s
E ? d:
s 5 ke 2 r̂ ? d :
r
Because the magnitude of r̂ is 1, the dot product r̂ ? d :
s 5 ds cos , where is
the angle between r̂ and d :
s . Furthermore, ds cos is the projection of d :
s onto
r̂; therefore, ds cos 5 dr. That is, any displacement d :
s along the path from point
𝖠 to point 𝖡 produces a change dr in the magnitude of :
r , the position vector of
the point relative to the charge creating the field. Making these substitutions, we
:
find that E ? d :
s 5 (ke q/r 2)dr ; hence, the expression for the potential difference
becomes
V𝖡 2 V𝖠 5 2keq
E
r𝖡
r𝖠
V𝖡 2 V𝖠 5 keq
훾
r̂
dr
u
S
ds
S
S
훽
r
r훾
S
r훽
q
q r𝖡
dr
5
k
e
r r𝖠
r2
u
3 1r 2 1r 4
𝖡
𝖠
:
:
20.10b
Equation 20.10 shows us that the integral of E ? d s is independent of the path
between points 𝖠 and 𝖡. Multiplying by a charge q0 that moves between points 𝖠
:
and 𝖡, we see that the integral of q 0 E ? d :
s is also independent of path. This latter
integral, which is the work done by the electric force on the charge q 0, shows that the
electric force is conservative (see Section 6.7). We define a field that is related to a
The two dashed circles represent
intersections of spherical equipotential surfaces with the page.
Figure 20.7 The potential
difference between points 𝖠 and
𝖡 due to a point charge q depends
only on the initial and final radial
coordinates r𝖠 and r𝖡.
662
CHAPTER 20 | Electric Potential and Capacitance
A potential k e q 2 /r 12
exists at point P due to
charge q 2 .
The potential energy of
the pair of charges is
given by k e q 1 q 2 /r 12.
r12
q2
q2
r12
P
q1
q
V ke r 2
a
b
12
conservative force as a conservative field. Therefore, Equation
20.10 tells us that the electric field of a fixed point charge q is
conservative. Furthermore, Equation 20.10 expresses the important result that the potential difference between any two points
𝖠 and 𝖡 in a field created by a point charge depends only on
the radial coordinates r𝖠 and r𝖡. It is customary to choose the
reference of electric potential for a point charge to be V 5 0 at
r𝖠 5 `. With this reference choice, the electric potential due to
a point charge at any distance r from the charge is
q
V 5 ke
20.11b
r
We obtain the electric potential resulting from two or more
point charges by applying the superposition principle. That is,
the total electric potential at some point P due to several point
charges is the sum of the potentials due to the individual charges. For a group of
point charges, we can write the total electric potential at P as
Active Figure 20.8 (a) Two point charges separated by a
distance r12. (b) Charge q1 is removed.
c Electric potential due to several
point charges
Pitfall Prevention | 20.4
Similar Equation Warning
Do not confuse Equation 20.11 for
the electric potential of a point
charge with Equation 19.5 for the
electric field of a point charge.
Potential is proportional to 1/r,
whereas the magnitude of the field
is proportional to 1/r 2. The effect of
a charge on the space surrounding
it can be described in two ways.
The charge sets up a vector electric
:
field E , which is related to the force
experienced by a test charge placed
in the field. It also sets up a scalar
potential V, which is related to the
potential energy of the two-charge
system when a test charge is placed
in the field.
qi
i ri
V 5 ke o
20.12b
where the potential is again taken to be zero at infinity and ri is the distance from the
point P to the charge q i. Notice that the sum in Equation 20.12 is an algebraic sum
of scalars rather than a vector sum (which we used to calculate the electric field of a
:
group of charges in Eq. 19.6). Therefore, it is often much easier to evaluate V than E.
Now consider the potential energy of a system of two charged particles. If V2 is
the electric potential at a point P due to charge q 2, the work an external agent must
do to bring a second charge q 1 from infinity to P without acceleration is q1V2. This
work represents a transfer of energy into the system, and the energy appears in the
system as potential energy U when the particles are separated by a distance r12 (Active Fig. 20.8a). Therefore, the potential energy of the system can be expressed as1
U 5 ke
q 1q 2
r12
20.13b
If the charges are of the same sign, then U is positive. Positive work must be done by
an external agent on the system to bring the two charges near each other (because
charges of the same sign repel). If the charges are of opposite sign, then U is negative. Negative work is done by an external agent against the attractive force between
the charges of opposite sign as they are brought near each other; a force must be
applied opposite the displacement to prevent q1 from accelerating toward q2.
In Active Figure 20.8b, we have removed the charge q1. At the position this charge
previously occupied, point P, Equations 20.2 and 20.13 can be used to define a potential
due to charge q2 as V 5 U/q1 5 keq2/r12. This expression is consistent with Equation 20.11.
If the system consists of more than two charged particles, we can obtain the total
potential energy of the system by calculating U for every pair of charges and summing the terms algebraically. The total electric potential energy of a system of point
charges is equal to the work required to bring the charges, one at a time, from an
infinite separation to their final positions.
QUI C K QU IZ 20.3 A spherical balloon contains a positively charged object at its
center. (i) As the balloon is inflated to a greater volume while the charged object
remains at the center, does the electric potential at the surface of the balloon (a) increase, (b) decrease, or (c) remain the same? (ii) Does the electric flux through the
surface of the balloon (a) increase, (b) decrease, or (c) remain the same?
1The expression for the electric potential energy of a system made up of two point charges, Equation 20.13, is of the
same form as the equation for the gravitational potential energy of a system made up of two point masses, 2Gm1m2/r
(see Chapter 11). The similarity is not surprising considering that both expressions are derived from an inverse-square
force law.
663
20.3 | Electric Potential and Potential Energy Due to Point Charges
Q U I CK QUI Z 20.4 In Active Figure 20.8a, take q1 to be a negative source charge
and q2 to be the test charge. (i) If q2 is initially positive and is changed to a charge of
the same magnitude but negative, what happens to the potential at the position of q2
due to q1? (a) It increases. (b) It decreases. (c) It remains the same. (ii) When q2 is
changed from positive to negative, what happens to the potential energy of the twocharge system? Choose from the same possibilities.
Exampl e 20.3 | The Electric Potential Due to Two Point Charges
As shown in Figure 20.9a, a charge q1 5 2.00 C is located
at the origin and a charge q2 5 26.00 C is located at
(0, 3.00) m.
y
y
6.00 mC
6.00 mC
3.00 m
(A) Find the total electric potential due to these charges
at the point P, whose coordinates are (4.00, 0) m.
3.00 m
2.00 mC
P
x
4.00 m
SOLUTION
3.00 mC
x
4.00 m
a
Conceptualize Recognize first that the 2.00-C and
26.00-C charges are source charges and set up an electric
field as well as a potential at all points in space, including
point P.
2.00 mC
b
Figure 20.9 (Example 20.3) (a) The electric potential at P due
to the two charges q1 and q2 is the algebraic sum of the potentials
due to the individual charges. (b) A third charge q3 5 3.00 C is
brought from infinity to point P.
Categorize The potential is evaluated using an equation de-
veloped in this chapter, so we categorize this example as a substitution problem.
q1
q2
1
2
1r 1 r 2
Use Equation 20.12 for the system of two source
charges:
VP 5 ke
Substitute numerical values:
VP 5 (8.99 3 109 N ? m2/C2)
26
26
2 6.00 3 10 C
3 10 C
1
12.004.00
2
m
5.00 m
5 2 6.29 3 103 V
(B) Find the change in potential energy of the system of two charges plus a third charge q 3 5 3.00 C as the latter charge
moves from infinity to point P (Fig. 20.9b).
SOLUTION
Assign Ui 5 0 for the system to the configuration in
which the charge q 3 is at infinity. Use Equation 20.2 to
evaluate the potential energy for the configuration in
which the charge is at P :
Uf 5 q 3VP
Substitute numerical values to evaluate DU :
DU 5 Uf 2 Ui 5 q 3VP 2 0 5 (3.00 3 1026 C)(2 6.29 3 103 V)
5 21.89 3 1022 J
Therefore, because the potential energy of the system has decreased, an external agent has to do positive work to remove
the charge q 3 from point P back to infinity.
What If? You are working through this example with a classmate and she says, “Wait a minute! In part (B), we ignored the
potential energy associated with the pair of charges q 1 and q 2!” How would you respond?
Answer Given the statement of the problem, it is not necessary to include this potential energy because part (B) asks for
the change in potential energy of the system as q 3 is brought in from infinity. Because the configuration of charges q 1 and q 2
does not change in the process, there is no DU associated with these charges.
664
CHAPTER 20 | Electric Potential and Capacitance
20.4 | Obtaining the Value of the Electric Field
from the Electric Potential
:
The electric field E and the electric potential V are related as shown in Equation 20.3,
:
which tells us how to find DV if the electric field E is known. We now show how to
calculate the value of the electric field if the electric potential is known in a certain
region.
From Equation 20.3, we can express the potential difference dV between two
points a distance ds apart as
:
s
dV 5 2E ? d :
20.14b
:
:
If the electric field has only one component Ex , then E ? d s 5 Ex dx. Therefore,
Equation 20.14 becomes dV 5 2Ex dx, or
dV
20.15b
dx
That is, the x component of the electric field is equal to the negative of the derivative
of the electric potential with respect to x. Similar statements can be made about the
y and z components. Equation 20.15 is the mathematical statement of the electric
field being a measure of the rate of change with position of the electric potential as
mentioned in Section 20.1.
When a test charge undergoes a displacement d :
s along an equipotential surface,
then dV 5 0 because the potential is constant along an equipotential surface. From
:
:
Equation 20.14, we see that dV 5 2E ? d :
s 5 0; therefore, E must be perpendicular to the displacement along the equipotential surface. This result shows that the
equipotential surfaces must always be perpendicular to the electric field lines passing through them.
As mentioned at the end of Section 20.2, the equipotential surfaces associated
with a uniform electric field consist of a family of planes perpendicular to the
field lines. Figure 20.10a shows some representative equipotential surfaces for this
situation.
If the charge distribution creating an electric field has spherical symmetry such that
the volume charge density depends only on the radial distance r, the electric field is
:
radial. In this case, E ? d :
s 5 Er dr, and we can express dV as dV 5 2Er dr. Therefore,
Ex 5 2
Er 5 2
A uniform electric field produced
by an infinite sheet of charge
dV
dr
A spherically symmetric electric
field produced by a point charge
20.16b
An electric field produced by an
electric dipole
q
⫹
S
E
a
b
c
Figure 20.10 Equipotential surfaces (the dashed blue lines are intersections of these surfaces with the page) and electric
field lines. In all cases, the equipotential surfaces are perpendicular to the electric field lines at every point.
20.4 | Obtaining the Value of the Electric Field from the Electric Potential
665
For example, the electric potential of a point charge is V 5 keq/r. Because V is a
function of r only, the potential function has spherical symmetry. Applying Equation 20.16, we find that the magnitude of the electric field due to the point charge
is Er 5 keq/r 2, a familiar result. Notice that the potential changes only in the radial
direction, not in any direction perpendicular to r. Therefore, V (like Er) is a function only of r, which is again consistent with the idea that equipotential surfaces are
perpendicular to field lines. In this case, the equipotential surfaces are a family of
spheres concentric with the spherically symmetric charge distribution (Fig. 20.10b).
The equipotential surfaces for an electric dipole are sketched in Figure 20.10c.
In general, the electric potential is a function of all three spatial coordinates. If
V(r) is given in terms of the Cartesian coordinates, the electric field components Ex ,
Ey , and Ez can readily be found from V(x, y, z) as the partial derivatives
Ex 5 2
0V
0x
Ey 5 2
0V
0y
Ez 5 2
0V
0z
20.17b
c Finding the electric field from
the potential
Q U I CK QUI Z 20.5 In a certain region of space, the electric potential is zero everywhere along the x axis. (i) From this information, you can conclude that the x component of the electric field in this region is (a) zero, (b) in the positive x direction,
or (c) in the negative x direction. (ii) Suppose the electric potential is 12 V everywhere along the x axis. From the same choices, what can you conclude about the
x component of the electric field now?
Exampl e 20.4 | The Electric Potential Due to a Dipole
y
An electric dipole consists of two charges of equal magnitude and opposite sign separated by a distance 2a as shown in Figure 20.11. The dipole is along the x axis and is
centered at the origin.
P
y
(A) Calculate the electric potential at point P on the y axis.
q
a
q
a
R
x
x
SOLUTION
Conceptualize Compare this situation to that in part (A) of Example 19.4. It is the same
situation, but here we are seeking the electric potential rather than the electric field.
Figure 20.11 (Example 20.4)
Categorize Because the dipole consists of only two source charges, the electric potential
An electric dipole located on the
x axis.
can be evaluated by summing the potentials due to the individual charges.
Analyze Use Equation 20.12 to find the electric potential
VP 5 ke o
i
at P due to the two charges:
q
qi
2q
5 ke
1
50
2
2
ri
"a 1 y
"a 2 1 y 2
1
2
(B) Calculate the electric potential at point R on the positive x axis.
SOLUTION
Use Equation 20.12 to find the electric potential at R due
to the two charges:
2q
qi
2keqa
q
5 ke
52 2
1
r
x
2
a
x
1
a
x
2 a2
i i
1
VR 5 ke o
2
(C) Calculate V and Ex at a point on the x axis far from the dipole.
SOLUTION
For point R far from the dipole such that x .. a, neglect
a 2 in the denominator of the answer to part (B) and write
V in this limit:
x
a
2keqa
2keqa
1 x 2a 2<2 x
VR 5 lim 2
2
2
2
(x
a)
continued
666
CHAPTER 20 | Electric Potential and Capacitance
20.4 cont.
Use Equation 20.15 and this result to calculate the x component of the electric field at a point on the x axis far
from the dipole:
Ex 5 2
2keqa
d
dV
52
2 2
dx
dx
x
1
5 2keqa
2
4keqa
d 1
5 2 3 (x
dx x 2
x
1 2
a)
Finalize The potentials in parts (B) and (C) are negative because points on the positive x axis are closer to the negative
charge than to the positive charge. For the same reason, the x component of the electric field is negative.
dq2
dq1
r2
r1
dq3
P
r3
Figure 20.12 The electric potential
at point P due to a continuous charge
distribution can be calculated by
dividing the charge distribution into
elements of charge dq and summing
the electric potential contributions
over all elements. Three sample
elements of charge are shown.
c Electric potential due to a
continuous charge distribution
20.5 | Electric Potential Due to Continuous
Charge Distributions
The electric potential due to a continuous charge distribution can be calculated
using two different methods. The first method is as follows. If the charge distribution
is known, we consider the potential due to a small charge element dq, treating this
element as a point charge (Fig. 20.12). From Equation 20.11, the electric potential
dV at some point P due to the charge element dq is
dV 5 ke
dq
r
20.18b
where r is the distance from the charge element to point P. To obtain the total potential at point P, we integrate Equation 20.18 to include contributions from all elements of the charge distribution. Because each element is, in general, a different
distance from point P and ke is constant, we can express V as
V 5 ke
E
dq
r
20.19b
In effect, we have replaced the sum in Equation 20.12 with an integral. In this expression for V, the electric potential is taken to be zero when point P is infinitely far from
the charge distribution.
The second method for calculating the electric potential is used if the electric
field is already known from other considerations such as Gauss’s law. If the charge
:
distribution has sufficient symmetry, we first evaluate E using Gauss’s law and then
substitute the value obtained into Equation 20.3 to determine the potential difference DV between any two points. We then choose the electric potential V to be zero
at some convenient point.
PROB LEM-SOLVING STRATEGY: Calculating Electric Potential
The following procedure is recommended for solving problems that involve the
determination of an electric potential due to a charge distribution.
1. Conceptualize Think carefully about the individual charges or the charge
distribution you have in the problem and imagine what type of potential
would be created. Appeal to any symmetry in the arrangement of charges
to help you visualize the potential.
2. Categorize Are you analyzing a group of individual charges or a continuous
charge distribution? The answer to this question will tell you how to proceed
in the Analyze step.
667
20.5 | Electric Potential Due to Continuous Charge Distributions
3. Analyze When working problems involving electric potential, remember
that it is a scalar quantity, so there are no components to consider. Therefore,
when using the superposition principle to evaluate the electric potential at a
point, simply take the algebraic sum of the potentials due to each charge. You
must keep track of signs, however.
As with potential energy in mechanics, only changes in electric potential
are significant; hence, the point where the potential is set at zero is arbitrary.
When dealing with point charges or a finite-sized charge distribution, we usually define V 5 0 to be at a point infinitely far from the charges. If the charge
distribution itself extends to infinity, however, some other nearby point must
be selected as the reference point.
(a) If you are analyzing a group of individual charges: Use the superposition principle, which states that when several point charges are present, the resultant
potential at a point P in space is the algebraic sum of the individual potentials at
P due to the individual charges (Eq. 20.12). Example 20.4 demonstrated this
procedure.
(b) If you are analyzing a continuous charge distribution: Replace the sums for
evaluating the total potential at some point P from individual charges by
integrals (Eq. 20.19). The charge distribution is divided into infinitesimal
elements of charge dq located at a distance r from the point P. An element
is then treated as a point charge, so the potential at P due to the element
is dV 5 ke dq/r. The total potential at P is obtained by integrating over the
entire charge distribution. For many problems, it is possible in performing
the integration to express dq and r in terms of a single variable. To simplify
the integration, give careful consideration to the geometry involved in the
problem. Examples 20.5 and 20.6 demonstrate such a procedure.
To obtain the potential from the electric field: Another method used to obtain
the potential is to start with the definition of the potential difference given by
:
Equation 20.3. If E is known or can be obtained easily (such as from Gauss’s
:
law), the line integral of E ? d :
s can be evaluated.
4. Finalize Check to see if your expression for the potential is consistent with
the mental representation and reflects any symmetry you noted previously.
Imagine varying parameters such as the distance of the observation point
from the charges or the radius of any circular objects to see if the mathematical result changes in a reasonable way.
Exampl e 20.5 | Electric Potential Due to a Uniformly Charged Ring
dq
(A) Find an expression for the electric potential at a point P located on the perpendicular central axis of a uniformly charged ring of radius a and total charge Q.
a 2 x 2
a
SOLUTION
Conceptualize Study Figure 20.13, in which the ring is oriented so that its plane is
x
perpendicular to the x axis and its center is at the origin. Notice that the symmetry of
the situation means that all the charges on the ring are the same distance from point P.
Categorize Because the ring consists of a continuous distribution of charge rather
than a set of discrete charges, we must use the integration technique represented by
Equation 20.19 in this example.
P
x
Figure 20.13 (Example 20.5) A
uniformly charged ring of radius a
lies in a plane perpendicular to the x
axis. All elements dq of the ring are the
same distance from a point P lying
on the x axis.
Analyze We take point P to be at a distance x from the center of the ring as shown
in Figure 20.13.
dq
dq
Use Equation 20.19 to express V in terms of the geometry:
5 ke
V 5 ke
r
"a 2 1 x 2
E
E
continued
668
CHAPTER 20 | Electric Potential and Capacitance
20.5 cont.
Noting that a and x are constants, bring "a 2 1 x 2 in
front of the integral sign and integrate over the ring:
(1) V 5
ke
"a 2 1 x 2
E
dq 5
keQ
"a 2 1 x 2
(B) Find an expression for the magnitude of the electric field at point P.
SOLUTION
:
From symmetry, notice that along the x axis E can have
only an x component. Therefore, apply Equation 20.15 to
Equation (1):
Ex 5 2
dV
d
5 2keQ
(a 2 1 x 2)21/2
dx
dx
5 2keQ(2 12)(a 2 1 x 2)23/2(2x)
Ex 5
ke x
(a 2
1 x 2 )3/2
Q
Finalize The only variable in the expressions for V and Ex is x. That is not surprising because our calculation is valid only
for points along the x axis, where y and z are both zero. This result for the electric field agrees with that obtained by direct
integration (see Example 19.6). For practice, use the result of part (B) in Equation 20.3 to verify that the potential is given
by the expression in part (A).
Example 20.6 | Electric Potential Due to a Uniformly Charged Disk
A uniformly charged disk has radius R and surface charge
density .
(A) Find the electric potential at a point P along the perpendicular central axis of the disk.
SOLUTION
Conceptualize If we consider the disk to be a set of concen-
Figure 20.14 (Example
20.6) A uniformly charged
disk of radius R lies in a
plane perpendicular to the x
axis. The calculation of the
electric potential at any point
P on the x axis is simplified by
dividing the disk into many
rings of radius r and width dr,
with area 2r dr.
R
r 2 x 2
r
x
P
x
dA 2pr dr
dr
tric rings, we can use our result from Example 20.5—which
gives the potential due to a ring of radius a—and sum the contributions of all rings making up the disk. Because point P is
on the central axis of the disk, symmetry again tells us that all points in a given ring are the same distance from P.
Categorize Because the disk is continuous, we evaluate the potential due to a continuous charge distribution rather than
a group of individual charges.
Analyze Find the amount of charge dq on a ring of radius
r and width dr as shown in Figure 20.14:
dq 5 dA 5 (2r dr) 5 2r dr
Use this result in Equation (1) in Example 20.5 (with a
replaced by r and Q replaced by dq) to find the potential
due to the ring:
dV 5
To obtain the total potential at P, integrate this expression
over the limits r 5 0 to r 5 R, noting that x is a constant:
V 5 ke This integral is of the common form eun du, where
n 5 2 12 and u 5 r 2 1 x 2, and has the value un11/(n 1 1).
Use this result to evaluate the integral:
(1) V 5 2ke [(R 2 1 x 2)1/2 2 x]
ke dq
"r 2 1 x 2
E
R
0
5
ke2r dr
"r 2 1 x 2
2r dr
"r 2 1 x 2
5 ke E
R
(r 2 1 x 2)21/2 2r dr
0
(B) Find the x component of the electric field at a point P along the perpendicular central axis of the disk.
SOLUTION
As in Example 20.5, use Equation 20.15 to find the
electric field at any axial point:
:
(2) Ex 5 2
dV
x
5 2ke 1 2
dx
(R 2 1 x 2)1/2
3
4
Finalize The calculation of V and E for an arbitrary point off the x axis is more difficult to perform because of the absence
of symmetry and we do not treat that situation in this book.
669
20.6 | Electric Potential Due to a Charged Conductor
20.6 | Electric Potential Due to a Charged Conductor
In Section 19.11, we found that when a solid conductor in equilibrium carries a net
charge, the charge resides on the conductor’s outer surface. Furthermore, the electric
field just outside the conductor is perpendicular to the surface and the field inside is zero.
We now generate another property of a charged conductor, related to electric potential. Consider two points 𝖠 and 𝖡 on the surface of a charged conductor as shown in
:
Figure 20.15. Along a surface path connecting these points, E is always perpendicular
:
to the displacement d :
s ; therefore, E ? d :
s 5 0. Using this result and Equation 20.3, we
conclude that the potential difference between 𝖠 and 𝖡 is necessarily zero:
E
𝖡
V𝖡 2 V𝖠 5 2
:
E ? d:
s 50
𝖠
Pitfall Prevention | 20.5
Potential May Not Be Zero
The electric potential inside the
conductor is not necessarily zero
in Figure 20.15, even though the
electric field is zero. Equation 20.14
shows that a zero value of the field
results in no change in the potential
from one point to another inside the
conductor. Therefore, the potential
everywhere inside the conductor,
including the surface, has the same
value, which may or may not be zero,
depending on where the zero of
potential is defined.
This result applies to any two points on the surface. Therefore, V is constant everywhere on the surface of a charged conductor in equilibrium. That is,
the surface of any charged conductor in electrostatic equilibrium is an equipotential surface: every point on the surface of a charged conductor in equilibrium is at the same electric potential. Furthermore, because the electric field is
zero inside the conductor, the electric potential is constant everywhere inside
the conductor and equal to its value at the surface.
Because of the constant value of the potential, no work is required to move a test
charge from the interior of a charged conductor to its surface.
Consider a solid metal conducting sphere of radius R and total positive charge Q
as shown in Figure 20.16a. As determined in part (A) of Example 19.10, the electric
field outside the sphere is keQ /r 2 and points radially outward. Because the field outside a spherically symmetric charge distribution is identical to that of a point charge,
we expect the potential to also be that of a point charge, keQ /r. At the surface of
the conducting sphere in Figure 20.16a, the potential must be keQ /R. Because the
entire sphere must be at the same potential, the potential at any point within the
sphere must also be keQ /R. Figure 20.16b is a plot of the electric potential as a function of r, and Figure 20.16c shows how the electric field varies with r.
When a net charge is placed on a spherical conductor, the surface charge density
is uniform as indicated in Figure 20.16a. If the conductor is nonspherical as in Figure 20.15, however, the surface charge density is high where the radius of curvature
is small (as noted in Section 19.11) and low where the radius of curvature is large.
Because the electric field immediately outside the conductor is proportional to the
surface charge density, the electric field is large near convex points having small radii
of curvature and reaches very high values at sharp points. In Example 20.7, the relationship between electric field and radius of curvature is explored mathematically.
a
R V
b
k eQ
r
k eQ
R
r
Notice from the spacing of the
positive signs that the surface
charge density is nonuniform.
k eQ
r2
E
c
훾
훽S
E
Figure 20.15 An arbitrarily
shaped conductor carrying a positive
charge. When the conductor is
in electrostatic equilibrium, all
the charge resides at the surface,
:
E 5 0 inside the conductor, and the
:
direction of E immediately outside
the conductor is perpendicular to
the surface. The electric potential is
constant inside the conductor and is
equal to the potential at the surface.
R
r
Figure 20.16 (a) The excess
charge on a conducting sphere of
radius R is uniformly distributed on
its surface. (b) Electric potential
versus distance r from the center
of the charged conducting sphere.
(c) Electric field magnitude versus
distance r from the center of the
charged conducting sphere.
670
CHAPTER 20 | Electric Potential and Capacitance
Example 20.7 | Two Connected Charged Spheres
Two spherical conductors of radii r 1 and r 2 are separated by a distance much
greater than the radius of either sphere. The spheres are connected by a conducting wire as shown in Figure 20.17. The charges on the spheres in equilibrium are q 1 and q 2, respectively, and they are uniformly charged. Find the ratio
of the magnitudes of the electric fields at the surfaces of the spheres.
SOLUTION
Conceptualize Imagine the spheres are much farther apart than shown in Figure 20.17. Because they are so far apart, the field of one does not affect the charge
distribution on the other. The conducting wire between them ensures that both
spheres have the same electric potential.
r1
q1
Figure 20.17
(Example 20.7) Two
charged spherical
conductors connected by a conducting wire. The
spheres are at the
same electric
potential V .
r2
q2
Categorize Because the spheres are so far apart, we model the charge distribution on them as spherically symmetric, and
we can model the field and potential outside the spheres to be that due to point charges.
Analyze Set the electric potentials at the surfaces of the
V 5 ke
spheres equal to each other:
Solve for the ratio of charges on the spheres:
Write expressions for the magnitudes of the electric fields
at the surfaces of the spheres:
Evaluate the ratio of these two fields:
Substitute for the ratio of charges from Equation (1):
q2
q1
5 ke
r1
r2
q1
r1
5
r2
q2
q1
E 1 5 ke 2 and
r1
(1)
E1
E2
(2)
5
E 2 5 ke
q2
r 22
q 1 r 22
q 2 r 12
E1
E2
5
r2
r 1 r 22
5
r 2 r 12
r1
Finalize The field is stronger in the vicinity of the smaller sphere even though the electric potentials at the surfaces of
both spheres are the same. If r 2 : 0, then E 2 : `, verifying the statement above that the electric field is very large at sharp
points.
THINKING PHYSICS 20.1
Why is the end of a lightning rod pointed?
Reasoning The role of a lightning rod is to serve as a loca-
tion at which the lightning strikes so that the charge delivered by the lightning will pass safely to the ground. If the
lightning rod is pointed, the electric field due to charges
moving between the rod and the ground is very strong
near the point because the radius of curvature of the conductor is very small. This large electric field will greatly increase the likelihood that the lightning strike will occur
near the tip of the lightning rod rather than elsewhere. b
A Cavity Within a Conductor
Suppose a conductor of arbitrary shape contains a cavity as shown in Figure 20.18.
Let’s assume no charges are inside the cavity. In this case, the electric field inside the
cavity must be zero regardless of the charge distribution on the outside surface of the
conductor as we mentioned in Section 19.11. Furthermore, the field in the cavity is
zero even if an electric field exists outside the conductor.
To prove this point, remember that every point on the conductor is at the same
electric potential; therefore, any two points 𝖠 and 𝖡 on the cavity’s surface must
20.7 | Capacitance
671
:
be at the same potential. Now imagine a field E exists in the cavity and evaluate the
potential difference V𝖡 2 V𝖠 defined by Equation 20.3:
V𝖡 2 V 𝖠 5 2
E
𝖡
The electric field in the cavity is
zero regardless of the charge on
the conductor.
:
E ? d:
s
𝖠
:
:
Because V𝖡 2 V𝖠 5 0, the integral of E ? d s must be zero for all paths between
any two points 𝖠 and 𝖡 on the conductor. The only way that can be true for all paths
:
is if E is zero everywhere in the cavity. Therefore, a cavity surrounded by conducting
walls is a field-free region as long as no charges are inside the cavity.
This result has some interesting applications. For example, it is possible to shield
an electronic device or even an entire laboratory from external fields by surrounding it with conducting walls. Shielding is often necessary during highly sensitive
electrical measurements. During a thunderstorm, the safest location is inside an automobile. Even if lightning strikes the car, the metal body guarantees that you will
:
not receive a shock inside the car, where E 5 0.
훾
훽
Figure 20.18 A conductor in
electrostatic equilibrium containing
a cavity.
When the capacitor is charged, the
conductors carry charges of equal
magnitude and opposite sign.
20.7 | Capacitance
As we continue with our discussion of electricity and, in later chapters, magnetism,
we shall build circuits consisting of circuit elements. A circuit generally consists of a
number of electrical components (circuit elements) connected together by conducting wires and forming one or more closed loops. These circuits can be considered as
systems that exhibit a particular type of behavior. The first circuit element we shall
consider is a capacitor.
In general, a capacitor consists of two conductors of any shape. Consider two conductors having a potential difference of DV between them. Let us assume that the
conductors have charges of equal magnitude and opposite sign as in Figure 20.19.
This situation can be accomplished by connecting two uncharged conductors to the
terminals of a battery. Once that is done and the battery is disconnected, the charges
remain on the conductors. We say that the capacitor stores charge.
The potential difference DV across the capacitor is the magnitude of the potential
difference between the two conductors. This potential difference is proportional to the
charge Q on the capacitor, which is defined as the magnitude of the charge on either of
the two conductors. The capacitance C of a capacitor is defined as the ratio of the charge
on the capacitor to the magnitude of the potential difference across the capacitor:
C;
Q
DV
20.20b
By definition, capacitance is always a positive quantity. Because the potential difference
is proportional to the charge, the ratio Q/DV is constant for a given capacitor. Equation 20.20 tells us that the capacitance of a system is a measure of the amount of
charge that can be stored on the capacitor for a given potential difference.
From Equation 20.20, we see that capacitance has the SI units coulombs per volt,
which is called a farad (F) in honor of Michael Faraday. The farad is a very large unit
of capacitance. In practice, typical devices have capacitances ranging from microfarads to picofarads.
Q U I CK QUI Z 20.6 A capacitor stores charge Q at a potential difference DV.
What happens if the voltage applied to a capacitor by a battery is doubled to 2 DV ?
(a) The capacitance falls to half its initial value, and the charge remains the same.
(b) The capacitance and the charge both fall to half their initial values. (c) The
capacitance and the charge both double. (d) The capacitance remains the same, and
the charge doubles.
The capacitance of a device depends on the geometric arrangement of the
conductors. To illustrate this point, let us calculate the capacitance of an isolated
Q
Q
Figure 20.19 A capacitor consists
of two conductors electrically
isolated from each other and their
surroundings.
c Definition of capacitance
Pitfall Prevention | 20.6
Capacitance Is a Capacity
To help you understand the concept
of capacitance, think of similar
notions that use a similar word.
The capacity of a milk carton is the
volume of milk it can store. The heat
capacity of an object is the amount
of energy an object can store per
unit of temperature difference.
The capacitance of a capacitor is the
amount of charge the capacitor can
store per unit of potential difference.
672
CHAPTER 20 | Electric Potential and Capacitance
Pitfall Prevention | 20.7
Potential Difference Is DV, Not V
We use the symbol DV for the
potential difference across a circuit
element or a device because this
notation is consistent with our
definition of potential difference and
with the meaning of the delta sign. It
is a common but confusing practice
to use the symbol V without the delta
sign for a potential difference. Keep
that in mind if you consult other
texts.
When the capacitor is connected
to the terminals of a battery,
electrons transfer between the
plates and the wires so that the
plates become charged.
Q
Q
Area A
spherical conductor of radius R and charge Q. (Based on the shape of the field lines
from a single spherical conductor, we can model the second conductor as a concentric spherical shell of infinite radius.) Because the potential of the sphere is simply
keQ/R (and V 5 0 for the shell of infinite radius), the capacitance of the sphere is
C5
Q
DV
5
Q
k e Q /R
R
5 4 0 R
ke
20.21b
(Remember from Section 19.4 that the Coulomb constant ke 5 1/40.) Equation 20.21
shows that the capacitance of an isolated charged sphere is proportional to the sphere’s
radius and is independent of both the charge and the potential difference.
The capacitance of a pair of oppositely charged conductors can be calculated in
the following manner. A convenient charge of magnitude Q is assumed, and the potential difference is calculated using the techniques described in Section 20.5. One
then uses C 5 Q/DV to evaluate the capacitance. As you might expect, the calculation is relatively straightforward if the geometry of the capacitor is simple.
Let us illustrate with two familiar geometries: parallel plates and concentric cylinders.
In these examples, we shall assume that the charged conductors are separated by a vacuum. (The effect of a material between the conductors will be treated in Section 20.10.)
The Parallel-Plate Capacitor
A parallel-plate capacitor consists of two parallel plates of equal area A separated
by a distance d as in Figure 20.20. If the capacitor is charged, one plate has charge
Q and the other, charge 2Q. The magnitude of the charge per unit area on either
plate is 5 Q/A. If the plates are very close together (compared with their length
and width), we adopt a simplification model in which the electric field is uniform
between the plates and zero elsewhere, as we discussed in Example 19.12. According
to Example 19.12, the magnitude of the electric field between the plates is
E5
d
5
Q
5
0
0A
Because the field is uniform, the potential difference across the capacitor can be
found from Equation 20.6. Therefore,
DV 5 Ed 5
Qd
0 A
Substituting this result into Equation 20.20, we find that the capacitance is
Figure 20.20 A parallel-plate
capacitor consists of two parallel
conducting plates, each of area A,
separated by a distance d.
C5
C5
Pitfall Prevention | 20.8
Too Many Cs
Be sure not to confuse italic C for
capacitance with regular C for the
unit coulomb.
Q
DV
0A
d
5
Q
Qd/ 0 A
20.22b
That is, the capacitance of a parallel-plate capacitor is proportional to the area of its
plates and inversely proportional to the plate separation.
As you can see from the definition of capacitance, C 5 Q/DV, the amount of
charge a given capacitor can store for a given potential difference across its plates
increases as the capacitance increases. It therefore seems reasonable that a capacitor
constructed from plates having large areas should be able to store a large charge.
A careful inspection of the electric field lines for a parallel-plate capacitor reveals
that the field is uniform in the central region between the plates, but is nonuniform
at the edges of the plates. Figure 20.21 shows a drawing of the electric field pattern
of a parallel-plate capacitor, showing the nonuniform field lines at the plates’ edges.
As long as the separation between the plates is small compared with the dimensions
of the plates, the edge effects can be ignored and we can use the simplification
model in which the electric field is uniform everywhere between the plates.
20.7 | Capacitance
673
Active Figure 20.22 shows a battery connected to a single parallel-plate
capacitor with a switch in the circuit. Let us identify the circuit as a system.
When the switch is closed, the battery establishes an electric field in the
Q
wires and charges flow between the wires and the capacitor. As that occurs,
energy is transformed within the system. Before the switch is closed, energy
is stored as chemical potential energy in the battery. This type of energy
Q
is associated with chemical bonds and is transformed during the chemical
reaction that occurs within the battery when it is operating in an electric
circuit. When the switch is closed, some of the chemical potential energy
Figure 20.21 The electric field between the
in the battery is converted to electric potential energy related to the sepa- plates of a parallel-plate capacitor is uniform
ration of positive and negative charges on the plates. As a result, we can near the center but nonuniform near the
describe a capacitor as a device that stores energy as well as charge. We will edges.
explore this energy storage in more detail in Section 20.9.
As a biological example of a parallel-plate capacitor, consider the plasma memCapacitance of cell
membranes
brane for a neuron. The plasma membrane is a lipid bilayer containing a variety of
types of molecules. This membrane contains a number of structures, including ion
channels and ion pumps, which control concentrations of various ions on either side
of the membrane. These ions include potassium, chlorine, calcium, and sodium. As
a result of differences in these concentrations, there is an effective sheet of negative
charge on the intracellular side of the membrane and a sheet of positive charge on
the extracellular side. This results in a voltage of about 70 to 80 mV across the membrane. The sheets of charge act as parallel plates, so that the membrane can be modeled as a parallel-plate capacitor. The capacitance of the plasma membrane is about
2 F for each cm2 of membrane area.
When a neuron is carrying a signal, an event called an action potential occurs. SpeAction potential
cial structures in the cell membrane, called voltage-gated ion channels, are normally
closed. If the voltage across the membrane capacitor falls in magnitude to a threshold value of about 50 mV, the ion channels open, allowing a flow of sodium ions into
the cell. This flow reduces the voltage even further, allowing more sodium ions to
enter the cell, thereby reversing the polarity of the voltage across the capacitor in a
time interval measured in milliseconds. The voltage-gated ion channels then close
and other channels open, allowing movement of ions until the neuron returns to its
resting state.
With the switch
open, the capacitor
remains uncharged.
Electrons move
from the plate
to the wire,
leaving the
plate positively
charged.
Electric
field in
wire
+
+
+
+
+
+
S
Electrons move
from the wire to
the plate.
–
–
–
–
–
–
E
Electric field
between plates
V
a
Separation
of charges
represents
potential
energy.
V
Electric
field in
wire
Chemical potential
energy in the
battery is reduced.
b
Active Figure 20.22 (a) A circuit consisting of a capacitor, a battery, and a switch. (b) When the
switch is closed, the battery establishes an electric field in the wire and the capacitor becomes charged.
674
CHAPTER 20 | Electric Potential and Capacitance
Q
b
a
Q
b
ᐉ
a
Q
Q
This process can disturb neighboring regions of the
plasma membrane so that the action potential is propagated along the neuron. In the next chapter, we will see
how the capacitance of the plasma membrane combines
with another electrical characteristic of the membrane to
provide an electrical model for the conduction of a signal
along the neuron.
r
The Cylindrical Capacitor
Gaussian
surface
a
b
Figure 20.23 (a) A cylindrical capacitor consists of a solid
cylindrical conductor of radius a and length / surrounded by a
coaxial cylindrical shell of radius b. (b) End view. The dashed line
represents the end of the cylindrical gaussian surface of radius r
and length /.
A cylindrical capacitor consists of a cylindrical conductor of
radius a and charge Q coaxial with a larger cylindrical shell
of radius b and charge 2Q (Fig. 20.23a). Let us find the capacitance of this device if its length is /. If we assume that /
is large compared with a and b, we can adopt a simplification
model in which we ignore end effects. In this case, the field
is perpendicular to the axis of the cylinders and is confined
to the region between them (Fig. 20.23b). We first calculate
the potential difference between the two cylinders, which is
given in general by
E
Vb 2 Va 5 2
b
:
E ? d:
s
a
:
where E is the electric field in the region a , r , b. In Chapter 19, using Gauss’s
law, we showed that the electric field of a cylinder with charge per unit length has the magnitude E 5 2ke /r. The same result applies here because the outer cylinder does not contribute to the electric field inside it. Using this result and noting
:
that the direction of E is radially away from the inner cylinder in Figure 20.23b,
we find that
E
Vb 2 Va 5 2
b
a
Er dr 5 22ke E
b dr
a
r
5 22ke ln
1ab 2
Substituting this result into Equation 20.20 and using that 5 Q//, we find that
C5
Q
DV
5
Q
2keQ
,
b
ln
a
5
12
,
b
2ke ln
a
12
20.23b
where the magnitude of the potential difference between the cylinders is DV 5
|V a 2 V b| 5 2ke ln(b/a), a positive quantity. Our result for C shows that the capacitance is proportional to the length of the cylinders. As you might expect, the capacitance also depends on the radii of the two cylindrical conductors. As an example, a
coaxial cable consists of two concentric cylindrical conductors of radii a and b separated by an insulator. The cable carries currents in opposite directions in the inner
and outer conductors. Such a geometry is especially useful for shielding an electrical
signal from external influences. From Equation 20.23, we see that the capacitance
per unit length of a coaxial cable is
C
5
,
1
2ke ln
b
a
12
20.24b
20.8 | Combinations of Capacitors
Two or more capacitors often are combined in electric circuits. We can calculate the
equivalent capacitance of certain combinations using methods described in this section. Throughout this section, we assume the capacitors to be combined are initially
uncharged.
20.8 | Combinations of Capacitors
In studying electric circuits, we use a simplified pictorial representation called
a circuit diagram. Such a diagram uses circuit symbols to represent various circuit
elements. The circuit symbols are connected by straight lines that represent the
wires between the circuit elements. The circuit symbols for capacitors, batteries,
and switches as well as the color codes used for them in this text are given in
Figure 20.24. The symbol for the capacitor reflects the geometry of the most common model for a capacitor, a pair of parallel plates. The positive terminal of the
battery is at the higher potential and is represented in the circuit symbol by the
longer line.
675
Capacitor
symbol
Battery
symbol
Switch
symbol
Open
Closed
Parallel Combination
Figure 20.24 Circuit symbols for
Two capacitors connected as shown in Active Figure 20.25a are known as a parallel
combination of capacitors. Active Figure 20.25b shows a circuit diagram for this combination of capacitors. The left plates of the capacitors are connected to the positive
terminal of the battery by a conducting wire and are therefore both at the same
electric potential as the positive terminal. Likewise, the right plates are connected
to the negative terminal and so are both at the same potential as the negative terminal. Therefore, the individual potential differences across capacitors connected
in parallel are the same and are equal to the potential difference applied across the
combination. That is,
capacitors, batteries, and switches.
Notice that capacitors are in blue,
batteries are in green, and switches
are in red. The closed switch can
carry current, whereas the open one
cannot.
DV 1 5 DV 2 5 DV
where DV is the battery terminal voltage.
After the battery is attached to the circuit, the capacitors quickly reach their maximum charge. Let’s call the maximum charges on the two capacitors Q 1 and Q 2. The
total charge Q tot stored by the two capacitors is the sum of the charges on the individual capacitors:
20.25b
Q tot 5 Q 1 1 Q 2
Suppose you wish to replace these two capacitors by one equivalent capacitor having a capacitance Ceq as in Active Figure 20.25c. The effect this equivalent capacitor
has on the circuit must be exactly the same as the effect of the combination of the
two individual capacitors. That is, the equivalent capacitor must store charge Q tot
when connected to the battery. Active Figure 20.25c shows that the voltage across
A pictorial
representation of two
capacitors connected in
parallel to a battery
A circuit diagram
showing the two
capacitors connected
in parallel to a battery
A circuit diagram
showing the equivalent
capacitance of the
capacitors in parallel
C1
C1
Q 1
Q 1
V1
Q1
C2
Q 2
Q 2
Q2
V2
V
a
C eq C 1 C 2
C2
b
V
V
c
Active Figure 20.25 Two capacitors connected in parallel. All three
diagrams are equivalent.
676
CHAPTER 20 | Electric Potential and Capacitance
the equivalent capacitor is DV because the equivalent capacitor is connected directly
across the battery terminals. Therefore, for the equivalent capacitor,
Q tot 5 Ceq DV
Substituting for the charges in Equation 20.25 gives
Ceq DV 5 Q 1 1 Q 2 5 C1 DV1 1 C 2 DV 2
Ceq 5 C1 1 C 2 (parallel combination)
where we have canceled the voltages because they are all the same. If this treatment
is extended to three or more capacitors connected in parallel, the equivalent capacitance is found to be
20.26b
Ceq 5 C1 1 C 2 1 C 3 1 ??? (parallel combination)
c Equivalent capacitance for
capacitors in parallel
Therefore, the equivalent capacitance of a parallel combination of capacitors
is (1) the algebraic sum of the individual capacitances and (2) greater than any
of the individual capacitances. Statement (2) makes sense because we are essentially combining the areas of all the capacitor plates when they are connected
with conducting wire, and capacitance of parallel plates is proportional to area
(Eq. 20.22).
Series Combination
Two capacitors connected as shown in Active Figure 20.26a and the equivalent circuit diagram in Active Figure 20.26b are known as a series combination of capacitors.
The left plate of capacitor 1 and the right plate of capacitor 2 are connected to the
terminals of a battery. The other two plates are connected to each other and to nothing else; hence, they form an isolated system that is initially uncharged and must
continue to have zero net charge. To analyze this combination, let’s first consider
the uncharged capacitors and then follow what happens immediately after a battery
is connected to the circuit. When the battery is connected, electrons are transferred
to the leftmost wire out of the left plate of C 1 and into the right plate of C 2 from
the rightmost wire. As this negative charge accumulates on the right plate of C 2,
an equivalent amount of negative charge is forced off the left plate of C 2, and this
left plate therefore has an excess positive charge. The negative charge leaving the
left plate of C 2 causes negative charges to accumulate on the right plate of C 1. As a
result, both right plates end up with a charge 2Q and both left plates end up with a
charge 1Q. Therefore, the charges on capacitors connected in series are the same:
Q1 5 Q2 5 Q
Active Figure 20.26 Two capaci-
A pictorial
representation of two
capacitors connected in
series to a battery
tors connected in series. All three
diagrams are equivalent.
V 1
C1
Q Q
A circuit diagram
showing the two
capacitors connected
in series to a battery
C2
V 2
Q Q
C2
V1
V2
1
C eq
1
V
V
b
c
1
C1 C2
V
a
C1
A circuit diagram
showing the equivalent
capacitance of the
capacitors in series
677
20.8 | Combinations of Capacitors
where Q is the charge that moved between a wire and the connected outside plate
of one of the capacitors.
Active Figure 20.26a shows that the total voltage DVtot across the combination is
split between the two capacitors:
20.27b
DVtot 5 DV1 1 DV2
where DV 1 and DV 2 are the potential differences across capacitors C1 and C 2, respectively. In general, the total potential difference across any number of capacitors connected in series is the sum of the potential differences across the individual capacitors.
Suppose the equivalent single capacitor in Active Figure 20.26c has the same effect
on the circuit as the series combination when it is connected to the battery. After it
is fully charged, the equivalent capacitor must have a charge of 2Q on its right plate
and a charge of 1Q on its left plate. Applying the definition of capacitance to the
circuit in Active Figure 20.26c gives
DVtot 5
Q
Ceq
Substituting for the voltages in Equation 20.27, we have
Q
Ceq
5 DV 1 1 DV 2 5
Q1
C1
1
Q2
C2
Canceling the charges because they are all the same gives
1
1
1
5
1
(series combination)
Ceq
C1
C2
When this analysis is applied to three or more capacitors connected in series, the
relationship for the equivalent capacitance is
1
1
1
1
5
1
1
1 ??? (series combination)
Ceq
C1
C2
C3
20.28b
c Equivalent capacitance for
capacitors in series
This expression shows that (1) the inverse of the equivalent capacitance is the algebraic
sum of the inverses of the individual capacitances and (2) the equivalent capacitance of
a series combination is always less than any individual capacitance in the combination.
Q U I CK QUI Z 20.7 Two capacitors are identical. They can be connected in series or
in parallel. If you want the smallest equivalent capacitance for the combination, how
should you connect them? (a) in series (b) in parallel (c) either way because both
combinations have the same capacitance
Exampl e 20.8 | Equivalent Capacitance
Find the equivalent capacitance between a and b for the
combination of capacitors shown in Figure 20.27a. All
capacitances are in microfarads.
SOLUTION
4.0
4.0
1.0
2.0
4.0
3.0
6.0
a
b a
b
b a 6.0 b
a
Conceptualize Study Figure 20.27a carefully and make
sure you understand how the capacitors are connected.
Categorize Figure 20.27a shows that the circuit contains
both series and parallel connections, so we use the rules for
series and parallel combinations discussed in this section.
Analyze Using Equations 20.26 and 20.28, we reduce the
combination step by step as indicated in the figure.
2.0
a
8.0
8.0
b
8.0
4.0
c
d
Figure 20.27 (Example 20.8) To find the equivalent capacitance
of the capacitors in (a), we reduce the various combinations in steps
as indicated in (b), (c), and (d), using the series and parallel rules
described in the text. All capacitances are in microfarads.
continued
678
CHAPTER 20 | Electric Potential and Capacitance
20.8 cont.
The 1.0-F and 3.0-F capacitors (upper red-brown circle
in Fig. 20.27a) are in parallel. Find the equivalent capacitance from Equation 20.26:
Ceq 5 C1 1 C2 5 4.0 F
The 2.0-F and 6.0-F capacitors (lower red-brown circle
in Fig. 20.27a) are also in parallel:
Ceq 5 C1 1 C2 5 8.0 F
The circuit now looks like Figure 20.27b. The two 4.0-F
capacitors (upper green circle in Fig. 20.27b) are in series.
Find the equivalent capacitance from Equation 20.28:
1
1
1
1
1
1
5
1
5
1
5
Ceq
C1
C2
4.0 F
4.0 F
2.0 F
Ceq 5 2.0 F
The two 8.0-F capacitors (lower green circle in Fig.
20.27b) are also in series. Find the equivalent capacitance
from Equation 20.28:
1
1
1
1
1
1
5
1
5
1
5
Ceq
C1
C2
8.0 F
8.0 F
4.0 F
The circuit now looks like Figure 20.27c. The 2.0-F and
4.0-F capacitors are in parallel:
Ceq 5 C1 1 C2 5 6.0 F
Ceq 5 4.0 F
Finalize This final value is that of the single equivalent capacitor shown in Figure 20.27d. For further practice in treating
circuits with combinations of capacitors, imagine a battery is connected between points a and b in Figure 20.27a so that
a potential difference DV is established across the combination. Can you find the voltage across and the charge on each
capacitor?
20.9 | Energy Stored in a Charged Capacitor
Almost everyone who works with electronic equipment has at some time verified that
a capacitor can store energy. If the plates of a charged capacitor are connected by a
conductor, such as a wire, charge transfers between the plates and the wire until the
two plates are uncharged. The discharge can often be observed as a visible spark. If
you accidentally touch the opposite plates of a charged capacitor, your fingers act as
pathways by which the capacitor discharges, resulting in an electric shock. The degree of shock depends on the capacitance and the voltage applied to the capacitor.
When high voltages are present, such as in the power supply of an electronic instrument, the shock can be fatal.
Consider a parallel-plate capacitor that is initially uncharged so that the initial
potential difference across the plates is zero. Now imagine that the capacitor is connected to a battery and develops a charge of Q. The final potential difference across
the capacitor is DV 5 Q/C.
To calculate the energy stored in the capacitor, we shall assume a charging process that is different from the actual process described in Section 20.7 but that gives
the same final result. This assumption is justified because the energy in the final
configuration does not depend on the actual charge-transfer process.2 Imagine the
plates are disconnected from the battery and you transfer the charge mechanically
through the space between the plates as follows. You grab a small amount of positive charge on the plate connected to the negative terminal and apply a force that
causes this positive charge to move over to the plate connected to the positive terminal. Therefore, you do work on the charge as it is transferred from one plate to the
other. At first, no work is required to transfer a small amount of charge dq from one
2This discussion is similar to that of state variables in thermodynamics. The change in a state variable such as tempera-
ture is independent of the path followed between the initial and final states. The potential energy of a capacitor (or any
system) is also a state variable, so its change does not depend on the process followed to charge the capacitor.
20.9 | Energy Stored in a Charged Capacitor
679
plate to the other,3 but once this charge has been transferred, a small potential difference exists between the plates. Therefore, work must be done to move additional
charge through this potential difference. As more and more charge is transferred
from one plate to the other, the potential difference increases in proportion and
more work is required.
The work required to transfer an increment of charge dq from one plate to the
other is
dW 5 DV dq 5
q
dq
C
Therefore, the total work required to charge the capacitor from q 5 0 to the final
charge q 5 Q is
Q q
Q2
W5
dq 5
2C
0 C
E
The capacitor can be modeled as a nonisolated system for this discussion. The
work done by the external agent on the system in charging the capacitor appears
as potential energy U stored in the capacitor. In reality, of course, this energy is not
the result of mechanical work done by an external agent moving charge from one
plate to the other, but is due to transformation of chemical energy in the battery.
We have used a model of work done by an external agent that gives us a result that
is also valid for the actual situation. Using Q 5 C DV, the energy stored in a charged
capacitor can be expressed in the following alternative forms:
U5
Q2
2C
5 12 Q DV 5 12C (DV )2
20.29b
c Energy stored in a charged
capacitor
This result applies to any capacitor, regardless of its geometry. In practice, the
maximum energy (or charge) that can be stored is limited because electric discharge ultimately occurs between the plates of the capacitor at a sufficiently large
value of DV. For this reason, capacitors are usually labeled with a maximum operating voltage.
For an object on an extended spring, the elastic potential energy can be modeled
as being stored in the spring. Internal energy of a substance associated with its temperature is located throughout the substance. Where is the energy in a capacitor located?
The energy stored in a capacitor can be modeled as being stored in the electric field
between the plates of the capacitor. For a parallel-plate capacitor, the potential difference
is related to the electric field through the relationship DV 5 Ed. Furthermore, the
capacitance is C 5 0A/d. Substituting these expressions into Equation 20.29 gives
U 5 12
0A
1 d 2(Ed) 5 ( Ad)E
2
1
2
0
2
20.30b
Because the volume of a parallel-plate capacitor that is occupied by the electric field
is Ad, the energy per unit volume uE 5 U/Ad, called the energy density, is
uE 5 12 0E 2
20.31b
Although Equation 20.31 was derived for a parallel-plate capacitor, the expression
is generally valid. That is, the energy density in any electric field is proportional to
the square of the magnitude of the electric field at a given point.
Q U I CK QUI Z 20.8 You have three capacitors and a battery. In which of the following combinations of the three capacitors is the maximum possible energy stored
when the combination is attached to the battery? (a) series (b) parallel (c) no difference because both combinations store the same amount of energy
3We shall use lowercase q for the time-varying charge on the capacitor while it is charging to distinguish it from uppercase
Q , which is the total charge on the capacitor after it is completely charged.
c Energy density in an electric
field
680
CHAPTER 20 | Electric Potential and Capacitance
THINKING PHYSICS 20.2
You charge a capacitor and then remove it from the battery. The capacitor consists of large movable plates, with air between them. You pull the plates farther
apart a small distance. What happens to the charge on the capacitor? To the
potential difference? To the energy stored in the capacitor? To the capacitance?
To the electric field between the plates? Is work done in pulling the plates apart?
Reasoning Because the capacitor is removed from the battery, charges on the
plates have nowhere to go. Therefore, the charge on the capacitor remains the
same as the plates are pulled apart. Because the electric field of large plates is
independent of distance for uniform fields, the electric field remains constant.
Because the electric field is a measure of the rate of change of potential with
distance, the potential difference between the plates increases as the separation
distance increases. Because the same charge is stored at a higher potential difference, the capacitance decreases. Because energy stored is proportional to both
charge and potential difference, the energy stored in the capacitor increases.
This energy must be transferred into the system from somewhere; the plates
attract each other, so work is done by you on the system of two plates when you
pull them apart. b
Example 20.9 | Rewiring Two Charged Capacitors
Q 1i C 1
Two capacitors C1 and C 2 (where C1 . C 2) are charged to the
same initial potential difference DVi . The charged capacitors
are removed from the battery, and their plates are connected
with opposite polarity as in Figure 20.28a. The switches S1 and
S2 are then closed as in Figure 20.28b.
Q 1f C 1
a
b
S1
(A) Find the final potential difference DVf between a and b
after the switches are closed.
S2
Q 2i
a
SOLUTION
Conceptualize Figure 20.28 helps us understand the initial and
a
b
S2
S1
Q 2f
C2
C2
b
Figure 20.28 (Example 20.9) (a) Two capacitors
are charged to the same initial potential difference and
connected together with plates of opposite sign to be in
contact when the switches are closed. (b) When the switches
are closed, the charges redistribute.
final configurations of the system. When the switches are closed,
the charge on the system will redistribute between the capacitors
until both capacitors have the same potential difference. Because
C1 . C 2, more charge exists on C1 than on C 2, so the final configuration will have positive charge on the left plates as shown in Figure 20.28b.
Categorize In Figure 20.28b, it might appear as if the capacitors are connected in parallel, but there is no battery in this
circuit to apply a voltage across the combination. Therefore, we cannot categorize this problem as one in which capacitors
are connected in parallel. We can categorize it as a problem involving an isolated system for electric charge. The left-hand
plates of the capacitors form an isolated system because they are not connected to the right-hand plates by conductors.
Analyze Write an expression for the total charge on
the left-hand plates of the system before the switches are
closed, noting that a negative sign for Q 2i is necessary
because the charge on the left plate of capacitor C 2 is
negative:
(1) Q i 5 Q 1i 1 Q 2i 5 C1 DVi 2 C 2 DVi 5 (C1 2 C 2)DVi
After the switches are closed, the charges on the individual
capacitors change to new values Q 1f and Q 2f such that
the potential difference is again the same across both capacitors, with a value of DVf . Write an expression for the
total charge on the left-hand plates of the system after the
switches are closed:
(2) Q f 5 Q 1f 1 Q 2f 5 C1 DVf 1 C2 DVf 5 (C1 1 C 2)DVf
20.10 | Capacitors with Dielectrics
681
20.9 cont.
Because the system is isolated, the initial and final
charges on the system must be the same. Use this
condition and Equations (1) and (2) to solve for
DVf :
Q f 5 Q i : (C1 1 C 2) DVf 5 (C1 2 C 2) DVi
(3) DVf 5
C1 2 C 2
1C 1 C 2 DV
i
1
2
(B) Find the total energy stored in the capacitors before and after the switches are closed and determine the ratio of the
final energy to the initial energy.
SOLUTION
Use Equation 20.29 to find an expression for the
total energy stored in the capacitors before the
switches are closed:
(4) Ui 5 12C 1(DVi )2 1 12C 2(DVi)2 5 12(C1 1 C 2)(DVi )2
Write an expression for the total energy stored in
the capacitors after the switches are closed:
Uf 5 12C 1(DVf )2 1 12C 2(DVf )2 5 12(C1 1 C 2)(DVf )2
Use the results of part (A) to rewrite this expression in terms of DVi :
(5) Uf 5 12(C 1 1 C 2)
Divide Equation (5) by Equation (4) to obtain the
ratio of the energies stored in the system:
C1 2 C 2
2
31C 1 C 2 4
Uf
Ui
5
(6)
1
DVi
2
5 12
(C 1 2 C 2)2(DVi )2
C1 1 C 2
1
2
2
2 (C 1 2 C 2) (DVi ) /(C 1 1 C 2)
1
2
2 (C 1 1 C 2)(DVi )
Uf
Ui
5
C1 2 C 2 2
1C 1 C 2
1
2
Finalize The ratio of energies is less than unity, indicating that the final energy is less than the initial energy. At first, you
might think the law of energy conservation has been violated, but that is not the case. The “missing” energy is transferred
out of the system by the mechanism of electromagnetic waves (TER in Eq. 7.2), as we shall see in Chapter 24. Therefore, this
system is isolated for electric charge, but nonisolated for energy.
What If? What if the two capacitors have the same capacitance? What would you expect to happen when the switches are
closed?
Answer Because both capacitors have the same initial potential difference applied to them, the charges on the capaci-
tors have the same magnitude. When the capacitors with opposite polarities are connected together, the equal-magnitude
charges cancel each other, leaving the capacitors uncharged.
Let’s test our results to see if that is the case mathematically. In Equation (1), because the capacitances are equal, the
initial charge Q i on the system of left-hand plates is zero. Equation (3) shows that DVf 5 0, which is consistent with uncharged capacitors. Finally, Equation (5) shows that Uf 5 0, which is also consistent with uncharged capacitors.
20.10 | Capacitors with Dielectrics
A dielectric is an insulating material such as rubber, glass, or waxed paper. When
a dielectric material is inserted between the plates of a capacitor, the capacitance
increases. If the dielectric completely fills the space between the plates, the capacitance increases by the dimensionless factor , called the dielectric constant of the
material.
The following experiment can be performed to illustrate the effect of a dielectric in a capacitor. Consider a parallel-plate capacitor of charge Q 0 and capacitance
C 0 in the absence of a dielectric. The potential difference across the capacitor as
measured by a voltmeter is DV 0 5 Q 0 /C 0 (Fig. 20.29a, page 682). Notice that the
capacitor circuit is open; that is, the plates of the capacitor are not connected to a
battery and charge cannot flow through an ideal voltmeter. Hence, there is no path
682
CHAPTER 20 | Electric Potential and Capacitance
The potential
difference across the
charged capacitor is
initially V0.
After the dielectric is inserted between
the plates, the charge remains the same,
but the potential difference decreases
and the capacitance increases.
Dielectric
C0
C
Q0
Q0
by which charge can flow and alter the charge on
the capacitor. If a dielectric is now inserted between the plates as in Figure 20.29b, it is found
that the voltmeter reading decreases by a factor of to the value DV, where
DV0
DV 5
Because DV , DV0, we see that . 1.
Because the charge Q 0 on the capacitor does
not change, we conclude that the capacitance must
change to the value
C5
DV
C 5 C0
V
V0
Q0
5
Q0
DV0/
5
Q0
DV0
20.32b
where C0 is the capacitance in the absence of the
dielectric. That is, the capacitance increases by the
factor when the dielectric completely fills the
Figure 20.29 A charged capacitor (a) before and (b) after insertion of a
region between the plates.4 For a parallel-plate
dielectric between the plates.
capacitor, where C0 5 0A/d, we can express the
capacitance when the capacitor is filled with a dielectric as
a
b
C5
Pitfall Prevention | 20.9
Is the Capacitor Connected to a
Battery?
In problems in which you are
modifying a capacitor (by insertion
of a dielectric, for example), you
must note whether modifications
to the capacitor are being made
while the capacitor is connected to
a battery or after it is disconnected.
If the capacitor remains connected
to the battery, the voltage across the
capacitor necessarily remains the
same. If you disconnect the capacitor
from the battery before making any
modifications to the capacitor, the
capacitor is an isolated system and its
charge remains the same.
0A
20.33b
d
From this result, it would appear that the capacitance could be made very large
by decreasing d, the distance between the plates. In practice, however, the lowest
value of d is limited by the electric discharge that could occur through the dielectric
medium separating the plates. For any given separation d, the maximum voltage that
can be applied to a capacitor without causing a discharge depends on the dielectric
strength (maximum electric field) of the dielectric, which for dry air is equal to
3 3 106 V/m. If the electric field in the medium exceeds the dielectric strength, the
insulating properties break down and the medium begins to conduct. Most insulating materials have dielectric strengths and dielectric constants greater than those of
air, as Table 20.1 indicates. Therefore, we see that a dielectric provides the following
advantages:
• An increase in capacitance
• An increase in maximum operating voltage
• Possible mechanical support between the plates, which allows the plates to be
close together without touching, thereby decreasing d and increasing C
We can understand the effects of a dielectric by considering the polarization
of molecules that we discussed in Section 19.3. Figure 20.30a shows polarized
molecules of a dielectric in random orientations in the absence of an electric
field. Figure 20.30b shows the polarization of the molecules when the dielectric is
placed between the plates of the charged capacitor and the polarized molecules
:
tend to line up parallel to the field lines. The plates set up an electric field E 0 in
a direction to the right in Figure 20.30b. In the body of the dielectric, a general
homogeneity of charge exists, but look along the edges. There is a layer of negative charge along the left edge of the dielectric and a layer of positive charge along
the right edge. These layers of charge can be modeled as additional charged parallel plates, as in Figure 20.30c. Because the polarity is opposite that of the real
4If another experiment is performed in which the dielectric is introduced while the potential difference is held constant
by means of a battery, the charge increases to the value Q 5 Q 0. The additional charge is transferred from the connecting wires, and the capacitance still increases by the factor .
20.10 | Capacitors with Dielectrics
683
TABLE 20.1 | Approximate Dielectric Constants and Dielectric Strengths of Various
Materials at Room Temperature
Material
Dielectric Constant ␬
Air (dry)
1.000 59
3
Bakelite
4.9
24
Fused quartz
3.78
8
Mylar
3.2
7
Neoprene rubber
6.7
12
Nylon
3.4
14
Paper
3.7
16
Paraffin-impregnated paper
3.5
11
Polystyrene
2.56
24
Polyvinyl chloride
3.4
40
Porcelain
6
12
Pyrex glass
5.6
14
2.5
15
Silicone oil
Strontium titanate
Dielectric Strengtha (106 V/m)
233
8
Teflon
2.1
60
Vacuum
1.000 00
—
Water
80
—
aThe dielectric strength equals the maximum electric field that can exist in a dielectric without electrical breakdown.
Note that these values depend strongly on the presence of impurities and flaws in the materials.
When an external
electric field is applied,
the molecules partially
align with the field.
Polar molecules are
randomly oriented in
the absence of an
external electric field.
S
The charged edges of the dielectric
can be modeled as an additional
pair of parallel plates establishing
S
an electric field E ind in the
S
direction opposite that of E 0.
E0
a
sind
S
E0
S
E ind
sind
Figure 20.30 (a) Polar molecules
in a dielectric. (b) An electric field is
applied to the dielectric. (c) Details
of the electric field inside the
dielectric.
c
b
:
Chris Vuille
plates, these charges set up an induced electric field E ind directed to the left in the
diagram that partially cancels the electric field due to the real plates. Therefore,
for the charged capacitor removed from a battery, the electric field and hence the
voltage between the plates is reduced by the introduction of the dielectric.
The charge on the plates is stored at a lower potential difference, so the capacitance increases.
Types of Capacitors
Many capacitors are built into integrated circuit chips, but some electrical devices
still use stand-alone capacitors. Commercial capacitors are often made using metal
A collection of capacitors used in a
variety of applications.
684
CHAPTER 20 | Electric Potential and Capacitance
Figure 20.31 Three commercial
capacitor designs.
A tubular capacitor
whose plates are
separated by paper
and then rolled into
a cylinder
A high-voltage
capacitor consisting
of many parallel
plates separated by
insulating oil
An electrolytic
capacitor
Plates
Case
Electrolyte
Paper
Contacts
Oil
Metal foil
a
The materials between the
plates of the capacitor are
the wallboard and air.
Capacitor
plates
Stud
finder
Stud
Wallboard
When the capacitor moves across
a stud in the wall, the materials
between the plates are the
wallboard and the wood stud.
The change in the dielectric
constant causes a signal light to
illuminate.
a
b
Figure 20.32 (Quick Quiz 20.9)
b
Metallic foil oxide layer
c
foil interlaced with a dielectric such as thin sheets of paraffin-impregnated paper.
These alternating layers of metal foil and dielectric are then rolled into the shape of
a cylinder to form a small package (Fig. 20.31a). High-voltage capacitors commonly
consist of interwoven metal plates immersed in silicone oil (Fig. 20.31b). Small capacitors are often constructed from ceramic materials. Variable capacitors (typically
10–500 pF), such as the one in the chapter-opening photograph, usually consist of
two interwoven sets of metal plates, one fixed and the other movable, with air as the
dielectric.
An electrolytic capacitor is often used to store large amounts of charge at relatively
low voltages. This device, shown in Figure 20.31c, consists of a metal foil in contact
with an electrolyte, a solution that conducts electricity by virtue of the motion of
ions contained in the solution. When a voltage is applied between the foil and the
electrolyte, a thin layer of metal oxide (an insulator) is formed on the foil, and this
layer serves as the dielectric. Very large capacitance values can be attained because
the dielectric layer is very thin.
When electrolytic capacitors are used in circuits, they must be installed with
the proper polarity. If the polarity of the applied voltage is opposite what is intended, the oxide layer will be removed and the capacitor will not be able to store
charge.
QUI C K QU IZ 20.9 If you have ever tried to hang a picture or a mirror, you know
it can be difficult to locate a wooden stud in which to anchor your nail or screw.
A carpenter’s stud finder is a capacitor with its plates arranged side by side instead
of facing each other as shown in Figure 20.32. When the device is moved over a stud,
does the capacitance (a) increase or (b) decrease?
An electric stud finder.
Example 20.10 | Energy Stored Before and After
A parallel-plate capacitor is charged with a battery to a charge Q 0. The battery is then removed, and a slab of material that
has a dielectric constant is inserted between the plates. Identify the system as the capacitor and the dielectric. Find the
energy stored in the system before and after the dielectric is inserted.
SOLUTION
Conceptualize Think about what happens when the dielectric is inserted between the plates. Because the battery has been
removed, the charge on the capacitor must remain the same. We know from our earlier discussion, however, that the capacitance must change. Therefore, we expect a change in the energy of the system.
Categorize Because we expect the energy of the system to change, we model it as a nonisolated system involving a capacitor and a dielectric.
20.11 | Context Connection: The Atmosphere as a Capacitor
685
20.10 cont.
Analyze From Equation 20.29, find the energy stored in
U0 5
the absence of the dielectric:
Find the energy stored in the capacitor after the dielectric is inserted between the plates:
U5
Use Equation 20.32 to replace the capacitance C :
U5
Q 02
2C 0
Q 02
2C
Q 02
2C 0
5
U0
Finalize Because . 1, the final energy is less than the initial energy. We can account for the decrease in energy of the
system by performing an experiment and noting that the dielectric, when inserted, is pulled into the device. To keep the
dielectric from accelerating, an external agent must do negative work (W in Eq. 7.2) on the dielectric, which is simply
the difference U 2 U0.
as a Capacitor
In the Context Connection of Chapter 19, we mentioned some processes occurring
on the surface of the Earth and in the atmosphere that result in charge distributions. These processes result in a negative charge on the Earth’s surface and positive
charges distributed throughout the air.
This separation of charge can be modeled as a capacitor. The surface of the
Earth is one plate and the positive charge in the air is the other plate. The
positive charge in the atmosphere is not all located at one height but is spread
throughout the atmosphere. Therefore, the single position of the upper plate
of the capacitor must be modeled, based on the charge distribution. Models of
the atmosphere show that an appropriate effective height of the upper plate
is about 5 km from the surface. The model atmospheric capacitor is shown in
Figure 20.33.
Considering the charge distribution on the surface of the Earth to be spherically
symmetric, we can use Figure 20.16 and its associated discussion in Section 20.6 to
argue that the potential at a point above the Earth’s surface is
V 5 ke
Q
1 Q
5
r
40 r
where Q is the charge on the surface. The potential difference between the plates of
our atmospheric capacitor is
DV 5
Q
1
1
2
1
2
r
4 r
surface
0
5
upper plate
Q
Q
1
1
h
2
5
40 RE
RE 1 h
40 RE(RE 1 h)
1
2
3
4
where RE is the radius of the Earth and h 5 5 km. From this expression, we can calculate the capacitance of the atmospheric capacitor:
C5
Q
DV
5
Q
Q
h
40 RE(RE 1 h)
3
5
4
40RE(RE 1 h)
h
Negative plate
(Earth’s surface)
⫹
⫹
⫺
⫺
⫹
⫺
⫹
⫺
Positive plate
(charges in
atmosphere)
⫹
⫺
⫹
⫺
⫹
⫺
⫹⫺
⫺⫹
⫹⫺
⫺⫹
⫹⫺
⫺⫹
⫹
⫺
⫹
⫺
⫺
⫹
⫹
⫺
20.11 | Context Connection: The Atmosphere
⫺
⫹
⫺
⫺
⫹
⫹
Figure 20.33 The atmospheric
capacitor.
686
CHAPTER 20 | Electric Potential and Capacitance
Substituting the numerical values, we have
40RE(RE 1 h)
C5
h
4(8.85 3 10212 C2/N ?m2)(6.4 3 103 km)(6.4 3 103 km 1 5 km) 1000 m
5
5 km
1 km
1
2
< 0.9 F
This result is extremely large, compared with the picofarads and microfarads that are
typical values for capacitors in electrical circuits, especially for a capacitor having
plates that are 5 km apart! We shall use this model of the atmosphere as a capacitor
in our Context Conclusion, in which we calculate the number of lightning strikes on
the Earth in one day.
SUMMARY |
When a positive test charge q0 is moved between points 𝖠
:
and 𝖡 in an electric field E , the change in potential energy
of the charge–field system is
DU 5 2q 0
E
𝖡
:
E ?d:
s
20.1b
𝖠
The potential difference DV between points 𝖠 and 𝖡 in
:
an electric field E is defined as the change in potential energy divided by the test charge q0:
DV 5
E
𝖡
DU
:
52
E ? d:
s
q0
𝖠
20.3b
which represents the work required to bring the charges
from an infinite separation to the separation r12. The potential energy of a distribution of point charges is obtained
by summing terms like Equation 20.13 over all pairs of
particles.
If the electric potential is known as a function of coordinates x, y, and z, the components of the electric field can be
obtained by taking the negative derivative of the potential
with respect to the coordinates. For example, the x component of an electric field is
Ex 5 2
0V
0x
20.17b
where electric potential V is a scalar and has the units joules
per coulomb, defined as 1 volt (V).
The potential difference between two points 𝖠 and 𝖡 in
:
a uniform electric field E is
The electric potential due to a continuous charge distribution is
dq
20.19b
V 5 ke
r
E
Every point on the surface of a charged conductor in electrostatic equilibrium is at the same potential. Furthermore,
the potential is constant everywhere inside the conductor
and is equal to its value at the surface.
A capacitor is a device for storing charge. A charged capacitor consists of two equal and oppositely charged conductors with a potential difference DV between them. The
capacitance C of any capacitor is defined as the ratio of the
magnitude of the charge Q on either conductor to the magnitude of the potential difference DV :
DV 5 2E
𝖡
ds 5 2Ed
20.6b
𝖠
where d is the magnitude of the displacement vector between
𝖠 and 𝖡.
Equipotential surfaces are surfaces on which the electric
potential remains constant. Equipotential surfaces are perpendicular to electric field lines.
The electric potential due to a point charge q at a distance
r from the charge is
q
20.11b
r
The electric potential due to a group of point charges is
obtained by summing the potentials due to the individual
charges. Because V is a scalar, the sum is a simple algebraic
operation.
The electric potential energy of a pair of point charges
separated by a distance r 12 is
V 5 ke
U 5 ke
q 1q 2
r 12
20.13b
E
C;
Q
20.20b
DV
The SI units of capacitance are coulombs per volt, or the
farad (F), so 1 F 5 1 C/V.
If two or more capacitors are connected in parallel, the
potential differences across them must be the same. The
equivalent capacitance of a parallel combination of capacitors is
Ceq 5 C 1 1 C 2 1 C 3 1 ? ? ?
20.26b
| Objective Questions
If two or more capacitors are connected in series, the
charges on them are the same and the equivalent capacitance of the series combination is given by
1
1
1
1
5
1
1
1???
Ceq
C1
C2
C3
20.28b
Energy is required to charge a capacitor because the
charging process is equivalent to transferring charges from
one conductor at a lower potential to another conductor at
a higher potential. The electric potential energy U stored in
the capacitor is
U5
Q2
2C
5 12 Q DV 5 12C (DV )2
2. The electric potential at x 5 3.00 m is 120 V, and the electric potential at x 5 5.00 m is 190 V. What is the x component of the electric field in this region, assuming the
field is uniform? (a) 140 N/C (b) 2140 N/C (c) 35.0 N/C
(d) 235.0 N/C (e) 75.0 N/C
3. A proton is released from rest at the origin in a uniform
electric field in the positive x direction with magnitude
850 N/C. What is the change in the electric potential
energy of the proton–field system when the proton travels
to x 5 2.50 m? (a) 3.40 3 10216 J (b) 23.40 3 10216 J
(c) 2.50 3 10216 J (d) 22.50 3 10216 J (e) 21.60 3 10219 J
C 5 C 0
6. Rank the potential energies of the four systems of particles
shown in Figure OQ20.6 from largest to smallest. Include
equalities if appropriate.
Q
r
2Q
Q
2r
Q
b
a
Q
r
Q
Q
c
2r
2Q
d
Figure OQ20.6
7. True or False? (a) From the definition of capacitance C 5
Q/DV, it follows that an uncharged capacitor has a
20.32b
where C 0 is the capacitance in the absence of the dielectric.
denotes answer available in Student
Solutions Manual/Study Guide
capacitance of zero. (b) As described by the definition of
capacitance, the potential difference across an uncharged
capacitor is zero.
8. In a certain region of space, a uniform electric field is in
the x direction. A particle with negative charge is carried
from x 5 20.0 cm to x 5 60.0 cm. (i) Does the electric potential energy of the charge-field system (a) increase, (b) remain constant, (c) decrease, or (d) change unpredictably?
(ii) Has the particle moved to a position where the electric potential is (a) higher than before, (b) unchanged,
(c) lower than before, or (d) unpredictable?
9. Rank the electric potentials at the four points shown in Figure OQ20.9 from largest to smallest.
A
B
d
4. By what factor is the capacitance of a metal sphere multiplied if its volume is tripled? (a) 3 (b) 31/3 (c) 1 (d) 321/3
(e) 13
5. An electron in an x-ray machine is accelerated through a
potential difference of 1.00 3 104 V before it hits the target.
What is the kinetic energy of the electron in electron volts?
(a) 1.00 3 104 eV (b) 1.60 3 10215 eV (c) 1.60 3 10222 eV
(d) 6.25 3 1022 eV (e) 1.60 3 10219 eV
20.29b
When a dielectric material is inserted between the
plates of a capacitor, the capacitance generally increases by
the dimensionless factor , called the dielectric constant.
That is,
OBJECTIVE QUESTIONS |
1. A parallel-plate capacitor is charged and then is disconnected from the battery. By what factor does the stored
energy change when the plate separation is then doubled?
(a) It becomes four times larger. (b) It becomes two times
larger. (c) It stays the same. (d) It becomes one-half as large.
(e) It becomes one-fourth as large.
687
C
D
Q
d
2Q
Figure OQ20.9
10. A capacitor with very large capacitance is in series with another capacitor with very small capacitance. What is the
equivalent capacitance of the combination? (a) slightly
greater than the capacitance of the large capacitor
(b) slightly less than the capacitance of the large capacitor (c) slightly greater than the capacitance of the small
capacitor (d) slightly less than the capacitance of the small
capacitor
11. Consider the equipotential surfaces shown in Figure 20.4.
In this region of space, what is the approximate direction of
the electric field? (a) It is out of the page. (b) It is into the
page (c) It is toward the top of the page. (d) It is toward the
bottom of the page. (e) The field is zero.
12. A parallel-plate capacitor is connected to a battery. What
happens to the stored energy if the plate separation is
688
CHAPTER 20 | Electric Potential and Capacitance
doubled while the capacitor remains connected to the battery? (a) It remains the same. (b) It is doubled. (c) It decreases by a factor of 2. (d) It decreases by a factor of 4.
(e) It increases by a factor of 4.
y 5 80.0 cm, carrying the same amount of charge with the
same uniform density. At the same point P, is the electric potential created by the pair of filaments (a) greater than 200 V,
(b) 200 V, (c) 100 V, (d) between 0 and 200 V, or (e) 0?
13. Rank the electric potential energies of the systems of
charges shown in Figure OQ20.13 from largest to smallest.
Indicate equalities if appropriate.
17. An electronics technician wishes to construct a parallel plate
capacitor using rutile ( 5 100) as the dielectric. The area
of the plates is 1.00 cm2. What is the capacitance if the rutile
thickness is 1.00 mm? (a) 88.5 pF (b) 177 pF (c) 8.85 F
(d) 100 F (e) 35.4 F
Q
d
Q
Q
d
d
d
Q
Q
a
d
d
Q
b
Q
Q
d
Q
Q
d
Q
c
d
Q
Q
d
Q
d
Figure OQ20.13
14. Four particles are positioned on the rim of a circle. The
charges on the particles are 10.500 C, 11.50 C, 21.00 C,
and 20.500 C. If the electric potential at the center of the circle due to the 10.500 C charge alone is
4.50 3 104 V, what is the total electric potential at the center
due to the four charges? (a) 18.0 3 104 V (b) 4.50 3 104 V
(c) 0 (d) 24.50 3 104 V (e) 9.00 3 104 V
15. In a certain region of space, the electric field is zero. From
this fact, what can you conclude about the electric potential
in this region? (a) It is zero. (b) It does not vary with position. (c) It is positive. (d) It is negative. (e) None of those
answers is necessarily true.
16. A filament running along the x axis from the origin to x 5
80.0 cm carries electric charge with uniform density. At the
point P with coordinates (x 5 80.0 cm, y 5 80.0 cm), this
filament creates electric potential 100 V. Now we add another filament along the y axis, running from the origin to
CONCEPTUAL QUESTIONS |
18. If three unequal capacitors, initially uncharged, are connected in series across a battery, which of the following
statements is true? (a) The equivalent capacitance is greater
than any of the individual capacitances. (b) The largest voltage appears across the smallest capacitance. (c) The largest
voltage appears across the largest capacitance. (d) The capacitor with the largest capacitance has the greatest charge.
(e) The capacitor with the smallest capacitance has the
smallest charge.
19. (i) What happens to the magnitude of the charge on each
plate of a capacitor if the potential difference between the
conductors is doubled? (a) It becomes four times larger.
(b) It becomes two times larger. (c) It is unchanged. (d) It
becomes one-half as large. (e) It becomes one-fourth as
large. (ii) If the potential difference across a capacitor is
doubled, what happens to the energy stored? Choose from
the same possibilities as in part (i).
20. A parallel-plate capacitor filled with air carries a charge
Q. The battery is disconnected, and a slab of material with
dielectric constant 5 2 is inserted between the plates.
Which of the following statements is true? (a) The voltage
across the capacitor decreases by a factor of 2. (b) The voltage across the capacitor is doubled. (c) The charge on the
plates is doubled. (d) The charge on the plates decreases by
a factor of 2. (e) The electric field is doubled.
21. Assume a device is designed to obtain a large potential difference by first charging a bank of capacitors connected in
parallel and then activating a switch arrangement that in
effect disconnects the capacitors from the charging source
and from each other and reconnects them all in a series
arrangement. The group of charged capacitors is then
discharged in series. What is the maximum potential difference that can be obtained in this manner by using ten
500-F capacitors and an 800-V charging source? (a) 500 V
(b) 8.00 kV (c) 400 kV (d) 800 V (e) 0
denotes answer available in Student
Solutions Manual/Study Guide
1. Assume you want to increase the maximum operating voltage of a parallel-plate capacitor. Describe how you can do
that with a fixed plate separation.
would take positive work to increase the plate separation.
What type of energy in the system changes due to the external work done in this process?
2. Distinguish between electric potential and electric potential
energy.
5. Explain why the work needed to move a particle with charge
Q through a potential difference DV is W 5 Q DV, whereas
the energy stored in a charged capacitor is U 5 12Q DV.
Where does the factor 12 come from?
3. Describe the motion of a proton (a) after it is released from
rest in a uniform electric field. Describe the changes (if any)
in (b) its kinetic energy and (c) the electric potential energy of the proton–field system.
6. Describe the equipotential surfaces for (a) an infinite line
of charge and (b) a uniformly charged sphere.
4. Because the charges on the plates of a parallel-plate capacitor are opposite in sign, they attract each other. Hence, it
7. When charged particles are separated by an infinite distance, the electric potential energy of the pair is zero. When
| Problems
the particles are brought close, the electric potential energy
of a pair with the same sign is positive, whereas the electric
potential energy of a pair with opposite signs is negative.
Give a physical explanation of this statement.
8. An air-filled capacitor is charged, then disconnected from
the power supply, and finally connected to a voltmeter.
Explain how and why the potential difference changes
when a dielectric is inserted between the plates of the
capacitor.
9. Why is it dangerous to touch the terminals of a high-voltage
capacitor even after the voltage source that charged the capacitor is disconnected from the capacitor? (b) What can be
done to make the capacitor safe to handle after the voltage
source has been removed?
689
10. Study Figure 19.4 and the accompanying text discussion of
charging by induction. When the grounding wire is touched
to the rightmost point on the sphere in Figure 19.4c, electrons are drained away from the sphere to leave the sphere
positively charged. Suppose the grounding wire is touched
to the leftmost point on the sphere instead. (a) Will electrons still drain away, moving closer to the negatively
charged rod as they do so? (b) What kind of charge, if any,
remains on the sphere?
11. If you were asked to design a capacitor in which small size
and large capacitance were required, what would be the two
most important factors in your design?
12. Explain why a dielectric increases the maximum operating
voltage of a capacitor even though the physical size of the
capacitor doesn’t change.
PROBLEMS |
The problems found in this chapter may be assigned
online in Enhanced WebAssign.
denotes Master It tutorial available in Enhanced
WebAssign
denotes asking for quantitative and conceptual reasoning
denotes symbolic reasoning problem
denotes “paired problems” that develop reasoning with
symbols and numerical values
denotes Watch It video solution available in Enhanced
WebAssign
1. denotes straightforward problem; 2. denotes intermediate
problem; 3. denotes challenging problem
1. denotes full solution available in the Student Solutions Manual/
Study Guide
1. denotes problems most often assigned in Enhanced
WebAssign.
denotes biomedical problem
denotes guided problem
is the change in the potential energy of the charge–field
system? (b) Through what potential difference does the
charge move?
Section 20.1 Electric Potential and Potential Difference
1.
A uniform electric field of
magnitude 325 V/m is directed
in the negative y direction in Figure P20.1. The coordinates of
point 𝖠 are (20.200, 20.300) m,
and those of point 𝖡 are (0.400,
0.500) m. Calculate the electric
potential difference V𝖡 2 V𝖠
using the dashed-line path.
y
훾
5.
An electron moving parallel to the x axis has an initial
speed of 3.70 3 106 m/s at the origin. Its speed is reduced
to 1.40 3 105 m/s at the point x 5 2.00 cm. (a) Calculate
the electric potential difference between the origin and that
point. (b) Which point is at the higher potential?
6.
Review. A block
m, Q
having mass m and charge
S
k
E
1Q is connected to an
insulating spring having
a force constant k. The
block lies on a frictionx 0
less, insulating, horizontal track, and the system
Figure P20.6
is immersed in a uniform
electric field of magnitude E directed as shown in Figure P20.6. The block is released from rest when the spring is unstretched (at x 5 0).
We wish to show that the ensuing motion of the block is
simple harmonic. (a) Consider the system of the block, the
spring, and the electric field. Is this system isolated or nonisolated? (b) What kinds of potential energy exist within
this system? (c) Call the initial configuration of the system
that existing just as the block is released from rest. The
final configuration is when the block momentarily comes
to rest again. What is the value of x when the block comes
to rest momentarily? (d) At some value of x we will call
x
훽
S
E
2.
How much work is done (by
Figure P20.1
a battery, generator, or some
other source of potential difference) in moving Avogadro’s
number of electrons from an initial point where the electric
potential is 9.00 V to a point where the electric potential is
25.00 V? (The potential in each case is measured relative to
a common reference point.)
3.
Calculate the speed of a proton that is accelerated from
rest through an electric potential difference of 120 V. Calculate the speed of an electron that is accelerated through the
same electric potential difference.
Section 20.2 Potential Difference in a Uniform
Electric Field
4. A uniform electric field of magnitude 250 V/m is directed
in the positive x direction. A 112.0-C charge moves from
the origin to the point (x, y) 5 (20.0 cm, 50.0 cm). (a) What
690
CHAPTER 20 | Electric Potential and Capacitance
x 5 x0, the block has zero net force on it. What analysis
model describes the particle in this situation? (e) What is
the value of x0? (f) Define a new coordinate system x9 such
that x9 5 x 2 x0. Show that x9 satisfies a differential equation for simple harmonic motion. (g) Find the period of the
simple harmonic motion. (h) How does the period depend
on the electric field magnitude?
of the atom, which we’ll study in Chapter 29.) Assume
an alpha particle, initially very far from a stationary gold
nucleus, is fired with a velocity of 2.00 3 107 m/s directly
toward the nucleus (charge 179e). What is the smallest
distance between the alpha particle and the nucleus before the alpha particle reverses direction? Assume the
gold nucleus remains stationary.
13.
Four identical charged partiy
cles (q 5 110.0 C) are located
q
on the corners of a rectangle q as shown in Figure P20.13. The
dimensions of the rectangle are
W
L 5 60.0 cm and W 5 15.0 cm.
Calculate the change in elec
x
q
q
L
tric potential energy of the system as the particle at the lower
Figure P20.13
left corner in Figure P20.13 is
brought to this position from infinitely far away. Assume
the other three particles in Figure P20.13 remain fixed in
position.
14.
Review. A light, unstressed spring has length d. Two identical particles, each with charge q, are connected to the opposite ends of the spring. The particles are held stationary
a distance d apart and then released at the same moment.
The system then oscillates on a frictionless, horizontal table.
The spring has a bit of internal kinetic friction, so the oscillation is damped. The particles eventually stop vibrating
when the distance between them is 3d. Assume the system
of the spring and two charged particles is isolated. Find the
increase in internal energy that appears in the spring during the oscillations.
15.
Review. Two insulating spheres have radii 0.300 cm
and 0.500 cm, masses 0.100 kg and 0.700 kg, and uniformly
distributed charges 22.00 C and 3.00 C. They are released from rest when their centers are separated by 1.00 m.
(a) How fast will each be moving when they collide?
(b) What If? If the spheres were conductors, would the
speeds be greater or less than those calculated in part (a)?
Explain.
16.
Review. Two insulating spheres have radii r 1 and r 2,
masses m 1 and m 2, and uniformly distributed charges 2q 1
and q 2. They are released from rest when their centers are
separated by a distance d. (a) How fast is each moving when
they collide? (b) What If? If the spheres were conductors,
would their speeds be greater or less than those calculated
in part (a)? Explain.
Section 20.3 Electric Potential and Potential Energy Due
to Point Charges
Note: Unless stated otherwise, assume that the reference level
of potential is V 5 0 at r 5 `.
7. (a) Find the potential at a distance of 1.00 cm from a proton. (b) What is the potential difference between two points
that are 1.00 cm and 2.00 cm from a proton? (c) Repeat
parts (a) and (b) for an electron.
8.
9.
Show that the amount of work required to assemble four
identical charged particles of magnitude Q at the corners of
a square of side s is 5.41keQ 2/s.
Given two particles with 2.00-C charges as shown in Figure P20.9 and a particle with charge q 5 1.28 3 10218 C at the
origin, (a) what is the net force exerted by the two 2.00-C
charges on the test charge q ? (b) What is the electric field
at the origin due to the two 2.00-C particles? (c) What is
the electric potential at the origin due to the two 2.00-C
particles?
y
2.00 mC
q
x 0.800 m
0
x 0.800 m
2.00 mC
x
Figure P20.9
10.
11.
Three particles with equal positive charges q are at the corners of
an equilateral triangle of side a as
shown in Figure P20.10. (a) At what
point, if any, in the plane of the particles is the electric potential zero?
(b) What is the electric potential at
the position of one of the particles
due to the other two particles in the
triangle?
The three charged particles in
Figure P20.11 are at the vertices of an
isosceles triangle (where d 5 2.00 cm).
Taking q 5 7.00 C, calculate the electric potential at point A, the midpoint
of the base.
q
a
a
a
q
q
Figure P20.10
q
2d
12. In 1911, Ernest Rutherford and his
A
q
assistants Geiger and Marsden con- q d
ducted an experiment in which they
scattered alpha particles (nuclei of
Figure P20.11
helium atoms) from thin sheets of
gold. An alpha particle, having charge 12e and mass
6.64 3 10227 kg, is a product of certain radioactive decays.
The results of the experiment led Rutherford to the idea
that most of an atom’s mass is in a very small nucleus, with
electrons in orbit around it. (This is the planetary model
17. Two particles each with charge 12.00 C are located on the
x axis. One is at x 5 1.00 m, and the other is at x 5 21.00 m.
(a) Determine the electric potential on the y axis at
y 5 0.500 m. (b) Calculate the change in electric potential
energy of the system as a third charged particle of 23.00 C
is brought from infinitely far away to a position on the y axis
at y 5 0.500 m.
18. Two charged particles create influences at the origin, described by the expressions
29
3 10 C
cos 70.08ˆi
3 7.00
(0.070 0 m)
8.99 3 109 N ? m2/C2 2
2
2
7.00 3 10 29 C
8.00 3 10 29 C ˆ
sin 70.08 ˆj 1
j
(0.070 0 m)2
(0.030 0 m)2
4
691
| Problems
and
8.99 3 109 N ? m2/C2
7.00 3 10 29 C
3 0.070 0 m
2
8.00 3 10 29 C
0.030 0 m
4
difference between the point at the center of the ring and a
point on its axis a distance 2R from the center?
26.
(a) Identify the locations of the particles and the charges
on them. (b) Find the force on a 216.0 nC charge placed
at the origin and (c) the work required to move this third
charge to the origin from a very distant point.
19. Two particles, with charges
of 20.0 nC and 220.0 nC,
are placed at the points with
coordinates (0, 4.00 cm) 20.0 nC
and (0, 24.00 cm) as shown
4.00 cm
in Figure P20.19. A particle
with charge 10.0 nC is lo40.0 nC
cated at the origin. (a) Find 10.0 nC 3.00 cm the electric potential energy of the configuration
4.00 cm
of the three fixed charges.
(b) A fourth particle, with
a mass of 2.00 3 10213 kg –20.0 nC and a charge of 40.0 nC,
is released from rest at the
point (3.00 cm, 0). Find its
Figure P20.19
speed after it has moved
freely to a very large distance away.
20.
Section 20.4 Obtaining the Value of the Electric Field
from the Electric Potential
22.
The electric potential inside a charged spherical conductor of radius R is given by V 5 keQ/R, and the potential
outside is given by V 5 keQ/r. Using Er 5 2dV/dr, derive
the electric field (a) inside and (b) outside this charge
distribution.
23.
Over a certain region of space, the electric potential is
V 5 5x 2 3x 2y 1 2yz 2. (a) Find the expressions for the x, y,
and z components of the electric field over this region.
(b) What is the magnitude of the field at the point P that
has coordinates (1.00, 0, 22.00) m?
24.
The potential in a region between x 5 0 and x 5 6.00 m
is V 5 a 1 bx, where a 5 10.0 V and b 5 27.00 V/m. Determine (a) the potential at x 5 0, 3.00 m, and 6.00 m and
(b) the magnitude and direction of the electric field at
x 5 0, 3.00 m, and 6.00 m.
Section 20.5 Electric Potential Due to Continuous
Charge Distributions
25.
Consider a ring of radius R with the total charge Q
spread uniformly over its perimeter. What is the potential
y
B
b
d
x
A
L
27.
For the arrangement described in Problem 26, calcu- Figure P20.26 Problems 26
late the electric potential at and 27.
point B, which lies on the perpendicular bisector of the rod a distance b above the x axis.
28.
A wire having a uniform linear charge density is
bent into the shape shown in Figure P20.28. Find the electric potential at point O.
R
2R
O
2R
Figure P20.28
29.
At a certain distance from a charged particle, the magnitude of the electric field is 500 V/m and the electric potential is 23.00 kV. (a) What is the distance to the particle?
(b) What is the magnitude of the charge?
21. A particle with charge 1q is at the origin. A particle with
charge 22q is at x 5 2.00 m on the x axis. (a) For what finite
value(s) of x is the electric field zero? (b) For what finite
value(s) of x is the electric potential zero?
A rod of length L (Fig.
P20.26) lies along the x axis
with its left end at the origin.
It has a nonuniform charge
density 5 x, where is a
positive constant. (a) What
are the units of ? (b) Calculate the electric potential at A.
A uniformly charged insulating rod of
length 14.0 cm is bent into the shape of a
semicircle as shown in Figure P20.29. The
rod has a total charge of 27.50 C. Find
the electric potential at O, the center of the
semicircle.
Section 20.6 Electric Potential Due
to a Charged Conductor
O
Figure P20.29
30. How many electrons should be removed from an initially
uncharged spherical conductor of radius 0.300 m to produce a potential of 7.50 kV at the surface?
31. Electric charge can accumulate on an airplane in flight.
You may have observed needle-shaped metal extensions
on the wing tips and tail of an airplane. Their purpose is
to allow charge to leak off before much of it accumulates.
The electric field around the needle is much larger than
the field around the body of the airplane and can become
large enough to produce dielectric breakdown of the air,
discharging the airplane. To model this process, assume two
charged spherical conductors are connected by a long conducting wire and a 1.20-C charge is placed on the combination. One sphere, representing the body of the airplane,
has a radius of 6.00 cm; the other, representing the tip of
the needle, has a radius of 2.00 cm. (a) What is the electric
potential of each sphere? (b) What is the electric field at the
surface of each sphere?
32.
A spherical conductor has a radius of 14.0 cm and charge
of 26.0 C. Calculate the electric field and the electric potential (a) r 5 10.0 cm, (b) r 5 20.0 cm, and (c) r 5 14.0 cm
from the center.
Section 20.7 Capacitance
33.
(a) How much charge is on each plate of a 4.00-F capacitor when it is connected to a 12.0-V battery? (b) If this
692
CHAPTER 20 | Electric Potential and Capacitance
same capacitor is connected to a 1.50-V battery, what charge
is stored?
34.
Two conductors having net charges of 110.0 C and
210.0 C have a potential difference of 10.0 V between
them. (a) Determine the capacitance of the system.
(b) What is the potential difference between the two conductors if the charges on each are increased to 1100 C
and 2100 C?
35. An isolated, charged conducting sphere of radius 12.0 cm
creates an electric field of 4.90 3 104 N/C at a distance
21.0 cm from its center. (a) What is its surface charge density? (b) What is its capacitance?
36.
41.
A spherical capacitor consists
of a spherical conducting shell
of radius b and charge 2Q that
is concentric with a smaller
conducting sphere of radius a
and charge 1Q (Fig. P20.36).
(a) Show that its capacitance is
Section 20.8 Combinations of Capacitors
42.
Two capacitors, C1 5 5.00 F and C2 5 12.0 F, are
connected in parallel, and the resulting combination is
connected to a 9.00-V battery. Find (a) the equivalent capacitance of the combination, (b) the potential difference
across each capacitor, and (c) the charge stored on each
capacitor.
43.
What If ? The two capacitors of Problem 42 (C1 5 5.00 F
and C2 5 12.0 F) are now connected in series and to a 9.00-V
battery. Find (a) the equivalent capacia
C1
C1
tance of the combination, (b) the potential difference across each capacitor, and
(c) the charge on each capacitor.
C
C
44.
(a) Find the equivalent capacitance
between points a and b for the group
of capacitors connected as shown
in Figure P20.44. Take C1 5 5.00 F,
C 2 5 10.0 F, and C 3 5 2.00 F.
(b) What charge is stored on C 3 if the
potential difference between points a
and b is 60.0 V?
Q
Q
a
b
ab
C5
ke(b 2 a)
(b) Show that as b approaches
infinity, the capacitance approaches the value a/ke 5 40a.
37.
38.
Figure P20.36
2
An air-filled capacitor consists of two parallel plates,
each with an area of 7.60 cm2, separated by a distance of
1.80 mm. A 20.0-V potential difference is applied to these
plates. Calculate (a) the electric field between the plates,
(b) the surface charge density, (c) the capacitance, and
(d) the charge on each plate.
A variable air capacitor
used in a radio tuning circuit
is made of N semicircular
plates, each of radius R and
positioned a distance d from its d
neighbors, to which it is elecu
trically connected. As shown in
Figure P20.38, a second idenR
tical set of plates is enmeshed
with the first set. Each plate in
the second set is halfway beFigure P20.38
tween two plates of the first set.
The second set can rotate as a
unit. Determine the capacitance as a function of the angle
of rotation , where 5 0 corresponds to the maximum
capacitance.
39.
A 50.0-m length of coaxial cable has an inner conductor that has a diameter of 2.58 mm and carries a charge of
8.10 C. The surrounding conductor has an inner diameter
of 7.27 mm and a charge of 28.10 C. Assume the region
between the conductors is air. (a) What is the capacitance of
this cable? (b) What is the potential difference between the
two conductors?
40.
Review. A small object of mass m carries a charge q and
is suspended by a thread between the vertical plates of a
parallel-plate capacitor. The plate separation is d. If the
thread makes an angle with the vertical, what is the potential difference between the plates?
(a) Regarding the Earth and a cloud layer 800 m
above the Earth as the “plates” of a capacitor, calculate the
capacitance of the Earth–cloud layer system. Assume the
cloud layer has an area of 1.00 km2 and the air between the
cloud and the ground is pure and dry. Assume charge builds
up on the cloud and on the ground until a uniform electric
field of 3.00 3 106 N/C throughout the space between them
makes the air break down and conduct electricity as a lightning bolt. (b) What is the maximum charge the cloud can
hold?
45.
Four capacitors are connected as shown in Figure P20.45.
(a) Find the equivalent capacitance between points a and b.
(b) Calculate the charge on each
capacitor, taking ΔVab 5 15.0 V.
C3
C2
2
C2
b
Figure P20.44
15.0 mF 3.00 mF
20.0 mF
a
b
6.00 mF
46. Why is the following situation impossible? A technician is testing a cirFigure P20.45
cuit that contains a capacitance
C. He realizes that a better design for the circuit would
include a capacitance 73C rather than C. He has three additional capacitors, each with capacitance C. By combining
these additional capacitors in a certain combination that
is then placed in parallel with the original capacitor, he
achieves the desired capacitance.
47. According to its design specification, the timer circuit delaying the closing of an elevator door is to have a capacitance
of 32.0 F between two points A and B. When one circuit
is being constructed, the inexpensive but durable capacitor installed between these two points is found to have capacitance 34.8 F. To meet the specification, one additional
capacitor can be placed between the two points. (a) Should
it be in series or in parallel with the 34.8-F capacitor?
(b) What should be its capacitance? (c) What If ? The next
circuit comes down the assembly line with capacitance
29.8 F between A and B. To meet the specification, what
additional capacitor should be installed in series or in parallel in that circuit?
| Problems
48. Two capacitors give an equivalent capacitance of Cp when
connected in parallel and an equivalent capacitance of Cs
when connected in series. What is the capacitance of each
capacitor?
49.
A group of identical capacitors is connected first in series and then in parallel. The combined capacitance in parallel is 100 times larger than for the series connection. How
many capacitors are in the group?
50.
Three capacitors are
C1
connected to a battery as shown
in Figure P20.50. Their capacitances are C1 5 3C, C 2 5 C, and C 3 5 5C. (a) What is the equiva- C3
C2
lent capacitance of this set of capacitors? (b) State the ranking of
the capacitors according to the
Figure P20.50
charge they store from largest to
smallest. (c) Rank the capacitors according to the potential
differences across them from largest to smallest. (d) What If?
Assume C 3 is increased. Explain
what happens to the charge 4.0 mF
stored by each capacitor.
55.
Two identical parallel-plate capacitors, each with capacitance 10.0 F, are charged to potential difference 50.0 V
and then disconnected from the battery. They are then connected to each other in parallel with plates of like sign connected. Finally, the plate separation in one of the capacitors
is doubled. (a) Find the total energy of the system of two
capacitors before the plate separation is doubled. (b) Find
the potential difference across each capacitor after the plate
separation is doubled. (c) Find the total energy of the system after the plate separation is doubled. (d) Reconcile the
difference in the answers to parts (a) and (c) with the law of
conservation of energy.
56.
Two identical parallel-plate capacitors, each with
capacitance C, are charged to potential difference DV and
then disconnected from the battery. They are then connected to each other in parallel with plates of like sign connected. Finally, the plate separation in one of the capacitors
is doubled. (a) Find the total energy of the system of two
capacitors before the plate separation is doubled. (b) Find
the potential difference across each capacitor after the plate
separation is doubled. (c) Find the total energy of the system after the plate separation is doubled. (d) Reconcile the
difference in the answers to parts (a) and (c) with the law of
conservation of energy.
7.0 mF
51. Find the equivalent capacitance a
between points a and b in the
combination of capacitors shown
in Figure P20.51.
52.
5.0 mF
b
6.0 mF
Figure P20.51
Consider the circuit
shown in Figure P20.52,
C1
C2
where C1 5 6.00 F, C 2 5 V
3.00 F, and DV 5 20.0 V.
Capacitor C1 is first charged
S1
S2
by closing switch S1. Switch
S1 is then opened, and the
Figure P20.52
charged capacitor is connected to the uncharged
capacitor by closing S2. Calculate (a) the initial charge acquired by C1 and (b) the final charge on each capacitor.
57. Two capacitors, C1 5 25.0 F and C 2 5 5.00 F, are connected in parallel and charged with a 100-V power supply.
(a) Draw a circuit diagram and (b) calculate the total energy stored in the two capacitors. (c) What If? What potential difference would be required across the same two
capacitors connected in series for the combination to store
the same amount of energy as in part (b)? (d) Draw a circuit
diagram of the circuit described in part (c).
58.
Consider two conducting spheres with radii R1 and
R2 separated by a distance much greater than either radius. A total charge Q is shared between the spheres. We
wish to show that when the electric potential energy of the
system has a minimum value, the potential difference between the spheres is zero. The total charge Q is equal to
q1 1 q 2, where q1 represents the charge on the first sphere
and q 2 the charge on the second. Because the spheres are
very far apart, you can assume the charge of each is uniformly distributed over its surface. (a) Show that the energy associated with a single conducting sphere of radius
R and charge q surrounded by a vacuum is U 5 keq 2/2R.
(b) Find the total energy of the system of two spheres in
terms of q1, the total charge Q, and the radii R1 and R2.
(c) To minimize the energy, differentiate the result to part
(b) with respect to q1 and set the derivative equal to zero.
Solve for q1 in terms of Q and the radii. (d) From the result
to part (c), find the charge q 2. (e) Find the potential of
each sphere. (f) What is the potential difference between
the spheres?
59.
(a) A 3.00-F capacitor is connected to a 12.0-V battery.
How much energy is stored in the capacitor? (b) Had the
capacitor been connected to a 6.00-V battery, how much energy would have been stored?
60.
The immediate cause of many deaths is ventricular fibrillation, which is an uncoordinated quivering of the heart.
An electric shock to the chest can cause momentary paralysis of the heart muscle, after which the heart sometimes
Section 20.9 Energy Stored in a Charged Capacitor
53.
As a person moves about in a dry environment, electric
charge accumulates on the person’s body. Once it is at high
voltage, either positive or negative, the body can discharge
via sparks and shocks. Consider a human body isolated
from ground, with the typical capacitance 150 pF. (a) What
charge on the body will produce a potential of 10.0 kV?
(b) Sensitive electronic devices can be destroyed by electrostatic discharge from a person. A particular device can be
destroyed by a discharge releasing an energy of 250 J. To
what voltage on the body does this situation correspond?
54.
A parallel-plate capacitor has a charge Q and plates of
area A. What force acts on one plate to attract it toward the
other plate? Because the electric field between the plates is
E 5 Q /A0, you might think the force is F 5 QE 5 Q 2/A0.
This conclusion is wrong because the field E includes contributions from both plates, and the field created by the positive plate cannot exert any force on the positive plate. Show
that the force exerted on each plate is actually F 5 Q 2/2A0.
Suggestion: Let C 5 0A/x for an arbitrary plate separation x
and note that the work done in separating the two charged
plates is W 5 E Fdx.
693
CHAPTER 20 | Electric Potential and Capacitance
61. A uniform electric field E 5 3 000 V/m exists within a certain region. What volume of space contains an energy equal
to 1.00 3 1027 J? Express your answer in cubic meters and
in liters.
Section 20.10 Capacitors with Dielectrics
62. The supermarket sells rolls of aluminum foil, plastic wrap,
and waxed paper. (a) Describe a capacitor made from such
materials. Compute order-of-magnitude estimates for (b) its
capacitance and (c) its breakdown voltage.
63.
Determine (a) the capacitance and (b) the maximum
potential difference that can be applied to a Teflon-filled
parallel-plate capacitor having a plate area of 1.75 cm2 and
plate separation of 0.040 0 mm.
64. A commercial capacitor
Aluminum
is to be constructed as
shown in Figure P20.64.
This particular capacitor is made from two
strips of aluminum foil
7.00 cm
separated by a strip of
Paper
paraffin-coated paper.
Each strip of foil and
Figure P20.64
paper is 7.00 cm wide.
The foil is 0.004 00 mm thick, and the paper is 0.025 0 mm
thick and has a dielectric constant of 3.70. What length
should the strips have if a capacitance of 9.50 3 1028 F is
desired before the capacitor is rolled up? (Adding a second
strip of paper and rolling the capacitor would effectively
double its capacitance by allowing charge storage on both
sides of each strip of foil.)
65.
(a) How much charge can be placed on a capacitor with
air between the plates before it breaks down if the area of each
of the plates is 5.00 cm2 ? (b) What If ? Find the maximum
charge if polystyrene is used between the plates instead of air.
66. A parallel-plate capacitor in air has a plate separation of
1.50 cm and a plate area of 25.0 cm2. The plates are charged
to a potential difference of 250 V and disconnected from
the source. The capacitor is then immersed in distilled
water. Assume the liquid is an insulator. Determine (a) the
charge on the plates before and after immersion, (b) the
capacitance and potential difference after immersion, and
(c) the change in energy of the capacitor.
Section 20.11 Context Connection: The Atmosphere
as a Capacitor
67.
Lightning can be studied with a Van de Graaff
generator, which consists
of a spherical dome on
which charge is continuously deposited by a moving belt. Charge can be
added until the electric
field at the surface of the
Figure P20.67
dome becomes equal to
the dielectric strength of air. Any more charge leaks off in
sparks as shown in Figure P20.67. Assume the dome has
a diameter of 30.0 cm and is surrounded by dry air with
a “breakdown” electric field of 3.00 3 106 V/m. (a) What
is the maximum potential of the dome? (b) What is the
maximum charge on the dome?
68. Review. A storm cloud and the ground represent the plates
of a capacitor. During a storm, the capacitor has a potential
difference of 1.00 3 108 V between its plates and a charge
of 50.0 C. A lightning strike delivers 1.00% of the energy
of the capacitor to a tree on the ground. How much sap in
the tree can be boiled away? Model the sap as water initially
at 30.08C. Water has a specific heat of 4 186 J/kg · 8C, a
boiling point of 1008C, and a latent heat of vaporization of
2.26 3 106 J/kg.
Additional Problems
69. Review. From a large distance away, a particle of mass 2.00 g
and charge 15.0 C is fired at 21.0 î m/s straight toward a
second particle, originally stationary but free to move, with
mass 5.00 g and charge 8.50 C. Both particles are constrained to move only along the x axis. (a) At the instant of
closest approach, both particles will be moving at the same
velocity. Find this velocity. (b) Find the distance of closest
approach. After the interaction, the particles will move far
apart again. At this time, find the velocity of (c) the 2.00-g
particle and (d) the 5.00-g particle.
70.
Review. From a large distance away, a particle of mass m1
and positive charge q1 is fired at speed v in the positive x direction straight toward a second particle, originally stationary
but free to move, with mass m2 and positive charge q2. Both
particles are constrained to move only along the x axis. (a) At
the instant of closest approach, both particles will be moving
at the same velocity. Find this velocity. (b) Find the distance of
closest approach. After the interaction, the particles will move
far apart again. At this time, find the velocity of (c) the particle of mass m1 and (d) the particle of mass m2.
71.
A model of a red blood cell portrays the cell as a capacitor with two spherical plates. It is a positively charged
conducting liquid sphere of area A, separated by an insulating membrane of thickness t from the surrounding negatively charged conducting fluid. Tiny electrodes introduced
into the cell show a potential difference of 100 mV across
the membrane. Take the membrane’s thickness as 100 nm
and its dielectric constant as 5.00. (a) Assume that a typical
red blood cell has a mass of 1.00 3 10212 kg and density
1 100 kg/m3. Calculate its volume and its surface area.
(b) Find the capacitance of the cell. (c) Calculate the
David Evison/Shutterstock.com
resumes its proper beating. One type of defibrillator (Fig. P20.60) applies
a strong electric shock to
the chest over a time interval of a few milliseconds.
This device contains a capacitor of several microfarads, charged to several
thousand volts. Electrodes
called paddles are held
against the chest on both
Figure P20.60
sides of the heart, and the
capacitor is discharged
through the patient’s chest. Assume an energy of 300 J is
to be delivered from a 30.0-F capacitor. To what potential
difference must it be charged?
Andrew Olney/Getty Images
694
| Problems
charge on the surfaces of the membrane. How many electronic charges does this charge represent? (Suggestion: The
chapter text models the Earth’s atmosphere as a capacitor
with two spherical plates.)
72. Why is the following situation impossible? In the Bohr model of
the hydrogen atom, an electron moves in a circular orbit
about a proton. The model states that the electron can exist
only in certain allowed orbits around the proton: those
whose radius r satisfies r 5 n2(0.052 9 nm), where n 5 1, 2,
3, . . . . For one of the possible allowed states of the atom, the
electric potential energy of the system is 213.6 eV.
74.
The liquid-drop model of the atomic nucleus suggests
high-energy oscillations of certain nuclei can split the nucleus into two unequal fragments plus a few neutrons. The
fission products acquire kinetic energy from their mutual
Coulomb repulsion. Assume the charge is distributed uniformly throughout the volume of each spherical fragment
and, immediately before separating, each fragment is at rest
and their surfaces are in contact. The electrons surrounding the nucleus can be ignored. Calculate the electric potential energy (in electron volts) of two spherical fragments
from a uranium nucleus having the following charges and
radii: 38e and 5.50 3 10215 m, and 54e and 6.20 3 10215 m.
A Geiger–Mueller tube is a radiation detector that consists of a closed, hollow, metal cylinder (the cathode) of
inner radius ra and a coaxial cylindrical wire (the anode) of
radius rb (Fig. P20.74a). The charge per unit length on the
anode is , and the charge per unit length on the cathode
is 2. A gas fills the space between the electrodes. When the
tube is in use (Fig. P20.74b) and a high-energy elementary
particle passes through this space, it can ionize an atom of
the gas. The strong electric field makes the resulting ion
and electron accelerate in opposite directions. They strike
other molecules of the gas to ionize them, producing an
avalanche of electrical discharge. The pulse of electric current between the wire and the cylinder is counted by an external circuit. (a) Show that the magnitude of the electric
potential difference between the wire and the cylinder is
DV 5 2ke ln
ra
1r 2
b
(b) Show that the magnitude of the electric field in the
space between cathode and anode is
E5
DV
1
ln (ra /rb) r
12
where r is the distance from the axis of the anode to the
point where the field is to be calculated.
Cathode
l
ra
rb
l
Anode
a
Figure P20.74 Problems 74 and 75.
b
© Hank Morgan/Photo Researchers, Inc.
73.
695
75. Assume that the internal diameter of the Geiger–Mueller
tube described in Problem 20.74 is 2.50 cm and that the
wire along the axis has a diameter of 0.200 mm. The dielectric strength of the gas between the central wire and the cylinder is 1.20 3 106 V/m. Use the result of Problem 20.74 to
calculate the maximum potential difference that can be applied between the wire and the cylinder before breakdown
occurs in the gas.
76.
Four balls, each with mass m, 1 2
are connected by four nonconducting
strings to form a square with side a as
a
shown in Figure P20.76. The assembly
is placed on a nonconducting, friction4
3
a
less, horizontal surface. Balls 1 and 2
each have charge q, and balls 3 and 4
Figure P20.76
are uncharged. After the string connecting balls 1 and 2 is cut, what is the
maximum speed of balls 3 and 4?
77. Calculate the work that must be done on charges brought
from infinity to charge a spherical shell of radius R 5 0.100 m
to a total charge Q 5 125 C.
78.
Calculate the work that must be done on charges
brought from infinity to charge a spherical shell of radius R
to a total charge Q.
79. A 2.00-nF parallel-plate capacitor is charged to an initial potential difference DVi 5 100 V and is then isolated. The dielectric material between the plates is mica, with a dielectric
constant of 5.00. (a) How much work is required to withdraw the mica sheet? (b) What is the potential difference
across the capacitor after the mica is withdrawn?
80. Why is the following situation Q
impossible? You set up an apparatus in your laboratory
R
as follows. The x axis is the
Q
symmetry axis of a stationary,
x
x
uniformly charged ring of radius R 5 0.500 m and charge
S
v
Q 5 50.0 C (Fig. P20.80).
You place a particle with
charge Q 5 50.0 C and
Figure P20.80
mass m 5 0.100 kg at the center of the ring and arrange
for it to be constrained to move only along the x axis. When
it is displaced slightly, the particle is repelled by the ring and
accelerates along the x axis. The particle moves faster than
you expected and strikes the opposite wall of your laboratory
at 40.0 m/s.
81.
A parallel-plate capacitor is constructed using a dielectric material whose dielectric constant is 3.00 and whose dielectric strength is 2.00 3 108 V/m. The desired capacitance
is 0.250 F, and the capacitor must withstand a maximum
potential difference of 4 000 V. Find the minimum area of
the capacitor plates.
82.
An electric dipole is located along the y axis as shown in
Figure P20.82 (page 696). The magnitude of its electric dipole moment is defined as p 5 2aq. (a) At a point P, which is
far from the dipole (r .. a), show that the electric potential is
V5
ke p cos
r2
696
CHAPTER 20 | Electric Potential and Capacitance
(b) Calculate the radial component
y
Er and the perpendicular component E of the associated electric
field. Note that E 5 2(1/r)(−V/−).
Do these results seem reasonable
for (c) 5 908 and 08? (d) For r 5 0? q a u
(e) For the dipole arrangement
shown in Figure P20.82, express
V in terms of Cartesian coordinates
a
using r 5 (x 2 1 y 2)1/2 and
cos
5
constant is inserted a distance x into the capacitor as
shown in Figure P20.85. Assume d is much smaller than x.
(a) Find the equivalent capacitance of the device. (b) Calculate the energy stored in the capacitor. (c) Find the direction and magnitude of the force exerted by the plates
on the dielectric. (d) Obtain a numerical value for the
force when x 5 //2, assuming / 5 5.00 cm, d 5 2.00 mm,
the dielectric is glass ( 5 4.50), and the capacitor was
charged to 2.00 3 103 V before the dielectric was inserted.
Suggestion: The system can be considered as two capacitors
connected in parallel.
Er
P
r1
Eu
r
r2
x
q y
(x 2 1 y 2)1/2
Figure P20.82
86.
(f) Using these results and again
taking r .. a, calculate the field components Ex and Ey .
84.
10.0 mF
5.00 mF
83. A 10.0-F capacitor is charged to
15.0 V. It is next connected in series
with an uncharged 5.00-F capacitor.
The series combination is finally connected across a 50.0-V battery as diagrammed in Figure P20.83. Find the
new potential differences across the
5.00-F and 10.0-F capacitors after
the switch is thrown closed.
Vi 15.0 V
50.0 V
Figure P20.83
Two large, parallel metal plates,
each of area A, are oriented horizontally and separated by
a distance 3d. A grounded conducting wire joins them, and
initially each plate carries no charge. Now a third identical
plate carrying charge Q is inserted between the two plates,
parallel to them and located a distance d from the upper
plate as shown in Figure P20.84. (a) What induced charge
appears on each of the two original plates? (b) What potential difference appears between the middle plate and each
of the other plates?
Two square plates of sides / are placed parallel to
each other with separation d as suggested in Figure P20.86.
You may assume d is much less than /. The plates carry uniformly distributed static charges 1Q 0 and 2Q 0. A block of
metal has width /, length /, and thickness slightly less than d.
It is inserted a distance x into the space between the plates.
The charges on the plates remain uniformly distributed as
the block slides in. In a static situation, a metal prevents
an electric field from penetrating inside it. The metal can
be thought of as a perfect dielectric, with : `. (a) Calculate the stored energy in the system as a function of x.
(b) Find the direction and magnitude of the force that acts
on the metallic block. (c) The area of the advancing front
face of the block is essentially equal to /d. Considering the
force on the block as acting on this face, find the stress
(force per area) on it. (d) Express the energy density in
the electric field between the charged plates in terms of Q 0,
/, d, and 0. (e) Explain how the answers to parts (c) and
(d) compare with each other.
ᐉ
x
Q 0
d
Q 0
d
Figure P20.86
87.
2d
Figure P20.84
85. A capacitor is constructed
from two square, metallic plates of sides / and
separation d. Charges 1Q
and 2Q are placed on
the plates, and the power
supply is then removed.
A material of dielectric
Q
x
k
d
C
Figure P20.85
Q
2C
3C
C
ᐉ
Determine the equivalent
capacitance of the combination shown in Figure P20.87.
Suggestion: Consider the symmetry involved.
2C
Figure P20.87
Chapter
21
Current and Direct
Current Circuits
Chapter Outline
21.1 Electric Current
21.2 Resistance and Ohm’s Law
21.3 Superconductors
21.4 A Model for Electrical
Conduction
21.5 Energy and Power in
Electric Circuits
21.6 Sources of emf
21.7 Resistors in Series
and Parallel
Trombax/Shutterstock.com
21.8 Kirchhoff ’s Rules
21.9 RC Circuits
21.10 Context Connection: The
Atmosphere as a Conductor
SUMMARY
T
hus far, our discussion of electrical phenomena has focused on charges
at rest, or the study of electrostatics. We shall now consider situations
involving electric charges in motion. The term electric current, or simply current, is used to describe the flow of charge through some region of
space. Most practical applications of electricity involve electric currents. For example, in a flashlight with an incandescent bulb, charges flow in the filament of
the lightbulb after the switch is turned on. In most common situations, the flow
of charge takes place in a conductor, such as a copper wire. Currents can also
exist outside a conductor. For instance, a beam of electrons in a particle accelerator
constitutes a current.
In Chapter 20, we introduced the notion of a circuit. As we continue our
investigations into circuits in this chapter, we introduce the resistor as a new circuit
element.
A technician repairs a connection on a
circuit board from a computer. In our
lives today, we use many items containing
electric circuits, including many with circuit
boards much smaller than the board
shown in the photograph, including MP3
players, cell phones, and digital cameras.
In this chapter, we study simple types of
circuits and learn how to analyze them.
697
698
CHAPTER 21 | Current and Direct Current Circuits
21.1 | Electric Current
Whenever charge is flowing, an electric current is said to exist. To define current
mathematically, suppose charged particles are moving perpendicular to a surface of
area A as in Figure 21.1. (This area could be the cross-sectional area of a wire, for
example.) The current is defined as the rate at which electric charge flows through
this surface. If DQ is the amount of charge that passes through this area in a time
interval Dt, the average current Iavg over the time interval is the ratio of the charge
to the time interval:
Figure 21.1 Charges in motion
through an area A. The time rate at
which charge flows through the area
is defined as the current I.
DQ
21.1b
Dt
It is possible for the rate at which charge flows to vary in time. We define the instantaneous current I as the limit of the preceding expression as Dt goes to zero:
Iavg 5
I ; lim
c Electric current
DQ
Dt:0 Dt
5
dQ
dt
21.2b
The SI unit of current is the ampere (A):
1 A 5 1 C/s
Pitfall Prevention | 21.1
“Current Flow” Is Redundant
The phrase current flow is commonly
used, although it is technically
incorrect because current is a flow
(of charge). This wording is similar to
the phrase heat transfer, which is also
redundant because heat is a transfer
(of energy). We will avoid this phrase
and speak of flow of charge or charge flow.
S
v
21.3b
That is, 1 A of current is equivalent to 1 C of charge passing through a surface in 1 s.
The particles flowing through a surface as in Figure 21.1 can be charged positively or negatively, or we can have two or more types of particles moving, with
charges of both signs in the flow. Conventionally, we define the direction of
the current as the direction of flow of positive charge, regardless of the sign of
the actual charged particles in motion.1 In a common conductor such as copper,
the current is physically due to the motion of the negatively charged electrons.
Therefore, when we speak of current in such a conductor, the direction of the
current is opposite the direction of flow of electrons. On the other hand, if one
considers a beam of positively charged protons in a particle accelerator, the current
is in the direction of motion of the protons. In some cases—gases and electrolytes,
for example—the current is the result of the flow of both positive and negative
charged particles. It is common to refer to a moving charged particle (whether it
is positive or negative) as a mobile charge carrier. For example, the charge carriers
in a metal are electrons.
We now build a structural model that will allow us to relate the macroscopic current to the motion of the charged particles. Consider identical charged particles
moving in a conductor of cross sectional area A (Fig. 21.2). The volume of a segment
of the conductor of length Dx (between the two circular cross sections shown in
Fig. 21.2) is A Dx. If n represents the number of mobile charge carriers per unit
volume (in other words, the charge carrier density), the number of carriers in the
segment is nA Dx. Therefore, the total charge DQ in this segment is
DQ 5 number of carriers in section 3 charge per carrier 5 (nA Dx)q
Figure 21.2 A segment of a
uniform conductor of cross-sectional
area A.
where q is the charge on each carrier. If the carriers move with an average velocity
component vd in the x direction (along the wire), the displacement they experience
in this direction in a time interval Dt is Dx 5 vd Dt. The speed vd of the charge carrier
along the wire is an average speed called the drift speed. Let us choose Dt to be the
time interval required for the charges in the segment to move through a displacement whose magnitude is equal to the length of the segment. This time interval is
also that required for all the charges in the segment to pass through the circular area
at one end. With this choice, we can write DQ in the form
DQ 5 (nAvd Dt)q
1Even though we discuss a direction for current, current is not a vector. As we shall see later in the chapter, currents add
algebraically and not vectorially.
699
21.1 | Electric Current
If we divide both sides of this equation by Dt, we see that the average current in the
conductor is
DQ
5 nqvdA
21.4b
Dt
Equation 21.4 relates a macroscopically measured average current to the microscopic origin of the current: the density of charge carriers n, the charge per carrier
q, and the drift speed vd .
Iavg 5
c Current in terms of microscopic
parameters
Q U I CK QUI Z 21.1 Consider positive and negative charges moving horizontally
through the four regions shown in Figure 21.3. Rank the currents in these four
regions, from highest to lowest.
Figure 21.3 (Quick Quiz 21.1) Four groups of charges move
through a region.
Let us investigate further the notion of drift speed. We have identified drift speed
as an average speed along the wire, but the charge carriers are by no means moving
in a straight line with speed vd. Consider a conductor in which the charge carriers are free electrons. In the absence of a potential difference across the conductor, these electrons undergo random motion similar to that of gas molecules in the
structural model of kinetic theory that we studied in Chapter 16. This random motion is related to the temperature of the conductor. The electrons undergo repeated
collisions with the metal atoms, and the result is a complicated zigzag motion. When
a potential difference is applied across the conductor, an electric field is established in the conductor. The electric field exerts an electric force on the electrons
(Eq. 19.4). This force accelerates the electrons and hence produces a current. The
motion of the electrons due to the electric force is superimposed on their random
motion to provide an average velocity whose magnitude is the drift speed as shown
in Active Figure 21.4.
When electrons make collisions with metal atoms during their motion, they transfer energy to the atoms. This energy transfer causes an increase in the vibrational
energy of the atoms and a corresponding increase in the temperature of the conductor.2 This process involves all three types of energy storage in the conservation
of energy equation, Equation 7.2. If we consider the system to be the electrons, the
metal atoms, and the electric field (which is established by an external source such
as a battery), the energy at the instant when the potential difference is applied across
the conductor is electric potential energy associated with the field and the electrons.
This energy is transformed by work done within the system by the field on the electrons to kinetic energy of electrons. When the electrons strike the metal atoms, some
of the kinetic energy is transferred to the atoms, which adds to the internal energy
of the system.
The current density J in the conductor is defined as the current per unit area.
From Equation 21.4, the current density is
J ;
I
5 nqvd
A
21.5b
where J has the SI units amperes per square meter.
2This increase in temperature is sometimes called Joule heating, but that term is a misnomer because there is no heat
involved. We will not use this wording.
Pitfall Prevention | 21.2
Batteries Do Not Supply Electrons
A battery does not supply electrons to
the circuit. It establishes the electric
field that exerts a force on electrons
already in the wires and elements of
the circuit.
S
E
S
v
Active Figure 21.4 A schematic
representation of the zigzag motion
of negative charge carriers in a conductor. Because of the acceleration
of the charge carriers due to the
electric force, the paths are actually
parabolic. The drift speed, however, is
much smaller than the average speed,
so the parabolic shape is not visible
on this scale.
700
CHAPTER 21 | Current and Direct Current Circuits
THINKING PHYSICS 21.1
In Chapter 19, we claimed that the electric field inside a conductor is zero. In
the preceding discussion, however, we have used the notion of an electric field
in a conducting wire that exerts electric forces on electrons, causing them to
move with a drift velocity. Is this notion inconsistent with Chapter 19?
Reasoning The electric field is zero only in a conductor in electrostatic equilib-
rium, that is, a conductor in which the charges are at rest after having moved to
equilibrium positions. In a current-carrying conductor, the charges are not at
rest, so the requirement for a zero field is not imposed. The electric field in a
conductor in a circuit is due to a distribution of charge over the surface of the
conductor that can be quite complicated.3 b
Example 21.1 | Drift Speed in a Copper Wire
The 12-gauge copper wire in a typical residential building has a cross-sectional area of 3.31 3 1026 m2. It carries a constant
current of 10.0 A. What is the drift speed of the electrons in the wire? Assume each copper atom contributes one free
electron to the current. The density of copper is 8.92 g/cm3.
SOLUTION
Conceptualize Imagine electrons following a zigzag motion with a drift velocity parallel to the wire superimposed on the
motion as in Active Figure 21.4. As mentioned earlier, the drift speed is small, and this example helps us quantify the speed.
Categorize We evaluate the drift speed using Equation 21.4. Because the current is constant, the average current during
any time interval is the same as the constant current: Iavg 5 I.
Analyze The periodic table of the elements in Appendix C shows that the molar mass of copper is M 5 63.5 g/mol. Recall
that 1 mol of any substance contains Avogadro’s number of atoms (NA 5 6.02 3 1023 mol21).
Use the molar mass and the density of copper to find the
volume of 1 mole of copper:
V5
From the assumption that each copper atom contributes
one free electron to the current, find the electron density
in copper:
n5
Solve Equation 21.4 for the drift speed and substitute for
the electron density:
vd 5
Substitute numerical values:
vd 5
M
NA
V
5
Iavg
nqA
NA M
5
I
IM
5
nqA
qANA (10.0 A)(0.063 5 kg/mol)
(1.60 3 10219 C)(3.31 3 1026 m2)(6.02 3 1023 mol21)(8 920 kg/m3)
5 2.23 3 1024 m/s
Finalize This result shows that typical drift speeds are very small. For instance, electrons traveling at 2.23 3 1024 m/s would
take about 75 min to travel 1 m! You might therefore wonder why a light turns on almost instantaneously when its switch
is thrown. In a conductor, changes in the electric field that drives the free electrons travel through the conductor with a
speed close to that of light. So, when you flip on a light switch, electrons already in the filament of an incandescent lightbulb
experience electric forces and begin moving after a time interval on the order of nanoseconds.
3See Chapter 18 in R. Chabay and B. Sherwood, Matter & Interactions II: Electric and Magnetic Interactions (Hoboken: Wiley,
2007) for details on this charge distribution.
21.2 | Resistance and Ohm’s Law
ᐉ
21.2 | Resistance and Ohm’s Law
The drift speed of electrons in a current-carrying wire is related to the electric field
in the wire. If the field is increased, the electric force on the electrons is stronger
and the drift speed increases. We shall show in Section 21.4 that this relationship is
linear and that the drift speed is directly proportional to the electric field. For a uniform field in a conductor of uniform cross-section, the potential difference across
the conductor is proportional to the electric field as in Equation 20.6. Therefore,
when a potential difference DV is applied across the ends of a metallic conductor
as in Figure 21.5, the current in the conductor is found to be proportional to the
applied voltage; that is, I ~ DV. We can write this proportionality as DV 5 IR, where
R is called the resistance of the conductor. We define this resistance according to
the equation we have just written, as the ratio of the voltage across the conductor to
the current it carries:
R ;
DV
I
701
21.6b
Resistance has the SI unit volt per ampere, called an ohm (V). Therefore, if a potential difference of 1 V across a conductor produces a current of 1 A, the resistance of
the conductor is 1 V. As another example, if an electrical appliance connected to a
120-V source carries a current of 6.0 A, its resistance is 20 V.
Resistance is the quantity that determines the current that results due to a voltage in a simple circuit. For a fixed voltage, if the resistance increases, the current
decreases. If the resistance decreases, the current increases.
It might be useful for you to build a mental representation for current, voltage,
and resistance by comparing these concepts to analogous concepts for the flow of
water in a river. As water flows downhill in a river of constant width and depth, the
rate of flow of water (analogous to current) depends on the total vertical distance
through which the water drops between two points (analogous to voltage) and on
the width and depth as well as on the effects of rocks, the riverbank, and other obstructions (analogous to resistance). Likewise, electric current in a uniform conductor depends on the applied voltage, and the resistance of the conductor is caused by
collisions of the electrons with atoms in the conductor.
For many materials, including most metals, experiments show that the resistance
is constant over a wide range of applied voltages. This behavior is known as Ohm’s
law after Georg Simon Ohm (1787–1854), who was the first to conduct a systematic
study of electrical resistance.
Many individuals call Equation 21.6 Ohm’s law, but this terminology is incorrect.
This equation is simply the definition of resistance, and it provides an important relationship between voltage, current, and resistance. Ohm’s law is not a fundamental
law of nature, but a behavior that is valid only for certain materials and devices, and
only over a limited range of conditions. Materials or devices that obey Ohm’s law,
and hence that have a constant resistance over a wide range of voltages, are said to be
ohmic (Fig. 21.6a). Materials or devices that do not obey Ohm’s law are nonohmic.
One common semiconducting device that is nonohmic is the diode, a circuit element
that acts like a one-way valve for current. Its resistance is small for currents in one
direction (positive DV) and large for currents in the reverse direction (negative DV )
Figure 21.6 (a) The current–
potential difference curve for
an ohmic material. The curve is
linear, and the slope is equal to
the inverse of the resistance of the
conductor. (b) A nonlinear current–
potential difference curve for a
semiconducting diode. This device
does not obey Ohm’s law.
S
E
S
E
Figure 21.5 A uniform conductor
of length O and cross-sectional area A.
c Definition of resistance
Pitfall Prevention | 21.3
We’ve Seen Something Like
Equation 21.6 Before
In Chapter 4, we introduced
Newton’s second law, oF 5 ma, for a
net force on an object of mass m. It
can be written as
oF
m5
a
In Chapter 4, we defined mass as
resistance to a change in motion in
response to an external force. Mass as
resistance to changes in motion is
analogous to electrical resistance to
charge flow, and Equation 21.6 is
analogous to the form of Newton’s
second law above. Each equation
states that the resistance (electrical or
mechanical) is equal to (1) DV, the
cause of current, or (2) oF, the cause
of changes in motion, divided by the
result, (1) a charge flow, quantified
by current I, or (2) a change in
motion, quantified by acceleration a.
702
CHAPTER 21 | Current and Direct Current Circuits
as shown in Figure 21.6b. Most modern electronic devices have nonlinear current–
voltage relationships; their operation depends on the particular ways they violate
Ohm’s law.
QUI C K QU IZ 21.2 In Figure 21.6b, as the applied voltage increases, does the resistance of the diode (a) increase, (b) decrease, or (c) remain the same?
A resistor is a simple circuit element that provides a specified resistance in an
electrical circuit. The symbol for a resistor in circuit diagrams is a zigzag red line
(
). We can express Equation 21.6 in the form
21.7b
DV 5 IR
This equation tells us that the voltage across a resistor is the product of the resistance
and the current in the resistor.
The resistance of an ohmic conducting wire such as that shown in Figure 21.5
is found to be proportional to its length / and inversely proportional to its crosssectional area A. That is,
c Resistance of a uniform material
of resistivity along a length ,
Pitfall Prevention | 21.4
Resistance and Resistivity
Resistivity is a property of a substance,
whereas resistance is a property of an
object. We have seen similar pairs of
variables before. For example, density
is a property of a substance, whereas
mass is a property of an object.
Equation 21.8 relates resistance
to resistivity, and we have seen a
previous equation (Eq. 1.1) that
relates mass to density.
,
21.8b
A
where the constant of proportionality is called the resistivity of the material,4 which
has the unit ohm ? meter (V ? m). To understand this relationship between resistance and resistivity, note that every ohmic material has a characteristic resistivity, a
parameter that depends on the properties of the material and on temperature. On
the other hand, as you can see from Equation 21.8, the resistance of a particular
conductor depends on its size and shape as well as on the resistivity of the material.
Table 21.1 provides a list of resistivities for various materials measured at 208C.
R5
TABLE 21.1 | Resistivities and Temperature Coefficients of Resistivity for
Various Materials
Material
Resistivitya (V ? m)
Temperature
Coefficientb [(8C)21]
Silver
1.59 3 1028
3.8 3 1023
Copper
1.7 3 1028
3.9 3 1023
Gold
2.44 3 1028
3.4 3 1023
Aluminum
2.82 3 1028
3.9 3 1023
Tungsten
5.6 3 1028
4.5 3 1023
Iron
10 3 1028
5.0 3 1023
Platinum
11 3 1028
3.92 3 1023
Lead
22 3 1028
3.9 3 1023
Nichromec
1.00 3 1026
0.4 3 1023
Carbon
3.5 3 1025
20.5 3 1023
Germanium
0.46
248 3 1023
Silicond
2.3 3 103
275 3 1023
Glass
1010 to 1014
Hard rubber
Sulfur
Quartz (fused)
, 1013
1015
75 3 1016
aAll values at 208C. All elements in this table are assumed to be free of impurities.
bThe temperature coefficient of resistivity will be discussed later in this section.
cA
nickel–chromium alloy commonly used in heating elements. The resistivity of Nichrome varies with
composition and ranges between 1.00 3 1026 and 1.50 3 1026 V ? m.
dThe resistivity of silicon is very sensitive to purity. The value can be changed by several orders of magnitude
when it is doped with other atoms.
4The symbol used for resistivity should not be confused with the same symbol used earlier in the text for mass density
and volume charge density.
21.2 | Resistance and Ohm’s Law
703
Douglas C. Johnson/Cal Poly Pomona
An assortment of resistors used in
electric circuits.
The inverse of the resistivity is defined5 as the conductivity . Hence, the resistance
of an ohmic conductor can be expressed in terms of its conductivity as
R5
,
A
21.9b
where 5 1/.
Equation 21.9 shows that the resistance of a conductor is proportional to its
length and inversely proportional to its cross-sectional area, similar to the flow
of liquid through a pipe. As the length of the pipe is increased and the pressure
difference between the ends of the pipe is held constant, the pressure difference
between any two points separated by a fixed distance decreases and there is less
force pushing the element of fluid between these points through the pipe. Therefore, less fluid flows for a given pressure difference between the ends of the pipe,
representing an increased resistance. As its cross-sectional area is increased, the
pipe can transport more fluid in a given time interval for a given pressure difference between the ends of the pipe, so its resistance drops.
As another analogy between electrical circuits and our previous studies, let us
combine Equations 21.6 and 21.9:
R5
q
,
DV
DV
DV
5
: I 5 A
:
5 A
A
I
,
Dt
,
where q is the amount of charge transferred in a time interval Dt. Let us compare this
equation to Equation 17.35 for conduction of energy through a slab of material of
area A, length /, and thermal conductivity k, which we reproduce below:
P 5 kA
(Th 2 Tc)
L
:
Q
Dt
5 kA
DT
L
In this equation, Q is the amount of energy transferred by heat in a time interval
Dt. Notice the striking similarity between these last two equations.
Another analogy arises in an example that is important in biochemical applications. Fick’s law describes the rate of transfer of a chemical solute through a solvent
by the process of diffusion. This transfer occurs because of a difference in concentration of the solute (mass of solute per volume) between the two locations. Fick’s law
is as follows:
DC
n
5 DA
Dt
L
where n/Dt is the rate of flow of the solute in moles per second, A is the area through
which the solute moves, and L is the length over which the concentration difference
is DC. The concentration is measured in moles per cubic meter. The parameter D
is a diffusion constant (with units of meters squared per second) that describes the
5Do not confuse the symbol for conductivity with the same symbol used earlier for the Stefan–Boltzmann constant and
surface charge density.
Diffusion in biological
systems
704
CHAPTER 21 | Current and Direct Current Circuits
dexns/Shutterstock.com
rate of diffusion of a solute through the solvent and is similar in nature to electrical or thermal conductivity. Fick’s law has important applications in describing the
transport of molecules across biological membranes.
All three of the preceding equations have exactly the same mathematical form.
Each has a time rate of change on the left, and each has the product of a conductivity,
an area, and a ratio of a difference in a variable to a length on the right. This type of
equation is a transport equation used when we transport energy, charge, or moles of
matter. The difference in the variable on the right side of each equation is what drives
the transport. A temperature difference drives energy transfer by heat, an electric
potential difference drives a transfer of charge, and a concentration difference drives
a transfer of matter.
Most electric circuits use resistors to control the current level in the various parts
of the circuit. Two common types of resistors are the composition resistor containFigure 21.7 A close-up view of a
circuit board shows the color coding
ing carbon and the wire-wound resistor, which consists of a coil of wire. Resistors
on a resistor.
are normally color-coded to give their values in ohms, as shown in Figure 21.7 and
Table 21.2. As an example, the four colors on the resistor at the bottom of Figure 21.7 are yellow (5 4), violet (5 7), black (5 100), and gold (5 5%), and so the
resistance value is 47 3 100 Ω 5 47 Ω with a tolerance value of 5% 5 2 Ω.
Let’s consider the role of electrical resistance in maintaining proper beating of
Electrical activity in the heart
the human heart. The right atrium of the heart contains a specialized set of muscle
fibers called the SA (sinoatrial) node that initiates the heartbeat. Electrical impulses
that originate in these fibers gradually spread from cell to cell throughout the right
and left atrial muscles, causing them to contract. When the impulses reach the atrioventricular (AV) node, the muscles of the atria begin to relax, and the impulses are
directed to the ventricular muscles by an assembly of heart muscle cells called the
bundle of His and the Purkinje fibers. After the resulting contraction of the ventricles,
the heartbeat is complete and the cycle begins again.
The heart can experience a number of arrhythmias, in which the normal
Catheter ablation for atrial
fibrillation
heartbeat rhythm is interrupted. Arrhythmias are generally caused by abnormal
electrical activity in the heart. The most common cardiac arrhythmia is atrial fibrillation (AF). In this condition, the two upper chambers of the heart, the atria,
undergo random quivering at a rate that can be greater than 300 per minute,
rather than the usual coordinated contractions. In paroxysmal AF, the patient
goes into episodes of atrial fibrillation that may last from a few minutes to a few
days. It some cases, the condition may even become chronic. With episodes longer than a few days, blood can pool in the atria, due to the inefficiency of the
quivering action in pumping blood out of the heart. This pooled blood can result
in clots, which can then travel to the brain and cause a stroke. Patients with longlasting episodes of AF are treated with anticoagulants to
prevent clots, rate control medications to slow the rate
TABLE 21.2 | Color Code for Resistors
of impulses conducted to the ventricles, and antiarrhythmics to return the heart to its normal rhythm. DefibrilColor
Number
Multiplier
Tolerance
lator device paddles are sometimes used to deliver an
Black
0
1
electric shock to a patient’s chest in order to restore norBrown
1
101
mal heart rhythm.
Red
2
102
In a large percentage of patients, the source of the
Orange
3
103
chaotic activity is found in the four pulmonary veins
leading into the left atrium. Atrial tissue has grown into
Yellow
4
104
these veins and can act as electrical triggers competing
5
Green
5
10
with the SA node. As a result, the atrial muscles receive
6
Blue
6
10
electrical signals from a variety of sources rather than
Violet
7
107
the SA node alone, leading to chaotic contractions.
Gray
8
108
Patients whose arrhythmias can not be controlled with
White
9
109
medications, as well as those who do not wish to take
medications, have an additional option. A procedure
Gold
1021
5%
known as cardiac catheter ablation may be performed by
22
10%
Silver
10
an electrophysiologist in an effort to restore normal sinus
Colorless
20%
rhythm. In this procedure, the patient is anesthetized
21.2 | Resistance and Ohm’s Law
© 2009 David Klemm
and catheters are inserted into a vein in the groin and
guided into the right atrium of the heart. The catheter
then punctures the septum and enters the left atrium.
Figure 21.8 shows an ablation catheter passing into
the heart from a vein. The electrophysiologist maps
the atrium and then stimulates the heart to determine
areas of abnormal electrical activity. Finally, the electrophysiologist ablates the tissue around the four pulmonary veins, usually with radiofrequency energy from
the tip of one of the catheters. The resulting scar tissue
represents a high-resistance path, through which the
electrical signals from the AF triggers in the pulmonary
veins cannot travel. As a result, the SA node alone again
controls the electrical activity of the heart. Because the
triggers in the pulmonary veins have been cut off electrically from the rest of the heart, this specific procedure is called a pulmonary vein isolation.
705
Figure 21.8 During a cardiac catheter ablation procedure,
catheters are guided into the left atrium through a vein from
the groin. Radiofrequency energy is used to ablate tissue
surrounding the pulmonary veins, where abnormal electrical
activity is happening.
Exampl e 21.2 | The Resistance of Nichrome Wire
The radius of 22-gauge Nichrome wire is 0.32 mm.
(A) Calculate the resistance per unit length of this wire.
SOLUTION
Conceptualize Table 21.1 shows that Nichrome has a resistivity two orders of magnitude larger than the best conductors in
the table. Therefore, we expect it to have some special practical applications that the best conductors may not have.
Categorize We model the wire as a cylinder so that a simple geometric analysis can be applied to find the resistance.
Analyze Use Equation 21.8 and the resistivity of Nichrome
from Table 21.1 to find the resistance per unit length:
R
1.0 3 1026 V ? m
5 5
5
5 3.1 V/m
,
A
r 2
(0.32 3 1023 m)2
(B) If a potential difference of 10 V is maintained across a 1.0-m length of the Nichrome wire, what is the current in the
wire?
SOLUTION
Analyze Use Equation 21.6 to find the current:
I5
10 V
DV
DV
5
5
5 3.2 A
R
(R/,),
(3.1 V/m)(1.0 m)
Finalize Because of its high resistivity and resistance to oxidation, Nichrome is often used for heating elements in toasters,
irons, and electric heaters.
What If? What if the wire were composed of copper instead of Nichrome? How would the values of the resistance per
unit length and the current change?
Answer Table 21.1 shows us that copper has a resistivity two orders of magnitude smaller than that for Nichrome. There-
fore, we expect the answer to part (A) to be smaller and the answer to part (B) to be larger. Calculations show that a copper
wire of the same radius would have a resistance per unit length of only 0.053 V/m. A 1.0-m length of copper wire of the
same radius would carry a current of 190 A with an applied potential difference of 10 V.
706
CHAPTER 21 | Current and Direct Current Circuits
Change in Resistivity with Temperature
Resistivity depends on a number of factors, one of which is temperature. For most
metals, resistivity increases approximately linearly with increasing temperature over
a limited temperature range according to the expression
c Variation of resistivity with
temperature
5 0[1 1 (T 2 T0)]
21.10b
where is the resistivity at some temperature T (in degrees Celsius), 0 is the resistivity
at some reference temperature T0 (usually 208C), and is called the temperature coefficient of resistivity (not to be confused with the average coefficient of linear expansion in Chapter 16). From Equation 21.10, we see that can be expressed as
c Temperature coefficient of
resistivity
5
1 D
0 DT
21.11b
where D 5 2 0 is the change in resistivity in the temperature interval DT 5 T 2 T0.
The resistivities and temperature coefficients of certain materials are listed in
Table 21.1. Note the enormous range in resistivities, from very low values for good
conductors, such as copper and silver, to very high values for good insulators, such
as glass and rubber. An ideal, or “perfect,” conductor would have zero resistivity, and
an ideal insulator would have infinite resistivity.
Because resistance is proportional to resistivity according to Equation 21.8, the
temperature variation of the resistance can be written as
c Variation of resistance with
temperature
R 5 R0[1 1 (T 2 T0)]
21.12b
Precise temperature measurements are often made using this property.
QUI C K QU IZ 21.3 When does an incandescent lightbulb carry more current?
(a) immediately after it is turned on and the glow of the metal filament is increasing
or (b) after it has been on for a few milliseconds and the glow is steady?
21.3 | Superconductors
For several metals, resistivity is nearly proportional to temperature as shown in Figure 21.9. In reality, however, there is always a nonlinear region at very low temperatures, and the resistivity usually approaches some finite value near absolute zero (see
the magnified inset in Fig. 21.9). This residual resistivity near absolute zero is due
primarily to collisions of electrons with impurities and to imperfections in the metal.
In contrast, the high temperature resistivity (the linear region) is dominated by collisions of electrons with the vibrating metal atoms. We shall describe this process in
more detail in Section 21.4.
There is a class of metals and compounds whose resistance decreases to zero
when they are below a certain temperature Tc , known as the critical temperature.
These materials are known as superconductors. The resistance–temperature graph
for a superconductor follows that of a normal metal at temperatures above Tc
(Fig. 21.10). When the temperature is at or below Tc , the resistivity drops suddenly
r
r
r
r
Figure 21.9 Resistivity versus
temperature for a normal metal, such
as copper. The curve is linear over a
wide range of temperatures, and increases with increasing temperature.
Figure 21.10 Resistance versus
temperature for a sample of mercury
(Hg). The graph follows that of
a normal metal above the critical
temperature Tc .
707
to zero. This phenomenon was discovered in 1911 by Dutch physicist Heike
TABLE 21.3 | Critical Temperatures
Kamerlingh-Onnes (1853–1926) as he worked with mercury, which is a sufor Various Superconductors
perconductor below 4.15 K. Measurements have shown that the resistivities
Material
Tc (K)
of superconductors below their Tc values are less than 4 3 10225 V ? m, or apHgBa
Ca
Cu
O
134
2
2
3 8
proximately 1017 times smaller than the resistivity of copper. In practice, these
Tl—Ba—Ca—Cu—O
125
resistivities are considered to be zero.
Bi—Sr—Ca—Cu—O
105
Today, thousands of superconductors are known, and as Table 21.3 illus92
YBa2Cu3O7
trates, the critical temperatures of recently discovered superconductors are
substantially higher than initially thought possible. Two kinds of supercon23.2
Nb3Ge
ductors are recognized. The more recently identified ones are essentially
18.05
Nb3Sn
ceramics with high critical temperatures, whereas superconducting mateNb
9.46
rials such as those observed by Kamerlingh-Onnes are metals. If a roomPb
7.18
temperature superconductor is ever identified, its effect on technology
Hg
4.15
could be tremendous.
Sn
3.72
The value of Tc is sensitive to chemical composition, pressure, and molecular structure. Copper, silver, and gold, which are excellent conductors at room
Al
1.19
temperature, do not exhibit superconductivity.
Zn
0.88
One truly remarkable feature of superconductors is that once a current is set
up in them, it persists without any applied potential difference (because R 5 0). Steady
currents have been observed to persist in superconducting loops for several years
with no apparent decay!
An important and useful application of superconductivity is in the development
of superconducting magnets, in which the magnitudes of the magnetic field are
approximately ten times greater than those produced by the best normal electromagnets. Such superconducting magnets are being considered as a means of
storing energy. Superconducting magnets are currently used in medical magnetic
resonance imaging, or MRI, units, which produce high-quality images of internal
organs without the need for excessive exposure of patients to x-rays or other harmful radiation.
21.4 | A Model for Electrical Conduction
In this section, we describe a classical model of electrical conduction in metals that
was first proposed by Paul Drude (1863–1906) in 1900. This structural model leads
to Ohm’s law and shows that resistivity can be related to the motion of electrons in
metals. Although the Drude model described here has limitations, it introduces concepts that are applied in more elaborate treatments.
Using the components of structural models introduced in Section 11.2, we can
describe the Drude model as follows.
1. A description of the physical components of the system: Consider a conductor as a
regular array of ionized atoms plus a collection of free electrons, which are
sometimes called conduction electrons. The conduction electrons, although
bound to their respective atoms when the atoms are not part of a solid, become
free when the atoms condense into a solid.
2. A description of where the components are located relative to one another and how
they interact: The conduction electrons fill the interior of the conductor.
In the absence of an electric field, they move in random directions through
the conductor. The situation is similar to the motion of gas molecules confined in a vessel. In fact, some scientists refer to conduction electrons in a
metal as an electron gas. The conduction electrons experience no interaction
with the array of ionized atoms except during a collision with one of those
atoms.
3. A description of the time evolution of the system: When an electric field is applied
to the conductor, the conduction electrons drift slowly in a direction opposite
that of the electric field (Active Fig. 21.4), with an average drift speed vd that is
much smaller (typically 1024 m/s) than their average speed between collisions
A small permanent magnet levitated
above a disk of the superconductor
YBa2Cu3O7, which is in liquid
nitrogen at 77 K.
Courtesy of IBM Research Laboratory
21.4 | A Model for Electrical Conduction
708
CHAPTER 21 | Current and Direct Current Circuits
(typically 106 m/s). An electron’s motion after a collision is independent of
its motion before the collision. The kinetic energy acquired by the electrons
in the electric field is transferred to the ionized atoms of the conductor when
the electrons and atoms collide. The energy transferred to the atoms increases
their vibrational energy, which causes the temperature of the conductor to
increase.
4. A description of the agreement between predictions of the model and actual observations
and, possibly, predictions of new effects that have not yet been observed: The test of
Drude’s model will be this: Can we generate an expression for the resistivity of
the conductor that agrees with experimental observations?
We begin to answer the question in (4) above by deriving an expression for
the drift velocity. When a free electron of mass me and charge q (5 2e) is sub:
:
:
jected to an electric field E , it experiences a force F 5 qE . The electron is a particle
under a net force, and its acceleration can be found from Newton’s second law,
:
a:
o F 5 m:
:
a 5
:
:
m
me
o F 5 qE
21.13b
Because the electric field is uniform, the electron’s acceleration is constant, so the
electron can be modeled as a particle under constant acceleration. If :
v i is the electron’s initial velocity the instant after a collision (which occurs at a time defined as
t 5 0), the velocity of the electron at a very short time t later (immediately before the
next collision occurs) is, from Equation 3.8,
:
:
vi 1 :
at 5 :
vi 1
vf 5 :
qE
t
me
21.14b
Let’s now take the average value of :
v f for all the electrons in the wire over all possible collision times t and all possible values of :
v i . Assuming the initial velocities are
randomly distributed over all possible directions, the average value of :
v i is zero.
:
The average value of the second term of Equation 21.14 is (q E/me), where is
the average time interval between successive collisions. Because the average value of :
v f is
equal to the drift velocity,
:
c Drift velocity in terms of
microscopic quantities
:
v f,avg 5 :
vd 5
qE
me
21.15b
Substituting the magnitude of this drift velocity (the drift speed) into Equation 21.4,
we have
2
I 5 nevd A 5 ne
1meE 2 A 5 nem E A
e
21.16b
e
According to Equation 21.6, the current is related to the macroscopic variables of
potential difference and resistance:
I5
DV
R
Incorporating Equation 21.8, we can write this expression as
I5
DV
1 A, 2
5
DV
A
,
In the conductor, the electric field is uniform, so we use Equation 20.6, DV 5 E/, to
substitute for the magnitude of the potential difference across the conductor:
I5
E,
E
A5 A
,
21.17b
21.4 | A Model for Electrical Conduction
709
Setting the two expressions for the current, Equations 21.16 and 21.17, equal, we
solve for the resistivity:
I5
me
E
ne 2E
A 5 A : 5 2
me
ne 21.18b
c Resistivity in terms of
microscopic parameters
According to this structural model, our prediction is that resistivity does not depend on the electric field or, equivalently, on the potential difference, but depends
only on fixed parameters associated with the material and the electron. This feature
is characteristic of a conductor obeying Ohm’s law. The model shows that the resistivity can be calculated from a knowledge of the density of the electrons, their charge
and mass, and the average time interval between collisions. This time interval is
related to the average distance between collisions /avg (the mean free path) and the
average speed vavg through the expression6
5
,avg
21.19b
vavg
EXAMPLE 21.3 | Electron Collisions in Copper
(A) Using the data from Example 21.1 and the structural model of electron conduction, estimate the average time interval between collisions for electrons in copper at 208C.
SOLUTION
Conceptualize Imagine the conduction electrons moving in the conductor and making collisions with the array of ionized
atoms. Because the speed of the electrons is high, we expect many collisions to occur per unit time interval, so the time
interval between collisions should be short.
Categorize We will be using the results of our structural model, so we categorize this problem as a substitution problem.
Solve Equation 21.18 for the average time interval
between collisions:
In Equation (1), is the resistivity of the conductor.
From Example 21.1, write the expression for the
electron density in a conductor:
(1) 5
(2) n 5
In Equation (2), is the density of the conductor
and M is the molecular mass of the conductor. Substitute numerical values in Equation (2):
n5
Substitute this result and other numerical values
into Equation (1):
5
me
ne 2
NA M
(6.022 3 1023 mol 21)(8 920 kg/m3)
0.063 5 kg/mol
5 8.46 3 1028 m 23
9.109 3 10 231 kg
(8.46 3 1028 m 23)(1.602 3 10 219 C)2(1.7 3 10 28 V ? m)
5 2.5 3 10214 s
Note that this result is a very short time interval so that the electrons make a very large number of collisions per second.
(B) Assuming that the average speed for free electrons in copper is 1.6 3 106 m/s and using the result from part (A),
calculate the mean free path for electrons in copper.
SOLUTION
Solve Equation 21.19 for the mean free path and
substitute numerical values:
,avg 5 vavg 5 (1.6 3 106 m/s)(2.5 3 10 214 s) 5 4.0 3 10 28 m
This result is equivalent to 40 nm (compared with atomic spacings of about 0.2 nm). Therefore, although the time interval between collisions is very short, the electrons travel about 200 atomic distances before colliding with an atom.
6Recall that the average speed of a group of particles depends on the temperature of the group (Chapter 16) and is not
the same as the drift speed vd .
710
CHAPTER 21 | Current and Direct Current Circuits
Although this structural model of conduction is consistent with Ohm’s law, it does
not correctly predict the values of resistivity or the behavior of the resistivity with
temperature. For example, the results of classical calculations for vavg using the ideal
gas model for the electrons are about a factor of ten smaller than the actual values,
which results in incorrect predictions of values of resistivity from Equation 21.18.
Furthermore, according to Equations 21.18 and 21.19, the resistivity is predicted to
vary with temperature as does vavg, which according to an ideal-gas model (Chapter
16, Eq. 16.22) is proportional to uT . This behavior is in disagreement with the linear
dependence of resistivity with temperature for pure metals (Fig. 21.9). Because of
these incorrect predictions, we must modify our structural model. We shall call the
model that we have developed so far the classical model for electrical conduction.
To account for the incorrect predictions of the classical model, we will develop it
further into a quantum mechanical model, which we shall describe briefly.
We discussed two important simplification models in earlier chapters, the particle
model and the wave model. Although we discussed these two simplification models
separately, quantum physics tells us that this separation is not so clear-cut. As we shall
discuss in detail in Chapter 28, particles have wave-like properties. The predictions
of some models can only be matched to experimental results if the model includes
the wave-like behavior of particles. The structural model for electrical conduction in
metals is one of these cases.
Let us imagine that the electrons moving through the metal have wave-like properties. If the array of atoms in a conductor is regularly spaced (that is, periodic), the wavelike character of the electrons makes it possible for them to move freely through the
conductor and a collision with an atom is unlikely. For an idealized conductor, no collisions would occur, the mean free path would be infinite, and the resistivity would be
zero. Electrons are scattered only if the atomic arrangement is irregular (not periodic),
as a result of structural defects or impurities, for example. At low temperatures, the resistivity of metals is dominated by scattering caused by collisions between the electrons
and impurities. At high temperatures, the resistivity is dominated by scattering caused
by collisions between the electrons and the atoms of the conductor, which are continuously displaced as a result of thermal agitation, destroying the perfect periodicity. The
thermal motion of the atoms makes the structure irregular (compared with an atomic
array at rest), thereby reducing the electron’s mean free path.
Although it is beyond the scope of this text to show this modification in detail, the
classical model modified with the wave-like character of the electrons results in predictions of resistivity values that are in agreement with measured values and predicts
a linear temperature dependence. When discussing the hydrogen atom in Chapter 11, we had to introduce some quantum notions to understand experimental observations such as atomic spectra. Likewise, we had to introduce quantum notions in
Chapter 17 to understand the temperature behavior of molar specific heats of gases.
Here we have another case in which quantum physics is necessary for the model to
agree with experiment. Although classical physics can explain a tremendous range
of phenomena, we continue to see hints that quantum physics must be incorporated
into our models. We shall study quantum physics in detail in Chapters 28 through 31.
21.5 | Energy and Power in Electric Circuits
Active Figure 21.11 A circuit
consisting of a resistor of resistance
R and a battery having a potential
difference DV across its terminals.
In Section 21.1, we discussed the energy transformations occurring when current
exists in a conductor. If a battery is used to establish an electric current in a conductor, there is a continuous transformation of chemical energy in the battery to kinetic
energy of the electrons to internal energy in the conductor, resulting in an increase
in the temperature of the conductor.
In typical electric circuits, energy is transferred from a source, such as a battery,
to some device, such as a lightbulb or a radio receiver by electrical transmission
(TET in Eq. 7.2). Let us determine an expression that will allow us to calculate the
rate of this energy transfer. First, consider the simple circuit in Active Figure 21.11,
21.5 | Energy and Power in Electric Circuits
where we imagine that energy is being delivered to a resistor. Because the connecting wires also have resistance, some energy is delivered to the wires and some energy
to the resistor. Unless noted otherwise, we will adopt a simplification model in which
the resistance of the wires is so small compared with the resistance of the circuit element that we ignore the energy delivered to the wires.
Let us now analyze the energetics of the circuit in which a battery is connected
to a resistor of resistance R as in Active Figure 21.11. Imagine following a positive
quantity of charge Q around the circuit from point a through the battery and resistor
and back to a. Point a is a reference point at which the potential is defined as zero.
We identify the entire circuit as our system. As the charge moves from a to b through
the battery whose potential difference is DV, the electrical potential energy of the
system increases by the amount Q DV, whereas the chemical energy in the battery
decreases by the same amount. (Recall from Chapter 20 that DU 5 q DV.) As the
charge moves from c to d through the resistor, however, the system loses this electrical potential energy during collisions with atoms in the resistor. In this process,
the energy is transformed to internal energy corresponding to increased vibrational
motion of the atoms in the resistor. Because we have neglected the resistance of the
interconnecting wires, no energy transformation occurs for paths bc and da. When
the charge returns to point a, the net result is that some of the chemical energy in
the battery has been delivered to the resistor and resides in the resistor as internal
energy associated with molecular vibration.
The resistor is normally in contact with air, so its increased temperature results in
a transfer of energy by heat into the air. In addition, there will be thermal radiation
from the resistor, representing another means of escape for the energy. After some
time interval has passed, the resistor remains at a constant temperature because the
input of energy from the battery is balanced by the output of energy by heat and
radiation. Some electrical devices include heat sinks7 connected to parts of the circuit
to prevent these parts from reaching dangerously high temperatures. Heat sinks are
pieces of metal with many fins. The high thermal conductivity of the metal provides
a rapid transfer of energy by heat away from the hot component and the large number of fins provides a large surface area in contact with the air, so energy can transfer
by radiation and into the air by heat at a high rate.
Let us consider now the rate at which the system loses electric potential energy as
the charge Q passes through the resistor:
dQ
dU
d
5 (Q DV ) 5
DV 5 I DV
dt
dt
dt
711
Pitfall Prevention | 21.5
Misconceptions about Current
Several common misconceptions are
associated with current in a circuit
like that in Active Figure 21.11. One
is that current comes out of one
terminal of the battery and is then
“used up” as it passes through the
resistor. According to this approach,
there is current in only one part
of the circuit. The current is actually
the same everywhere in the circuit.
A related misconception has the
current coming out of the resistor
being smaller than that going in
because some of the current is “used
up.” Yet another misconception has
current coming out of both terminals
of the battery, in opposite directions,
and then “clashing” in the resistor,
delivering the energy in this manner.
We know that is not the case; charges
flow in the same rotational sense at
all points in the circuit.
Pitfall Prevention | 21.6
Charges Do Not Move All the Way
Around a Circuit
Because of the very small magnitude
of the drift velocity, it might take
hours for a single electron to make
one complete trip around the circuit.
In terms of understanding the energy
transfer in a circuit, however, it is
useful to imagine a charge moving all
the way around the circuit.
where I is the current in the circuit. Of course, the system regains this potential energy when the charge passes through the battery, at the expense of chemical energy
in the battery. The rate at which the system loses potential energy as the charge
passes through the resistor is equal to the rate at which the system gains internal
energy in the resistor. Therefore, the power P, representing the rate at which energy
is delivered to the resistor, is
P 5 I DV
21.20b
We have developed this result by considering a battery delivering energy to a resistor. Equation 21.20, however, can be used to determine the power transferred from
a voltage source to any device carrying a current I and having a potential difference
DV between its terminals.
Using Equation 21.20 and that DV 5 IR for a resistor, we can express the power
delivered to the resistor in the alternative forms
P 5 I 2R 5
(DV )2
R
7This terminology is another misuse of the word heat that is ingrained in our common language.
21.21b
c Power delivered to a device
Pitfall Prevention | 21.7
Energy Is Not “Dissipated”
In some books, you may see
Equation 21.21 described as the
power “dissipated in” a resistor,
suggesting that energy disappears.
Instead, we say energy is “delivered
to” a resistor. The notion of dissipation
arises because a warm resistor expels
energy by radiation and heat, and
energy delivered by the battery leaves
the circuit. (It does not disappear!)
712
CHAPTER 21 | Current and Direct Current Circuits
The SI unit of power is the watt, introduced in Chapter 7. If you analyze the units in
Equations 21.20 and 21.21, you will see that the result of the calculation provides a
watt as the unit. The power delivered to a conductor of resistance R is often referred
to as an I 2R loss.
As we learned in Section 7.6, the unit of energy your electric company uses to calculate energy transfer, the kilowatt-hour, is the amount of energy transferred in 1 h
at the constant rate of 1 kW. We learned there that 1 kWh 5 3.6 3 106 J.
QUI C K QU IZ 21.4 For the two incandescent lightbulbs shown in Figure 21.12, rank
the currents at points a through f, from greatest to least.
THINKING PHYSICS 21.2
Two incandescent lightbulbs A and B are connected across the same potential
difference as in Figure 21.12. The electric input powers to the lightbulbs are
shown. Which lightbulb has the higher resistance? Which carries the greater
current?
Reasoning Because the voltage across each lightbulb is the same and the rate
Figure 21.12 (Quick Quiz 21.4
and Thinking Physics 21.2) Two
incandescent lightbulbs connected
across the same potential difference.
of energy delivered to a resistor is P 5 (DV )2/R, the lightbulb with the higher
resistance exhibits the lower rate of energy transfer. In this case, the resistance
of A is larger than that for B. Furthermore, because P 5 I DV, we see that the
current carried by B is larger than that of A. b
THINKING PHYSICS 21.3
When is an incandescent lightbulb more likely to fail, just
after it is turned on or after it has been on for a while?
Reasoning When the switch is closed, the source voltage is immediately applied across the lightbulb. As the
voltage is applied across the cold filament when the lightbulb is first turned on, the resistance of the filament is
low. Therefore, the current is high and a relatively large
amount of energy is delivered to the bulb per unit time
interval. This causes the temperature of the filament to
rise rapidly, resulting in thermal stress on the filament
that makes it likely to fail at that moment. As the filament
warms up in the absence of failure, its resistance rises and
the current falls. As a result, the rate of energy delivered
to the lightbulb falls. The thermal stress on the filament is
reduced so that the failure is less likely to occur after the
bulb has been on for a while. b
Example 21.4 | Linking Electricity and Thermodynamics
An immersion heater must increase the temperature of 1.50 kg of water from 10.08C to 50.08C in 10.0 min while operating at 110 V.
(A) What is the required resistance of the heater?
SOLUTION
Conceptualize An immersion heater is a resistor that is inserted into a container of water. As energy is delivered to the
immersion heater, raising its temperature, energy leaves the surface of the resistor by heat, going into the water. When the
immersion heater reaches a constant temperature, the rate of energy delivered to the resistance by electrical transmission
(TET) is equal to the rate of energy delivered by heat (Q ) to the water.
Categorize This example allows us to link our new understanding of power in electricity with our experience with specific
heat in thermodynamics (Chapter 17). The water is a nonisolated system. Its internal energy is rising because of energy
transferred into the water by heat from the resistor: DEint 5 Q. In our model, we assume the energy that enters the water
from the heater remains in the water.
713
21.6 | Sources of emf
21.4 cont.
Analyze To simplify the analysis, let’s ignore the initial period during which the temperature of the resistor increases and
also ignore any variation of resistance with temperature. Therefore, we imagine a constant rate of energy transfer for the
entire 10.0 min.
Set the rate of energy delivered to the resistor equal
to the rate of energy Q entering the water by heat:
Use Equation 17.3, Q 5 mc DT, to relate the energy
input by heat to the resulting temperature change
of the water and solve for the resistance:
Substitute the values given in the statement of the
problem:
P5
Q
(DV )2
5
R
Dt
(DV )2
(DV )2 Dt
mc DT
5
:R5
R
Dt
mc DT
R5
(110 V)2(600 s)
5 28.9 V
(1.50 kg)(4 186 J/kg ? 8C)(50.08C 2 10.08C)
(B) Estimate the cost of heating the water.
SOLUTION
Multiply the power by the time interval to find the
amount of energy transferred:
TET 5 P Dt 5
(110 V)2
(DV )2
1h
Dt 5
(10.0 min)
R
28.9 V
60.0 min
1
2
5 69.8 Wh 5 0.069 8 kWh
Find the cost knowing that energy is purchased at an
estimated price of 11. per kilowatt-hour:
Cost 5 (0.069 8 kWh)($0.11/kWh) 5 $0.008 5 0.8.
Finalize The cost to heat the water is very low, less than one cent. In reality, the cost is higher because some energy is transferred from the water into the surroundings by heat and electromagnetic radiation while its temperature is increasing. If
you have electrical devices in your home with power ratings on them, use this power rating and an approximate time interval of use to estimate the cost for one use of the device.
21.6 | Sources of emf
The entity that maintains the constant voltage in Figure 21.13 is called a source of
emf.8 Sources of emf are any devices (such as batteries and generators) that increase
the potential energy of a circuit system by maintaining a potential difference between points in the circuit while charges move through the circuit. One can think of
a source of emf as a “charge pump.” The emf of a source describes the work done
per unit charge, and hence the SI unit of emf is the volt.
At this point, you may wonder why we need to define a second quantity, emf, with
the volt as a unit when we have already defined the potential difference. To see the
need for this new quantity, consider the circuit shown in Figure 21.13, consisting of
a battery connected to a resistor. We shall assume that the connecting wires have no
resistance. We might be tempted to claim that the potential difference across the
battery terminals (the terminal voltage) equals the emf of the battery. A real battery,
however, always has some internal resistance r. As a result, the terminal voltage is not
equal to the emf, as we shall show.
e
8The term emf was originally an abbreviation for electromotive force, but it is not a force, so the long form is discouraged.
The name electromotive force was used early in the study of electricity before the understanding of batteries was as
sophisticated as it is today.
Figure 21.13 A circuit consisting of
a resistor connected to the terminals
of a battery.
714
CHAPTER 21 | Current and Direct Current Circuits
e
The circuit shown in Figure 21.13 can be described by the circuit diagram in
Active Figure 21.14a. The battery within the dashed rectangle is modeled as an
ideal, zero-resistance source of emf in series with the internal resistance r. Now
imagine moving from a to d in Active Figure 21.14a. As you pass from the negative to the positive terminal within the source of emf the potential increases by .
As you move through the resistance r, however, the potential decreases by an amount
Ir, where I is the current in the circuit. Therefore, the terminal voltage DV 5 Vd 2 Va
of the battery is9
e
e
DV 5
e
e 2 Ir
21.22b
e
Note from this expression that is equivalent to the open-circuit voltage, that is,
the terminal voltage when the current is zero. Active Figure 21.14b is a graphical
representation of the changes in potential as the circuit is traversed clockwise. By
inspecting Active Figure 21.14a, we see that the terminal voltage DV must also equal
the potential difference across the external resistance R, often called the load resistance; that is, DV 5 IR. Combining this expression with Equation 21.22, we see that
e
e5 IR 1 Ir
21.23b
Solving for the current gives
Active Figure 21.14 (a) Circuit
I5
e
diagram of a source of emf (in
this case, a battery) with internal resistance r, connected to an external
resistor of resistance R. (b) Graphical
representation showing how the potential changes as the circuit in (a) is
traversed clockwise.
What Is Constant in a Battery?
It is a common misconception that
a battery is a source of constant
current. Equation 21.24 shows that
this is not true. The current in the
circuit depends on the resistance R
connected to the battery. It is also
not true that a battery is a source of
constant terminal voltage, as shown
by Equation 21.22. A battery is a
source of constant emf.
21.24b
which shows that the current in this simple circuit depends on both the resistance
R external to the battery and the internal resistance r. If R is much greater than r,
we can adopt a simplification model in which we neglect r in our analysis. In many
circuits, we shall adopt this simplification model.
If we multiply Equation 21.23 by the current I, we have
I
Pitfall Prevention | 21.8
e
R1r
e 5 I 2R 1 I 2r
e
This equation tells us that the total power output I of the source of emf is equal to
the rate I 2R at which energy is delivered to the load resistance plus the rate I 2r at which
energy is delivered to the internal resistance. If r ,, R, much more of the energy from
the battery is delivered to the load resistance than stays in the battery, although the
amount of energy is relatively small because the load resistance is large, resulting in
a small current. If r .. R, a significant fraction of the energy from the source of emf
stays in the battery package because it is delivered to the internal resistance. For example, if a wire is simply connected between the terminals of a flashlight battery, the battery becomes warm. This warming represents the transfer of energy from the source
of emf to the internal resistance, where it appears as internal energy associated with
temperature. Problem 73 explores the conditions under which the largest amount of
energy is transferred from the battery to the load resistor.
Example 21.5 | Terminal Voltage of a Battery
A battery has an emf of 12.0 V and an internal resistance of 0.050 0 V. Its terminals are connected to a load resistance of
3.00 V.
(A) Find the current in the circuit and the terminal voltage of the battery.
SOLUTION
Conceptualize Study Active Figure 21.14a, which shows a circuit consistent with the problem statement. The battery delivers energy to the load resistor.
Categorize This example involves simple calculations from this section, so we categorize it as a substitution problem.
9The terminal voltage in this case is less than the emf by the amount I r. In some situations, the terminal voltage may exceed
the emf by the amount Ir. Such a situation occurs when the direction of the current is opposite that of the emf, as when a
battery is being charged by another source of emf.
21.7 | Resistors in Series and Parallel
715
21.5 cont.
e
12.0 V
5 3.93 A
3.00 V 1 0.050 0 V
Use Equation 21.24 to find the current in the circuit:
I5
Use Equation 21.22 to find the terminal voltage:
DV 5 e 2 Ir 5 12.0 V 2 (3.93 A)(0.050 0 V) 5 11.8 V
To check this result, calculate the voltage across the load
resistance R:
R1r
5
DV 5 IR 5 (3.93 A)(3.00 V) 5 11.8 V
(B) Calculate the power delivered to the load resistor, the power delivered to the internal resistance of the battery, and the
power delivered by the battery.
SOLUTION
Use Equation 21.21 to find the power delivered to the
load resistor:
PR 5 I 2R 5 (3.93 A)2(3.00 V) 5 46.3 W
Find the power delivered to the internal resistance:
Pr 5 I 2r 5 (3.93 A)2(0.050 0 V) 5 0.772 W
Find the power delivered by the battery by adding these
quantities:
P 5 PR 1 Pr 5 46.3 W 1 0.772 W 5 47.1 W
What If? As a battery ages, its internal resistance increases. Suppose the internal resistance of this battery rises to 2.00 V
toward the end of its useful life. How does that alter the battery’s ability to deliver energy?
Answer Let’s connect the same 3.00-V load resistor to the battery.
e
12.0 V
5 2.40 A
3.00 V 1 2.00 V
Find the new current in the battery:
I5
Find the new terminal voltage:
DV 5 e 2 Ir 5 12.0 V 2 (2.40 A)(2.00 V) 5 7.2 V
Find the new powers delivered to the load resistor and
internal resistance:
PR 5 I 2R 5 (2.40 A)2(3.00 V) 5 17.3 W
R1r
5
Pr 5 I 2r 5 (2.40 A)2(2.00 V) 5 11.5 W
The terminal voltage is only 60% of the emf. Notice that 40% of the power from the battery is delivered to the internal
resistance when r is 2.00 V. When r is 0.050 0 V as in part (B), this percentage is only 1.6%. Consequently, even though
the emf remains fixed, the increasing internal resistance of the battery significantly reduces the battery’s ability to deliver energy to an external load.
21.7 | Resistors in Series and Parallel
When two or more resistors are connected together as are the incandescent lightbulbs in Active Figure 21.15a (page 716), they are said to be in a series combination.
Active Figure 21.15b is the circuit diagram for the lightbulbs, shown as resistors, and
the battery. In a series connection, if an amount of charge Q exits resistor R 1, charge
Q must also enter the second resistor R 2. Otherwise, charge would accumulate on
the wire between the resistors. Therefore, the same amount of charge passes through
both resistors in a given time interval and the currents are the same in both resistors:
I 5 I1 5 I2
where I is the current leaving the battery, I 1 is the current in resistor R 1, and I 2 is the
current in resistor R 2.
716
CHAPTER 21 | Current and Direct Current Circuits
Active Figure 21.15 Two incandescent lightbulbs with resistances R1 and R2 connected in series. All three diagrams are
equivalent.
The potential difference applied across the series combination of resistors divides
between the resistors. In Active Figure 21.15b, because the voltage drop10 from a to
b equals I1R1 and the voltage drop from b to c equals I2R2, the voltage drop from a
to c is
DV 5 DV 1 1 DV 2 5 I 1R 1 1 I 2R 2
The potential difference across the battery is also applied to the equivalent resistance Req in Active Figure 21.15c:
DV 5 IReq
where the equivalent resistance has the same effect on the circuit as the series combination because it results in the same current I in the battery. Combining these
equations for DV gives
DV 5 IReq 5 I 1R 1 1 I 2R 2
:
R eq 5 R 1 1 R 2
21.25b
where we have canceled the currents I, I 1, and I 2 because they are all the same. We
see that we can replace the two resistors in series with a single equivalent resistance
whose value is the sum of the individual resistances.
The equivalent resistance of three or more resistors connected in series is
c The equivalent resistance of a
series combination of resistors
Pitfall Prevention | 21.9
Lightbulbs Don’t Burn
We will describe the end of the life
of an incandescent lightbulb by
saying the filament fails rather than
by saying the lightbulb “burns out.”
The word burn suggests a combustion
process, which is not what occurs in a
lightbulb. The failure of a lightbulb
results from the slow sublimation of
tungsten from the very hot filament
over the life of the lightbulb. The
filament eventually becomes very
thin because of this process. The
mechanical stress from a sudden
temperature increase when the
lightbulb is turned on causes the thin
filament to break.
R eq 5 R 1 1 R 2 1 R 3 1 ? ? ?
21.26b
This relationship indicates that the equivalent resistance of a series combination of
resistors is the numerical sum of the individual resistances and is always greater than
any individual resistance.
Looking back at Equation 21.24, we see that the denominator of the right-hand
side is the simple algebraic sum of the internal and external resistances. That is
consistent with the internal and external resistances being in series in Active
Figure 21.14a.
If the filament of one lightbulb in Active Figure 21.15a were to fail, the circuit would no longer be complete (resulting in an open-circuit condition) and
the second lightbulb would also go out. This fact is a general feature of a series circuit: if one device in the series creates an open circuit, all devices are
inoperative.
10The term voltage drop is synonymous with a decrease in electric potential across a resistor. It is often used by individuals
working with electric circuits.
21.7 | Resistors in Series and Parallel
717
Q U I CK QUI Z 21.5 With the switch in the circuit of Figure 21.16a closed, there is no
current in R2 because the current has an alternate zero-resistance path through the
switch. There is current in R1, and this current is measured with the ammeter
(a device for measuring current) at the bottom of the circuit. If the switch is opened
(Figure 21.16b), there is current in R2. What happens to the reading on the ammeter when the switch is opened? (a) The reading goes up. (b) The reading goes down.
(c) The reading does not change.
Now consider two resistors in a parallel combination as shown in Active Figure 21.17. Notice that both resistors are connected directly across the terminals
of the battery. Therefore, the potential differences across the resistors are the
same:
DV 5 DV1 5 DV2
where DV is the terminal voltage of the battery.
When charges reach point a in Active Figure 21.17b, they split into two parts, with
some going toward R1 and the rest going toward R2. A junction is any such point in a
circuit where a current can split. This split results in less current in each individual
resistor than the current leaving the battery. Because electric charge is conserved,
the current I that enters point a must equal the total current leaving that point:
Figure 21.16 (Quick Quiz 21.5)
DV2
DV1
1
I 5 I1 1 I2 5
R1
R2
What happens when the switch is
opened?
where I1 is the current in R1 and I2 is the current in R2.
The current in the equivalent resistance Req in Active Figure 21.17c is
Pitfall Prevention | 21.10
DV
I5
Req
Local and Global Changes
where the equivalent resistance has the same effect on the circuit as the two resistors
in parallel; that is, the equivalent resistance draws the same current I from the battery. Combining these equations for I, we see that the equivalent resistance of two
resistors in parallel is given by
I5
DV2
DV1
DV
1
1
1
5
1
:
5
1
R eq
R1
R2
R eq
R1
R2
21.27b
A local change in one part of a
circuit may result in a global change
throughout the circuit. For example,
if a single resistor is changed in a
circuit containing several resistors
and batteries, the currents in all
resistors and batteries, the terminal
voltages of all batteries, and the
voltages across all resistors may
change as a result.
where we have canceled DV, DV1, and DV2 because they are all the same.
Pitfall Prevention | 21.11
Current Does Not Take the Path
of Least Resistance
You may have heard the phrase
“current takes the path of least
resistance” (or similar wording) in
reference to a parallel combination
of current paths such that there are
two or more paths for the current
to take. Such wording is incorrect.
The current takes all paths. Those
paths with lower resistance have
larger currents, but even very high
resistance paths carry some of the
current. In theory, if current has a
choice between a zero-resistance
path and a finite resistance path, all
the current takes the path of zero
resistance; a path with zero resistance,
however, is an idealization.
Active Figure 21.17 Two incandescent lightbulbs with resistances R 1 and R 2 connected in parallel.
All three diagrams are equivalent.
718
CHAPTER 21 | Current and Direct Current Circuits
An extension of this analysis to three or more resistors in parallel gives
c The equivalent resistance
of a parallel combination
of resistors
1
1
1
1
5
1
1
1 ???
Req
R1
R2
R3
21.28b
This expression shows that the inverse of the equivalent resistance of two or more resistors in a parallel combination is equal to the sum of the inverses of the individual
resistances. Furthermore, the equivalent resistance is always less than the smallest
resistance in the group.
A circuit consisting of resistors can often be reduced to a simple circuit containing only one resistor. To do so, examine the initial circuit and replace any resistors in series or any in parallel with equivalent resistances using Equations 21.26
and 21.28. Draw a sketch of the new circuit after these changes have been made.
Examine the new circuit and replace any new series or parallel combinations
that now exist. Continue this process until a single equivalent resistance is found
for the entire circuit. (That may not be possible; if not, see the techniques of
Section 21.8.)
If the current in or the potential difference across a resistor in the initial circuit
is to be found, start with the final circuit and gradually work your way back through
the equivalent circuits. Find currents and voltages across resistors using DV 5 IR and
your understanding of series and parallel combinations.
Household circuits are always wired so that the electrical devices are connected in parallel as in Active Figure 21.17a. In this manner, each device operates independently of the others so that if one is switched off, the others remain
on. For example, if one of the lightbulbs in Active Figure 21.17a were removed
from its socket, the other would continue to operate. Equally important, each
device operates on the same voltage. If devices were connected in series, the
voltage applied to the combination would divide among the devices, so the voltage applied to any one device would depend on how many devices were in the
combination.
In many household circuits, circuit breakers are used in series with other circuit elements for safety purposes. A circuit breaker is designed to switch off and
open the circuit at some maximum current (typically 15 A or 20 A) whose value
depends on the nature of the circuit. If a circuit breaker were not used, large currents caused by turning on many devices could result in excessive temperatures in
wires and, perhaps, cause a fire. In older home construction, fuses were used in
place of circuit breakers. When the current in a circuit exceeds some value, the
conductor in a fuse melts and opens the circuit. The disadvantage of fuses is that
they are destroyed in the process of opening the circuit, whereas circuit breakers
can be reset.
QUI C K QU IZ 21.6 With the switch in the circuit of Figure 21.18a open, there is no
current in R2. There is current in R1, however, and it is measured with the ammeter at the right side of the circuit. If the switch is closed (Fig. 21.18b), there is current in R2. What happens to the reading on the ammeter when the switch is closed?
(a) The reading increases. (b) The reading decreases. (c) The reading does not
change.
Figure 21.18 (Quick Quiz 21.6)
What happens when the switch is
closed?
QUI C K QU IZ 21.7 Consider the following choices: (a) increases, (b) decreases,
(c) remains the same. From these choices, choose the best answer for the following
situations. (i) In Active Figure 21.15, a third resistor is added in series with the first
two. What happens to the current in the battery? (ii) What happens to the terminal
voltage of the battery? (iii) In Active Figure 21.17, a third resistor is added in parallel
with the first two. What happens to the current in the battery? (iv) What happens to
the terminal voltage of the battery?
21.7 | Resistors in Series and Parallel
719
THINKING PHYSICS 21.4
Compare the brightnesses of the four identical lightbulbs in Figure 21.19. What happens if bulb A fails so that it cannot conduct? What if bulb C fails? What if bulb D fails?
Reasoning Bulbs A and B are connected in series across the battery, whereas
bulb C is connected by itself. Therefore, the terminal voltage of the battery is
split between bulbs A and B. As a result, bulb C will be brighter than bulbs A and
B, which should be equally as bright as each other. Bulb D has a wire connected
across it. Therefore, there is no potential difference across bulb D and it does
not glow at all. If bulb A fails, bulb B goes out but bulb C stays lit. If bulb C fails,
there is no effect on the other bulbs. If bulb D fails, the event is undetectable
because bulb D was not glowing initially. b
Figure 21.19 (Thinking Physics
21.4) What happens to the lightbulbs
if one fails?
THINKING PHYSICS 21.5
Figure 21.20 illustrates how a three-way incandescent lightbulb is constructed to
provide three levels of light intensity. The socket of the lamp is equipped with a
three-way switch for selecting different light intensities. The bulb contains two
filaments. Why are the filaments connected in parallel? Explain how the two filaments are used to provide three different light intensities.
Reasoning If the filaments were connected in series and one of them were to
fail, there would be no current in the bulb and the bulb would give no illumination, regardless of the switch position. When the filaments are connected in
parallel, however, and one of them (say the 75-W filament) fails, the bulb still
operates in some switch positions because there is current in the other (100-W)
filament. The three light intensities are made possible by selecting one of three
values of filament resistance, using a single value of 120 V for the applied voltage. The 75-W filament offers one value of resistance, the 100-W filament offers
a second value, and the third resistance is obtained by combining the two filaments in parallel. When switch S1 is closed and switch S2 is opened, only the 75-W
filament carries current. When switch S1 is open and switch S2 is closed, only the
100-W filament carries current. When both switches are closed, both filaments
carry current, and a total illumination corresponding to 175 W is obtained. b
Figure 21.20 (Thinking Physics
21.5) A three-way incandescent
lightbulb.
Exampl e 21.6 | Find the Equivalent Resistance
Four resistors are connected as shown in Figure 21.21a.
(A) Find the equivalent resistance between points a and c.
SOLUTION
Conceptualize Imagine charges flowing into this combination from the left. All
charges must pass through the first two resistors, but the charges split into two different
paths when encountering the combination of the 6.0-V and the 3.0-V resistors.
Categorize Because of the simple nature of the combination of resistors in Figure
21.21, we categorize this example as one for which we can use the rules for series and
parallel combinations of resistors.
Analyze The combination of resistors can be reduced in steps as shown in Figure 21.21.
Find the equivalent resistance between a and b of
the 8.0-V and 4.0-V resistors, which are in series
(left-hand red-brown circles):
Req 5 8.0 V 1 4.0 V 5 12.0 V
Figure 21.21 (Example 21.6) The
original network of resistors is reduced
to a single equivalent resistance.
continued
720
CHAPTER 21 | Current and Direct Current Circuits
21.6 cont.
Find the equivalent resistance between b and c of the
6.0-V and 3.0-V resistors, which are in parallel (righthand red-brown circles):
The circuit of equivalent resistances now looks like Figure 21.21b. The 12.0-V and 2.0-V resistors are in series
(green circles). Find the equivalent resistance from a to c:
3
1
1
1
1
5
5
R eq
6.0 V
3.0 V
6.0 V
6.0 V
5 2.0 V
3
R eq 5
R eq 5 12.0 V 1 2.0 V 5 14.0 V
This resistance is that of the single equivalent resistor in Figure 21.21c.
(B) What is the current in each resistor if a potential difference of 42 V is maintained between a and c?
SOLUTION
The currents in the 8.0-V and 4.0-V resistors are the same because they are in series. In addition, they carry the same current that would exist in the 14.0-V equivalent resistor subject to the 42-V potential difference.
DVac
42 V
5 3.0 A
14.0 V
Use Equation 21.6 (R 5 DV/I) and the result from part
(A) to find the current in the 8.0-V and 4.0-V resistors:
I5
Set the voltages across the resistors in parallel in Figure
21.21a equal to find a relationship between the currents:
DV1 5 DV2 : (6.0 V)I1 5 (3.0 V)I2 : I2 5 2I1
Use I1 1 I2 5 3.0 A to find I1:
I1 1 I 2 5 3.0 A : I1 1 2I1 5 3.0 A : I1 5 1.0 A
Find I2:
I2 5 2I1 5 2(1.0 A) 5 2.0 A
Req
5
Finalize As a final check of our results, note that DVbc 5 (6.0 V)I1 5 (3.0 V)I2 5 6.0 V and DVab 5 (12.0 V)I 5 36 V; there-
fore, DVac 5 DVab 1 DVbc 5 42 V, as it must.
Example 21.7 | Three Resistors in Parallel
Three resistors are connected in parallel as shown in Figure 21.22a. A potential difference of 18.0 V is maintained
between points a and b.
(A) Calculate the equivalent resistance of the circuit.
SOLUTION
Conceptualize Figure 21.22a shows that we are dealing with
a simple parallel combination of three resistors. Notice that
the current I splits into three currents I1, I2, and I3 in the
three resistors.
Categorize Because the three resistors are connected in
parallel, we can use Equation 21.28 to evaluate the equivalent resistance.
Analyze Use Equation 21.28 to find Req:
Figure 21.22 (Example 21.7) (a) Three resistors connected in
parallel. The voltage across each resistor is 18.0 V. (b) Another circuit
with three resistors and a battery. Is it equivalent to the circuit in (a)?
1
1
1
1
11.0
5
1
1
5
Req
3.00 V
6.00 V
9.00 V
18.0 V
Req 5
18.0 V
5 1.64 V
11.0
21.8 | Kirchhoff ’s Rules
721
21.7 cont.
(B) Find the current in each resistor.
SOLUTION
The potential difference across each resistor is 18.0 V.
Apply the relationship DV 5 IR to find the currents:
I1 5
18.0 V
DV
5
5 6.00 A
R1
3.00 V
I2 5
18.0 V
DV
5
5 3.00 A
R2
6.00 V
I3 5
DV
18.0 V
5 2.00 A
5
R3
9.00 V
(C) Calculate the power delivered to each resistor and the total power delivered to the combination of resistors.
SOLUTION
Apply the relationship P 5 I 2R to each resistor using the
currents calculated in part (B):
3.00-V: P 1 5 I 12R 1 5 (6.00 A)2(3.00 V) 5 108 W
6.00-V: P 2 5 I 22R 2 5 (3.00 A)2(6.00 V) 5 54 W
9.00-V: P 3 5 I 32R 3 5 (2.00 A)2(9.00 V) 5 36 W
Finalize Part (C) shows that the smallest resistor receives the most power. Summing the three quantities gives a total
power of 198 W. We could have calculated this final result from part (A) by considering the equivalent resistance as follows:
P 5 (DV)2/Req 5 (18.0 V)2/1.64 V 5 198 W.
What If? What if the circuit were as shown in Figure 21.22b instead of as in Figure 21.22a? How would that affect the calculation?
Answer There would be no effect on the calculation. The physical placement of the battery is not important. Only the
electrical arrangement is important. In Figure 21.22b, the battery still maintains a potential difference of 18.0 V between
points a and b, so the two circuits in the figure are electrically identical.
21.8 | Kirchhoff ’s Rules
As we saw in the preceding section, combinations of resistors can be simplified
and analyzed using the expression DV 5 IR and the rules for series and parallel combinations of resistors. Very often, however, it is not possible to reduce a circuit to a
single loop using these rules. The procedure for analyzing more complex circuits is
made possible by using the two following principles, called Kirchhoff’s rules.
1. Junction rule. At any junction, the sum of the currents must equal zero:
o
I50
21.29b
junction
2. Loop rule. The sum of the potential differences across all elements around
any closed circuit loop must be zero:
o DV 5 0
21.30b
loop
Kirchhoff’s first rule is a statement of conservation of electric charge. All charges
that enter a given point in a circuit must leave that point because charge cannot
build up at a point. Currents directed into the junction are entered into the junction
rule as 1I, whereas currents directed out of a junction are entered as 2I. Applying
this rule to the junction in Figure 21.23a gives
I1 2 I2 2 I3 5 0
Figure 21.23b represents a mechanical analog of this situation, in which water flows
through a branched pipe having no leaks. Because water does not build up anywhere
Figure 21.23 (a) Kirchhoff’s
junction rule. (b) A mechanical
analog of the junction rule.
722
CHAPTER 21 | Current and Direct Current Circuits
e
e
e
e
Figure 21.24 Rules for determining the potential differences across
a resistor and a battery. (The battery
is assumed to have no internal
resistance.)
in the pipe, the flow rate into the pipe on the left equals the total flow rate out of the
two branches on the right.
Kirchhoff’s second rule follows from the law of conservation of energy. Let’s
imagine moving a charge around a closed loop of a circuit. When the charge returns to the starting point, the charge–circuit system must have the same total
energy as it had before the charge was moved. The sum of the increases in energy as the charge passes through some circuit elements must equal the sum of
the decreases in energy as it passes through other elements. The potential energy
decreases whenever the charge moves through a potential drop 2IR across a resistor or whenever it moves in the reverse direction through a source of emf. The
potential energy increases whenever the charge passes through a battery from the
negative terminal to the positive terminal.
When applying Kirchhoff’s second rule, imagine traveling around the loop and consider changes in electric potential rather than the changes in potential energy described in
the preceding paragraph. Imagine traveling through the circuit elements in Figure 21.24
toward the right. The following sign conventions apply when using the second rule:
• Charges move from the high-potential end of a resistor toward the low-potential
end, so if a resistor is traversed in the direction of the current, the potential difference DV across the resistor is 2IR (Fig. 21.24a).
• If a resistor is traversed in the direction opposite the current, the potential difference DV across the resistor is 1IR (Fig. 21.24b).
• If a source of emf (assumed to have zero internal resistance) is traversed in the
direction of the emf (from negative to positive), the potential difference DV is
1 (Fig. 21.24c).
• If a source of emf (assumed to have zero internal resistance) is traversed in the
direction opposite the emf (from positive to negative), the potential difference
DV is 2 (Fig. 21.24d).
e
e
Prints & Photographs Division, Library of Congress,
LC-USZ62-133715
There are limits on the number of times you can usefully apply Kirchhoff’s rules
in analyzing a circuit. You can use the junction rule as often as you need as long as
you include in it a current that has not been used in a preceding junction-rule equation. In general, the number of times you can use the junction rule is one fewer than
the number of junction points in the circuit. You can apply the loop rule as often
as needed as long as a new circuit element (resistor or battery) or a new current
appears in each new equation. In general, to solve a particular circuit problem, the
number of independent equations you need to obtain from the two rules equals the
number of unknown currents.
Gustav Kirchhoff
German Physicist (1824–1887)
Kirchhoff, a professor at Heidelberg, and
Robert Bunsen invented the spectroscope
and founded the science of spectroscopy,
which we discussed in Chapter 11. They
discovered the elements cesium and
rubidium and invented astronomical
spectroscopy.
P ROB LEM-SOLVING STRATEGY: Kirchhoff ’s Rules
The following procedure is recommended for solving problems that involve circuits
that cannot be reduced by the rules for combining resistors in series or parallel.
1. Conceptualize Study the circuit diagram and make sure you recognize all
elements in the circuit. Identify the polarity of each battery and try to imagine
the directions in which the current would exist in the batteries.
2. Categorize Determine whether the circuit can be reduced by means of combining series and parallel resistors. If so, use the techniques of Section 21.7. If
not, apply Kirchhoff’s rules according to the Analyze step below.
3. Analyze Assign labels to all known quantities and symbols to all unknown
quantities. You must assign directions to the currents in each part of the circuit.
Although the assignment of current directions is arbitrary, you must adhere
rigorously to the directions you assign when you apply Kirchhoff’s rules.
Apply the junction rule (Kirchhoff’s first rule) to all junctions in the
circuit except one. Now apply the loop rule (Kirchhoff’s second rule) to as
many loops in the circuit as are needed to obtain, in combination with the
21.8 | Kirchhoff ’s Rules
723
equations from the junction rule, as many equations as there are unknowns.
To apply this rule, you must choose a direction in which to travel around the
loop (either clockwise or counterclockwise) and correctly identify the change
in potential as you cross each element. Be careful with signs!
Solve the equations simultaneously for the unknown quantities.
4. Finalize Check your numerical answers for consistency. Do not be alarmed if any
of the resulting currents have a negative value. That only means you have guessed
the direction of that current incorrectly, but its magnitude will be correct.
Exampl e 21.8 | A Multiloop Circuit
Find the currents I1, I2, and I3 in the circuit shown in Figure 21.25.
SOLUTION
Conceptualize Imagine physically rearranging the circuit while keeping it electrically the
same. Can you rearrange it so that it consists of simple series or parallel combinations of
resistors? You should find that you cannot.
Categorize We cannot simplify the circuit by the rules associated with combining resistances in series and in parallel. (If the 10.0-V battery were removed and replaced by a wire
from b to the 6.0-V resistor, we could reduce the remaining circuit.) Because the circuit is
not a simple series and parallel combination of resistances, this problem is one in which we
must use Kirchhoff’s rules.
Analyze We arbitrarily choose the directions of the currents as labeled in Figure 21.25.
Figure 21.25 (Example 21.8)
A circuit containing different
branches.
Apply Kirchhoff’s junction rule to junction c :
(1) I1 1 I2 2 I3 5 0
We now have one equation with three unknowns: I1, I2,
and I3. There are three loops in the circuit: abcda, befcb,
and aefda. We need only two loop equations to determine
the unknown currents. (The third equation for loop aefda
would give no new information.) Let’s choose to traverse
these loops in the clockwise direction. Apply Kirchhoff’s
loop rule to loops abcda and befcb:
abcda: (2) 10.0 V 2 (6.0 V)I1 2 (2.0 V)I3 5 0
Solve Equation (1) for I3 and substitute into
Equation (2):
Multiply each term in Equation (3) by 4 and each term in
Equation (4) by 3:
befcb: 2(4.0 V)I2 2 14.0 V 1 (6.0 V)I1 2 10.0 V 5 0
(3) 224.0 V 1 (6.0 V)I1 2 (4.0 V)I2 5 0
10.0 V 2 (6.0 V)I1 2 (2.0 V)(I1 1 I2) 5 0
(4) 10.0 V 2 (8.0 V)I1 2 (2.0 V)I2 5 0
(5) 296.0 V 1 (24.0 V)I1 2 (16.0 V)I2 5 0
(6) 30.0 V 2 (24.0 V)I1 2 (6.0 V)I2 5 0
Add Equation (6) to Equation (5) to eliminate I1 and
find I2:
266.0 V 2 (22.0 V)I2 5 0
Use this value of I2 in Equation (3) to find I1:
224.0 V 1 (6.0 V)I1 2 (4.0 V)(23.0 A) 5 0
I2 5 23.0 A
224.0 V 1 (6.0 V)I1 1 12.0 V 5 0
I1 5 2.0 A
Use Equation (1) to find I3:
I3 5 I1 1 I2 5 2.0 A 2 3.0 A 5 21.0 A
Finalize Because our values for I2 and I3 are negative, the directions of these currents are opposite those indicated in Figure
21.25. The numerical values for the currents are correct. Despite the incorrect direction, we must continue to use these negative
values in subsequent calculations because our equations were established with our original choice of direction. What would have
happened had we left the current directions as labeled in Figure 21.25 but traversed the loops in the opposite direction?
724
CHAPTER 21 | Current and Direct Current Circuits
21.9 | RC Circuits
So far, we have analyzed direct-current circuits in which the current is constant. In
DC circuits containing capacitors, the current is always in the same direction but may
vary in time. A circuit containing a series combination of a resistor and a capacitor
is called an RC circuit.
Charging a Capacitor
e
Active Figure 21.26 shows a simple series RC circuit. Let’s assume the capacitor
in this circuit is initially uncharged. There is no current while the switch is open
(Active Fig. 21.26a). If the switch is thrown to position a at t 5 0 (Active Fig.
21.26b), however, charge begins to flow, setting up a current in the circuit, and
the capacitor begins to charge.11 Notice that during charging, charges do not
jump across the capacitor plates because the gap between the plates represents an
open circuit. Instead, charge is transferred between each plate and its connecting
wires due to the electric field established in the wires by the battery until the capacitor is fully charged. As the plates are being charged, the potential difference
across the capacitor increases. The value of the maximum charge on the plates
depends on the voltage of the battery. Once the maximum charge is reached, the
current in the circuit is zero because the potential difference across the capacitor
matches that supplied by the battery.
To analyze this circuit quantitatively, let’s apply Kirchhoff’s loop rule to the circuit
after the switch is thrown to position a. Traversing the loop in Active Figure 21.26b
clockwise gives
q
e 2 C 2 IR 5 0
21.31b
where q/C is the potential difference across the capacitor and IR is the potential difference across the resistor. We have used the sign conventions discussed earlier for
the signs on and IR. The capacitor is traversed in the direction from the positive
plate to the negative plate, which represents a decrease in potential. Therefore, we
use a negative sign for this potential difference in Equation 21.31. Note that q and
I are instantaneous values that depend on time (as opposed to steady-state values) as
the capacitor is being charged.
We can use Equation 21.31 to find the initial current in the circuit and the maximum charge on the capacitor. At the instant the switch is thrown to position a (t 5
0), the charge on the capacitor is zero. Equation 21.31 shows that the initial current
Ii in the circuit is a maximum and is given by
e
e
Ii 5
e (current at t 5 0)
Q5C
e (maximum charge)
21.32b
R
At this time, the potential difference from the battery terminals appears entirely
across the resistor. Later, when the capacitor is charged to its maximum value
Q, charges cease to flow, the current in the circuit is zero, and the potential
difference from the battery terminals appears entirely across the capacitor.
Substituting I 5 0 into Equation 21.31 gives the maximum charge on the
capacitor:
e
Active Figure 21.26 A capacitor
in series with a resistor, switch, and
battery.
21.33b
To determine analytical expressions for the time dependence of the charge and
current, we must solve Equation 21.31, a single equation containing two variables
q and I. The current in all parts of the series circuit must be the same. Therefore,
11In previous discussions of capacitors, we assumed a steady-state situation, in which no current was present in any branch
of the circuit containing a capacitor. Now we are considering the case before the steady-state condition is realized; in this
situation, charges are moving and a current exists in the wires connected to the capacitor.
725
21.9 | RC Circuits
the current in the resistance R must be the same as the
current between each capacitor plate and the wire connected to it. This current is equal to the time rate of
change of the charge on the capacitor plates. Therefore,
we substitute I 5 dq/dt into Equation 21.31 and rearrange
the equation:
dq
dt
e q
5 2
R
RC
e
e
e
e
e
To find an expression for q, we solve this separable differential equation as follows. First combine the terms on the
right-hand side:
dq
dt
5
q
q 2 Ce
Ce
2
52
RC
RC
RC
t
t
e
e
Multiply this equation by dt and divide by q 2 C :
dq
q 2 Ce
52
1
dt
RC
Figure 21.27 (a) Plot of capacitor charge versus time for the
circuit shown in Active Figure 21.26b. (b) Plot of current versus time
for the circuit shown in Active Figure 21.26b.
Integrate this expression, using q 5 0 at t 5 0:
E
q
dq
52
1
RC
E
t
dt
0
q 2 Ce
ln
1 2Ce 2 5 2 RCt
0
q 2 Ce
From the definition of the natural logarithm, we can write this expression as
e
q(t) 5 C (1 2 e2t/RC ) 5 Q(1 2 e2t/RC )
21.34b
c Charge as a function of time for
a capacitor being charged
where e is the base of the natural logarithm and we have made the substitution from
Equation 21.33.
We can find an expression for the charging current by differentiating Equation 21.34 with respect to time. Using I 5 dq/dt, we find that
I(t) 5
e e2t/RC
R
21.35b
a capacitor being charged
Plots of capacitor charge and circuit current versus time are shown in Figure 21.27.
Notice that the charge is zero at t 5 0 and approaches the maximum value C as
t : `. The current has its maximum value Ii 5 /R at t 5 0 and decays exponentially to zero as t : `. The quantity RC, which appears in the exponents of Equations 21.34 and 21.35, is called the time constant of the circuit:
e
5 RC
e
21.36b
The time constant represents the time interval during which the current decreases
to 1/e of its initial value; that is, after a time interval , the current decreases to I 5
e21Ii 5 0.368Ii. After a time interval 2, the current decreases to I 5 e22Ii 5 0.135Ii ,
and so forth. Likewise, in a time interval , the charge increases from zero to
C [1 2 e21] 5 0.632C .
The energy supplied by the battery during the time interval required to fully
charge the capacitor is Q 5 C 2. After the capacitor is fully charged, the energy stored in the capacitor is 12 Q 5 12 C 2, which is only half the energy output
of the battery. It is left as a problem (Problem 68) to show that the remaining
half of the energy supplied by the battery appears as internal energy in the
resistor.
e
e
e
e
e
e
c Current as a function of time for
726
CHAPTER 21 | Current and Direct Current Circuits
Discharging a Capacitor
Imagine that the capacitor in Active Figure 21.26b is completely charged. A potential
difference Q/C exists across the capacitor, and there is zero potential difference across
the resistor because I 5 0. If the switch is now thrown to position b at t 5 0 (Active
Fig. 21.26c), the capacitor begins to discharge through the resistor. At some time t during the discharge, the current in the circuit is I and the charge on the capacitor is q.
The circuit in Active Figure 21.26c is the same as the circuit in Active Figure 21.26b except for the absence of the battery. Therefore, we eliminate the emf from Equation
21.31 to obtain the appropriate loop equation for the circuit in Active Figure 21.26c:
e
q
2 IR 5 0
C
When we substitute I 5 dq/dt into this expression, it becomes
2
2R
dq
5
21.37b
q
C
dt
dq
1
52
dt
q
RC
Integrating this expression using q 5 Q at t 5 0 gives
E
q dq
Q
q
ln
c Charge as a function of time
52
1
RC
E
t
dt
0
q
1Q2 5 2 RCt
q(t) 5 Qe2t/RC
21.38b
for a discharging capacitor
c Current as a function of time
for a discharging capacitor
Cable theory for propagation
of an action potential along
a nerve
Figure 21.28 Modeling the cell
membrane of a neuron using cable
theory. Four small patches of the
cell membrane are shown, with each
patch being modeled electrically as
an RC circuit consisting of resistance
Rm and capacitance Cm. Adjacent
patches are connected electrically by
a resistance Rl in the cytoplasm of the
cell interior.
Differentiating Equation 21.38 with respect to time gives the instantaneous current
as a function of time:
Q 2t/RC
I(t) 5 2
e
21.39b
RC
where Q/RC 5 Ii is the initial current. The negative sign indicates that as the capacitor discharges, the current direction is opposite its direction when the capacitor was
being charged. (Compare the current directions in Active Figs. 21.26b and 21.26c.)
Both the charge on the capacitor and the current decay exponentially at a rate characterized by the time constant 5 RC.
In Section 20.7, we discussed the modeling of a patch of cell membrane as a
capacitor. Let us call the capacitance of a given patch of membrane Cm. We also
discussed the flow of ions through various ion channels and ion pumps in the membrane. This flow represents a current. The ions cannot move across the membrane
unimpeded, so there is a resistance to the current, called the membrane resistance Rm.
As a result, each small patch of the cell membrane can be modeled as an RC circuit
as shown in Figure 21.28.
21.9 | RC Circuits
A given long structure in a neuron (such as a dendrite or an axon) can be modeled as a series of RC circuit modules connected by a longitudinal resistance as shown
in Figure 21.28. The longitudinal resistance R l represents resistance to the current
along the axis of the neuron through the cytoplasm. This model of a neuron can be
analyzed using cable theory, first used by Kelvin in the 1850s to analyze the decay of
signals in underwater telegraphic cables. In a neuron, we consider the decay of the
propagation of an action potential along the neuron.
Using cable theory, we can model the propagation of an action potential along a
nerve cell and relate this model to the transfer of information within the human nervous system. The propagation of the action potential is governed by two primary parameters: the time constant and the length constant. The time constant 5 RmCm for
the RC circuit associated with each patch of membrane is similar to the time constant
discussed above and determines how rapidly the membrane capacitor can charge
and discharge. For a given input at a point on the neuron, the membrane voltage
along the neuron decays exponentially. The length constant 5 (Rm /Rl )1/2 determines a characteristic length along the neuron through which the voltage decays
to e21 of its original value. Together, these two parameters describe how efficient the
neuron is at transmitting a signal along its length.
The axons of some nerves are wrapped with sections of myelin, with each section separated from the next by intervals called the nodes of Ranvier. The myelin has the effect of
shutting off the transfer of ions across the cell membrane. As a result, the relatively slow
patch-to-patch propagation of an action potential as described above does not occur. Instead, the signal is carried primarily within the cell interior, such that an action potential
at one node rapidly causes another action potential at the next node. As a result, the
signal travels much faster along the neuron, in a process called saltatory conduction.
Some diseases cause damage to the myelin sheath around nerve cells, degrading the process of saltatory conduction. As a result, patients suffering from these
diseases experience impaired movement, due to the slowness of signals traveling to
the muscles. For example, transverse myelitis is an autoimmune disease, in which the
body attacks the spinal cord, with the resulting inflammation damaging the myelin.
In severe cases, patients are left wheelchair-bound and require assistance with daily
activities. If the damage to the myelin occurs within the white matter of the brain,
the disease is called multiple sclerosis, a highly debilitating disease.
Q U I CK QUI Z 21.8 Consider the circuit in Figure 21.29 and assume the battery has
no internal resistance. (i) Just after the switch is closed, what is the current in the
battery? (a) 0 (b) /2R (c) 2 /R (d) /R (e) impossible to determine (ii) After a
very long time, what is the current in the battery? Choose from the same choices.
e
e
727
The role of myelin in nerve
conduction
e
Figure 21.29 (Quick Quiz 21.8)
How does the current vary after the
switch is closed?
e
THINKING PHYSICS 21.6
Many roadway construction sites have flashing yellow lights to warn motorists of
possible dangers. What causes the lightbulbs to flash?
Reasoning A typical circuit for such a flasher is shown in Figure 21.30. The
lamp L is a gas-filled lamp that acts as an open circuit until a large potential difference causes an electrical discharge in the gas, which gives off a bright light.
During this discharge, charges flow through the gas between the electrodes of
the lamp. After switch S is closed, the battery charges up the capacitor of capacitance C. At the beginning, the current is high and the charge on the capacitor
is low, so most of the potential difference appears across the resistance R. As
the capacitor charges, more potential difference appears across it, reflecting
the lower current and therefore lower potential difference across the resistor.
Eventually, the potential difference across the capacitor reaches a value at which
the lamp will conduct, causing a flash. This discharges the capacitor through the
lamp and the process of charging begins again. The period between flashes can
be adjusted by changing the time constant of the RC circuit. b
Figure 21.30 (Thinking Physics
21.6) The RC circuit in a roadway
construction flasher. When the switch
is closed, the charge on the capacitor
increases until the voltage across the
capacitor (and across the flash lamp)
is high enough for the lamp to flash,
discharging the capacitor.
728
CHAPTER 21 | Current and Direct Current Circuits
Example 21.9 | Charging a Capacitor in an RC Circuit
e
An uncharged capacitor and a resistor are connected in series to a battery as shown in Active Figure 21.26, where 5
12.0 V, C 5 5.00 F, and R 5 8.00 3 105 V. The switch is thrown to position a. Find the time constant of the circuit, the
maximum charge on the capacitor, the maximum current in the circuit, and the charge and current as functions of time.
SOLUTION
Conceptualize Study Active Figure 21.26 and imagine throwing the switch to position a as shown in Active Figure 21.26b.
Upon doing so, the capacitor begins to charge.
Categorize We evaluate our results using equations developed in this section, so we categorize this example as a substitution problem.
5 RC 5 (8.00 3 105 V)(5.00 3 1026 F) 5 4.00 s
Evaluate the time constant of the circuit from
Equation 21.36:
Evaluate the maximum charge on the capacitor from
Equation 21.33:
Q 5 C e 5 (5.00 F)(12.0 V) 5 60.0 C
Evaluate the maximum current in the circuit from
Equation 21.32:
Ii 5
Use these values in Equations 21.34 and 21.35 to find the
charge and current as functions of time:
(1) q(t) 5 60.0(1 2 e2t/4.00)
␧5
R
12.0 V
5 15.0 A
8.00 3 105 V
(2) I(t) 5 15.0e2t/4.00
In Equations (1) and (2), q is in microcoulombs, I is in microamperes, and t is in seconds.
Example 21.10 | Discharging a Capacitor in an RC Circuit
Consider a capacitor of capacitance C that is being discharged through a resistor of resistance R as shown in Active Figure 21.26c.
(A) After how many time constants is the charge on the capacitor one-fourth its initial value?
SOLUTION
Conceptualize Study Active Figure 21.26 and imagine throwing the switch to position b as shown in Active Figure 21.26c.
Upon doing so, the capacitor begins to discharge.
Categorize We categorize the example as one involving a discharging capacitor and use the appropriate equations.
Analyze Substitute q(t) 5 Q/4 into Equation 21.38:
Q
4
5 Qe2t/RC
1
2t/RC
4 5e
Take the logarithm of both sides of the equation and
solve for t :
t
RC
t 5 RC ln 4 5 1.39RC 5 1.39
2ln 4 5 2
(B) The energy stored in the capacitor decreases with time as the capacitor discharges. After how many time constants is
this stored energy one-fourth its initial value?
SOLUTION
Use Equations 20.29 and 21.38 to express the energy
stored in the capacitor at any time t :
Substitute U(t) 5 14(Q 2/2C) into Equation (1):
(1) U(t) 5
Q2
Q2
q2
2C
5
1
22t/RC
4 2C 5 2C e
1
22t/RC
4 5e
Q 2 22t/RC
e
2C
729
21.10 | Context Connection: The Atmosphere as a Conductor
21.10 cont.
Take the logarithm of both sides of the equation and
solve for t :
2ln 4 5 2
2t
RC
t 5 12RC ln 4 5 0.693RC 5 0.693
Finalize Notice that because the energy depends on the square of the charge, the energy in the capacitor drops more rapidly than the charge on the capacitor.
21.10 | Context Connection: The Atmosphere
as a Conductor
When discussing capacitors with air between the plates in Chapter 20, we adopted
the simplification model that air was a perfect insulator. Although that was a good
model for typical potential differences encountered in capacitors, we know that it is
possible for a current to exist in air. Lightning is a dramatic example of this possibility, but a more mundane example is the common spark that you might receive upon
bringing your finger near a doorknob after rubbing your feet across a carpet.
Let us analyze the process that occurs in electrical discharge, which is the same
for lightning and the doorknob spark except for the size of the current. Whenever a
strong electric field exists in air, it is possible for the air to undergo electrical breakdown in which the effective resistivity of the air drops dramatically and the air becomes a conductor. At any given time, due to cosmic ray collisions and other events,
air contains a number of ionized molecules (Fig. 21.31a). For a relatively weak electric
field, such as the fair-weather electric field, these ions and freed electrons accelerate
slowly due to the electric force. They collide with other molecules with no effect and
eventually neutralize as a freed electron ultimately finds an ion and combines with
it. In a strong electric field such as that associated with a thunderstorm, however,
the freed electrons can accelerate to very high speeds (Fig. 21.31b) before making
a collision with a molecule (Fig. 21.31c). If the field is strong enough, the electron
may have enough energy to ionize the molecule in this collision (Fig. 21.31d). Now
there are two electrons to be accelerated by the field, and each can strike another
molecule at high speed (Fig. 21.31e). The result is a very rapid increase in the number of charge carriers available in the air and a corresponding decrease in resistance
of the air. Therefore, there can be a large current in the air that tends to neutralize
the charges that established the initial potential difference, such as the charges in
the cloud and on the ground. When that happens, we have lightning.
Typical currents during lightning strikes can be very high. While the stepped
leader is making its way toward the ground, the current is relatively modest, in the
range of 200 to 300 A. This current is large compared with typical household currents but small compared with peak currents in lightning discharges. Once the connection is made between the stepped leader and the return stroke, the current rises
rapidly to a typical value of 5 3 104 A. Considering that typical potential differences
between cloud and ground in a thunderstorm can be measured in hundreds of thousands of volts, the power during a lightning stroke is measured in billions of watts.
Much of the energy in the stroke is delivered to the air, resulting in a rapid temperature increase and the resultant flash of light and sound of thunder.
Even in the absence of a thundercloud, there is a flow of charge through the air.
The ions in the air make the air a conductor, although not a very good one. Atmospheric measurements indicate a typical potential difference across our atmospheric
capacitor (Section 20.11) of about 3 3 105 V. As we shall show in the Context Conclusion, the total resistance of the air between the plates in the atmospheric capacitor is
about 300 V. Therefore, the average fair-weather current in the air is
I5
DV
3 3 105 V
5
< 1 3 103 A
R
300 V
S
E
S
v
S
E
S
E
S
E
S
E
Figure 21.31 The anatomy of a
spark.
730
CHAPTER 21 | Current and Direct Current Circuits
A number of simplifying assumptions were made in these calculations, but this
result is on the right order of magnitude for the global current. Although the result
might seem surprisingly large, remember that this current is spread out over the entire surface area of the Earth. Therefore, the average fair-weather current density is
J5
I
I
1 3 103 A
5
5
< 2 3 10212 A/m2
2
A
4RE
4(6.4 3 106 m)2
In comparison, the current density in a lightning strike is on the order of 105 A/m2.
The fair-weather current and the lightning current are in opposite directions.
The fair-weather current delivers positive charge to the ground, whereas lightning
delivers negative charge. These two effects are in balance,12 which is the principle
that we shall use to estimate the average number of lightning strikes on the Earth in
the Context Conclusion.
SUMMARY |
(on the average) with a drift velocity :
vd , given by
The electric current I in a conductor is defined as
I ;
:
dQ
21.2b
dt
where dQ is the charge that passes through a cross-section of
the conductor in the time interval dt. The SI unit of current
is the ampere (A); 1 A 5 1 C/s.
The current in a conductor is related to the motion of the
charge carriers through the relationship
21.4b
Iavg 5 nqvd A
where n is the density of charge carriers, q is their charge, vd is
the drift speed, and A is the cross-sectional area of the conductor.
The resistance R of a conductor is defined as the ratio of
the potential difference across the conductor to the current:
DV
21.6b
I
The SI units of resistance are volts per ampere, defined as
ohms (V); 1 V 5 1 V/A.
If the resistance is independent of the applied voltage, the
conductor obeys Ohm’s law, and conductors that have a constant
resistance over a wide range of voltages are said to be ohmic.
If a conductor has a uniform cross-sectional area A and a
length /, its resistance is
,
R5
21.8b
A
where is called the resistivity of the material from which the
conductor is made. The inverse of the resistivity is defined as
the conductivity 5 1/.
The resistivity of a conductor varies with temperature in
an approximately linear fashion; that is,
R ;
5 0[1 1 (T 2 T0)]
21.10b
where 0 is the resistivity at some reference temperature T0
and is the temperature coefficient of resistivity.
In a classical model of electronic conduction in a metal,
the electrons are treated as molecules of a gas. In the absence
of an electric field, the average velocity of the electrons is
zero. When an electric field is applied, the electrons move
:
vd 5
qE
me
21.15b
where is the average time interval between collisions with
the atoms of the metal. The resistivity of the material according to this model is
me
21.18b
5 2
ne where n is the number of free electrons per unit volume.
If a potential difference DV is maintained across a circuit
element, the power, or the rate at which energy is delivered
to the circuit element, is
P 5 I DV
21.20b
Because the potential difference across a resistor is DV 5 IR,
we can express the power delivered to a resistor in the form
(DV )2
21.21b
R
The emf of a battery is the voltage across its terminals when
the current is zero. Because of the voltage drop across the internal resistance r of a battery, the terminal voltage of the battery is
less than the emf when a current exists in the battery.
The equivalent resistance of a set of resistors connected
in series is
P 5 I 2R 5
Req 5 R1 1 R2 1 R3 1 ? ? ?
21.26b
The equivalent resistance of a set of resistors connected in
parallel is given by
1
1
1
1
5
1
1
1 ???
R eq
R1
R2
R3
21.28b
Circuits involving more than one loop are analyzed using
two simple rules called Kirchhoff’s rules:
• At any junction, the sum of the currents must equal zero:
o I50
21.29b
junction
12There are a number of other effects, too, but we will adopt a simplification model in which these are the only two effects. For more information, see E. A. Bering, A. A. Few, and J. R. Benbrook, “The Global Electric Circuit,” Physics Today,
October 1998, pp. 24–30.
| Objective Questions
• The sum of the potential differences across each element
around any closed circuit loop must be zero:
o DV 5 0
21.30b
loop
OBJECTIVE QUESTIONS |
1. If the terminals of a battery with zero internal resistance are
connected across two identical resistors in series, the total
power delivered by the battery is 8.00 W. If the same battery
is connected across the same resistors in parallel, what is the
total power delivered by the battery? (a) 16.0 W (b) 32.0 W
(c) 2.00 W (d) 4.00 W (e) none of those answers
2. Wire B has twice the length and twice the radius of wire A.
Both wires are made from the same material. If wire A has a
resistance R, what is the resistance of wire B? (a) 4R (b) 2R
(c) R (d) 12R (e) 14R
3. The current-versus-voltage behavior of a certain electrical device is shown in Figure OQ21.3.
When the potential difference across the device is 2 V,
what is its resistance? (a) 1 V
(b) 34 V (c) 43 V (d) undefined
(e) none of those answers
If a capacitor is charged with a battery of emf e through a
resistance R, the charge on the capacitor and the current in
the circuit vary in time according to the expressions
q(t) 5 Q (1 2 e2t/RC )
For the junction rule, current in a direction into a junction is 1I, whereas current with a direction away from a junction is 2I.
For the loop rule, when a resistor is traversed in the direction of the current, the change in potential DV across the
resistor is 2IR. If a resistor is traversed in the direction opposite the current, DV 5 1IR.
If a source of emf is traversed in the direction of the emf
(negative to positive), the change in potential is 1e. If it is
traversed opposite the emf (positive to negative), the change
in potential is 2e.
Figure OQ21.3
4. Several resistors are connected in parallel. Which of the following statements are correct? Choose all that are correct.
(a) The equivalent resistance is greater than any of the resistances in the group. (b) The equivalent resistance is less
than any of the resistances in the group. (c) The equivalent
resistance depends on the voltage applied across the group.
(d) The equivalent resistance is equal to the sum of the resistances in the group. (e) None of those statements is correct.
5. A potential difference of 1.00 V is maintained across a 10.0-V resistor for a period of 20.0 s. What total charge passes by a point
in one of the wires connected to the resistor in this time interval? (a) 200 C (b) 20.0 C (c) 2.00 C (d) 0.005 00 C (e) 0.050 0 C
6. Several resistors are connected in series. Which of the following statements is correct? Choose all that are correct.
(a) The equivalent resistance is greater than any of the resistances in the group. (b) The equivalent resistance is less
than any of the resistances in the group. (c) The equivalent
resistance depends on the voltage applied across the group.
(d) The equivalent resistance is equal to the sum of the resistances in the group. (e) None of those statements is correct.
731
21.34b
e
I(t) 5 e2t/RC
21.35b
R
where Q 5 C e is the maximum charge on the capacitor. The
product RC is called the time constant of the circuit.
If a charged capacitor is discharged through a resistance
R, the charge and current decrease exponentially in time according to the expressions
q(t) 5 Qe 2t/RC
Q 2t/RC
e
I(t) 5 2
RC
where Q is the initial charge on the capacitor.
21.38b
21.39b
denotes answer available in Student
Solutions Manual/Study Guide
7. A metal wire of resistance R is cut into three equal pieces that
are then placed together side by side to form a new cable with
a length equal to one-third the original length. What is the
resistance of this new cable? (a) 19R (b) 13R (c) R (d) 3R (e) 9R
8. The terminals of a battery are connected across two resistors in parallel. The resistances of the resistors are not the
same. Which of the following statements is correct? Choose
all that are correct. (a) The resistor with the larger resistance carries more current than the other resistor. (b) The
resistor with the larger resistance carries less current than
the other resistor. (c) The potential difference across each
resistor is the same. (d) The potential difference across the
larger resistor is greater than the potential difference across
the smaller resistor. (e) The potential difference is greater
across the resistor closer to the battery.
9. A cylindrical metal wire at room temperature is carrying
electric current between its ends. One end is at potential
VA 5 50 V, and the other end is at potential VB 5 0 V. Rank
the following actions in terms of the change that each one
separately would produce in the current from the greatest
increase to the greatest decrease. In your ranking, note any
cases of equality. (a) Make VA 5 150 V with VB 5 0 V. (b) Adjust VA to triple the power with which the wire converts electrically transmitted energy into internal energy. (c) Double
the radius of the wire. (d) Double the length of the wire.
(e) Double the Celsius temperature of the wire.
10. Two conducting wires A and B of the same length and radius are connected across the same potential difference.
Conductor A has twice the resistivity of conductor B. What is
the ratio of the power delivered to A to the power delivered
to B? (a) 2 (b) u2 (c) 1 (d) 1/u2 (e) 12
11. When resistors with different resistances are connected in
series, which of the following must be the same for each resistor? Choose all correct answers. (a) potential difference
(b) current (c) power delivered (d) charge entering each
resistor in a given time interval (e) none of those answers
12. When operating on a 120-V circuit, an electric heater receives
1.30 3 103 W of power, a toaster receives 1.00 3 103 W, and an
732
CHAPTER 21 | Current and Direct Current Circuits
electric oven receives 1.54 3 103 W. If all three appliances are
connected in parallel on a 120-V circuit and turned on, what
is the total current drawn from an external source? (a) 24.0 A
(b) 32.0 A (c) 40.0 A (d) 48.0 A (e) none of those answers
13. Car batteries are often rated in ampere-hours. Does this information designate the amount of (a) current, (b) power,
(c) energy, (d) charge, or (e) potential the battery can supply?
14. The terminals of a battery are connected across two resistors
in series. The resistances of the resistors are not the same.
Which of the following statements are correct? Choose all that
are correct. (a) The resistor with the smaller resistance carries
more current than the other resistor. (b) The resistor with the
larger resistance carries less current than the other resistor.
(c) The current in each resistor is the same. (d) The potential
difference across each resistor is the same. (e) The potential
CONCEPTUAL QUESTIONS |
1. Suppose a parachutist lands on a high-voltage wire and grabs
the wire as she prepares to be rescued. (a) Will she be electrocuted? (b) If the wire then breaks, should she continue to
hold onto the wire as she falls to the ground? Explain.
2. What factors affect the resistance of a conductor?
3. Newspaper articles often contain statements such as “10 000
volts of electricity surged through the victim’s body.’’ What
is wrong with this statement?
4. Referring to Figure CQ21.4, describe what happens to the lightbulb after the switch is closed.
Assume the capacitor has a large
capacitance and is initially uncharged. Also assume the light
illuminates when connected directly across the battery terminals.
difference is greatest across the resistor closest to the positive
terminal.
15. In
the
circuit
shown in Figure
OQ21.15,
each
battery is delivering energy to the
circuit by electrical
transmission. All
the resistors have
Figure OQ21.15
equal resistance.
(i) Rank the electric potentials at points a, b, c, d, and e from
highest to lowest, noting any cases of equality in the ranking. (ii) Rank the magnitudes of the currents at the same
points from greatest to least, noting any cases of equality.
denotes answer available in Student
Solutions Manual/Study Guide
room. Your grandmother complains that she has had the radio
for many years and nobody has ever gotten a shock from it. You
end up having to buy a new plastic radio. (a) Why is your grandmother’s old radio dangerous in a hospital room? (b) Will the
old radio be safe back in her bedroom?
8. (a) What advantage does 120-V operation offer over 240 V?
(b) What disadvantages does it have?
9. How does the resistance for copper and for silicon change
with temperature? Why are the behaviors of these two materials different?
10. If charges flow very slowly through a metal, why does it not
require several hours for a light to come on when you throw
a switch?
Figure CQ21.4
5. When the potential difference across a certain conductor is
doubled, the current is observed to increase by a factor of 3.
What can you conclude about the conductor?
6. Use the atomic theory of matter to explain why the resistance
of a material should increase as its temperature increases.
7. So that your grandmother can listen to A Prairie Home Companion, you take her bedside radio to the hospital where she is staying. You are required to have a maintenance worker test the
radio for electrical safety. Finding that it develops 120 V on one
of its knobs, he does not let you take it to your grandmother’s
11. If you were to design an electric heater using Nichrome wire
as the heating element, what parameters of the wire could
you vary to meet a specific power output such as 1 000 W?
12. Is the direction of current in a battery always from the negative terminal to the positive terminal? Explain.
13. Given three lightbulbs and a battery, sketch as many different electric circuits as you can.
14. A student claims that the second of two lightbulbs in series
is less bright than the first because the first lightbulb uses up
some of the current. How would you respond to this statement?
15. Why is it possible for a bird to sit on a high-voltage wire without being electrocuted?
PROBLEMS |
The problems found in this chapter may be assigned
online in Enhanced WebAssign.
1. denotes straightforward problem; 2. denotes intermediate
problem; 3. denotes challenging problem
1. denotes full solution available in the Student Solutions Manual/
Study Guide
1. denotes problems most often assigned in Enhanced
WebAssign.
denotes biomedical problem
denotes guided problem
denotes Master It tutorial available in Enhanced
WebAssign
denotes asking for quantitative and conceptual reasoning
denotes symbolic reasoning problem
denotes “paired problems” that develop reasoning with
symbols and numerical values
denotes Watch It video solution available in Enhanced
WebAssign
| Problems
one free electron per atom. (a) Use the information in
Table 21.1 to determine the resistivity of aluminum at this
temperature. (b) What is the current density in the wire?
(c) What is the total current in the wire? (d) What is the
drift speed of the conduction electrons? (e) What potential
difference must exist between the ends of a 2.00-m length of
the wire to produce the stated electric field?
Section 21.1 Electric Current
1. In a particular cathode-ray tube, the measured beam current is 30.0 A. How many electrons strike the tube screen
every 40.0 s?
2.
3.
Suppose the current in a conductor decreases exponentially with time according to the equation I(t) 5 I0e2t/,
where I0 is the initial current (at t 5 0) and is a constant
having dimensions of time. Consider a fixed observation
point within the conductor. (a) How much charge passes
this point between t 5 0 and t 5 ? (b) How much charge
passes this point between t 5 0 and t 5 10? (c) What If ?
How much charge passes this point between t 5 0 and
t 5 `?
The quantity of charge q (in coulombs) that has passed
through a surface of area 2.00 cm2 varies with time according to the equation q 5 4t 3 1 5t 1 6, where t is in seconds.
What is the instantaneous current through the surface at
t 5 1.00 s? (b) What is the value of the current density?
4.
A small sphere that carries a charge q is whirled in a circle at the end of an insulating string. The angular frequency
of revolution is . What average current does this revolving
charge represent?
5.
The electron beam emerging from a certain high-energy
electron accelerator has a circular cross section of radius
1.00 mm. (a) The beam current is 8.00 A. Find the current density in the beam assuming it is uniform throughout.
(b) The speed of the electrons is so close to the speed of
light that their speed can be taken as 300 Mm/s with negligible error. Find the electron density in the beam. (c) Over
what time interval does Avogadro’s number of electrons
emerge from the accelerator?
6.
7.
Figure P21.6
represents a section of
a conductor of nonuniform diameter carrying
a current of I 5 5.00 A.
The radius of cross-section A1 is r1 5 0.400 cm.
(a) What is the magnitude of the current denFigure P21.6
sity across A1? The radius
r2 at A2 is larger than the radius r1 at A1. (b) Is the current at
A2 larger, smaller, or the same? (c) Is the current density at
A2 larger, smaller, or the same? Assume A2 5 4A1. Specify the
(d) radius, (e) current, and (f) current density at A2.
An aluminum wire having a cross-sectional area equal
to 4.00 3 1026 m2 carries a current of 5.00 A. The density
of aluminum is 2.70 g/cm3. Assume each aluminum atom
supplies one conduction electron per atom. Find the drift
speed of the electrons in the wire.
733
10.
A lightbulb has a resistance of 240 V when operating
with a potential difference of 120 V across it. What is the
current in the lightbulb?
11.
Suppose you wish to fabricate a uniform wire out of
1.00 g of copper. If the wire is to have a resistance of R 5
0.500 V and all the copper is to be used, what must be
(a) the length and (b) the diameter of this wire?
12.
Suppose you wish to fabricate a uniform wire from a mass
m of a metal with density m and resistivity . If the wire is to
have a resistance of R and all the metal is to be used, what
must be (a) the length and (b) the diameter of this wire?
13. While taking photographs in Death Valley on a day when the
temperature is 58.08C, Bill Hiker finds that a certain voltage applied to a copper wire produces a current of 1.000 A.
Bill then travels to Antarctica and applies the same voltage
to the same wire. What current does he register there if the
temperature is 288.08C? Assume that no change occurs
in the wire’s shape and size.
14. Review. An aluminum rod has a resistance of 1.234 V at
20.08C. Calculate the resistance of the rod at 1208C by accounting for the changes in both the resistivity and the dimensions of the rod. The coefficient of linear expansion for
aluminum is 24.0 3 1026 (8C)21.
Section 21.4 A Model for Electrical Conduction
15. If the current carried by a conductor is doubled, what happens to (a) the charge carrier density, (b) the current density, (c) the electron drift velocity, and (d) the average time
interval between collisions?
16.
An iron wire has a cross-sectional area equal to
5.00 3 1026 m2. Carry out the following steps to determine
the drift speed of the conduction electrons in the wire if
it carries a current of 30.0 A. (a) How many kilograms are
there in 1.00 mole of iron? (b) Starting with the density of
iron and the result of part (a), compute the molar density of
iron (the number of moles of iron per cubic meter). (c) Calculate the number density of iron atoms using Avogadro’s
number. (d) Obtain the number density of conduction electrons given that there are two conduction electrons per iron
atom. (e) Calculate the drift speed of conduction electrons
in this wire.
17.
If the magnitude of the drift velocity of free electrons in
a copper wire is 7.84 3 1024 m/s, what is the electric field in
the conductor?
Section 21.2 Resistance and Ohm’s Law
8. A 0.900-V potential difference is maintained across a 1.50-m
length of tungsten wire that has a cross-sectional area of
0.600 mm2. What is the current in the wire?
9.
An aluminum wire with a diameter of 0.100 mm has a
uniform electric field of 0.200 V/m imposed along its entire length. The temperature of the wire is 50.08C. Assume
Section 21.5 Energy and Power in Electric Circuits
18.
The potential difference across a resting neuron in the
human body is about 75.0 mV and carries a current of about
0.200 mA. How much power does the neuron release?
19. In a hydroelectric installation, a turbine delivers 1 500 hp to
a generator, which in turn transfers 80.0% of the mechanical
734
CHAPTER 21 | Current and Direct Current Circuits
wire connected to a 120-V power supply. Assume the wire is
at 1008C throughout the 4.00-min time interval. Specify a
relationship between a diameter and a length that the wire
can have. (b) Can it be made from less than 0.500 cm3 of
Nichrome?
energy out by electrical transmission. Under these conditions, what current does the generator deliver at a terminal
potential difference of 2 000 V?
20.
21.
Residential building codes typically require the use
of 12-gauge copper wire (diameter 0.205 cm) for wiring receptacles. Such circuits carry currents as large as 20.0 A. If
a wire of smaller diameter (with a higher gauge number)
carried that much current, the wire could rise to a high temperature and cause a fire. (a) Calculate the rate at which
internal energy is produced in 1.00 m of 12-gauge copper
wire carrying 20.0 A. (b) What If ? Repeat the calculation for
a 12-gauge aluminum wire. (c) Explain whether a 12-gauge
aluminum wire would be as safe as a copper wire.
A certain toaster has a heating element made of Nichrome wire. When the toaster is first connected to a 120-V
source (and the wire is at a temperature of 20.08C), the
initial current is 1.80 A. The current decreases as the heating element warms up. When the toaster reaches its final
operating temperature, the current is 1.53 A. (a) Find the
power delivered to the toaster when it is at its operating temperature. (b) What is the final temperature of the heating
element?
22. Why is the following situation impossible? A politician is decrying wasteful uses of energy and decides to focus on energy
used to operate plug-in electric clocks in the United States.
He estimates there are 270 million of these clocks, approximately one clock for each person in the population. The
clocks transform energy taken in by electrical transmission
at the average rate 2.50 W. The politician gives a speech in
which he complains that, at today’s electrical rates, the nation is losing $100 million every year to operate these clocks.
23. An 11.0-W energy-efficient fluorescent lightbulb is designed
to produce the same illumination as a conventional 40.0-W
incandescent lightbulb. Assuming a cost of $0.110/kWh for
energy from the electric company, how much money does
the user of the energy-efficient bulb save during 100 h of use?
24. Make an order-of-magnitude estimate of the cost of one person’s routine use of a handheld hair dryer for 1 year. If you
do not use a hair dryer yourself, observe or interview someone who does. State the quantities you estimate and their
values.
25. A 100-W lightbulb connected to a 120-V source experiences
a voltage surge that produces 140 V for a moment. By what
percentage does its power output increase? Assume its resistance does not change.
26. The cost of energy delivered to residences by electrical
transmission varies from $0.070/kWh to $0.258/kWh
throughout the United States; $0.110/kWh is the average
value. At this average price, calculate the cost of (a) leaving a 40.0-W porch light on for two weeks while you are on
vacation, (b) making a piece of dark toast in 3.00 min with
a 970-W toaster, and (c) drying a load of clothes in 40.0 min
in a 5.20 3 103-W dryer.
27.
Assuming the cost of energy from the electric company
is $0.110/kWh, compute the cost per day of operating a
lamp that draws a current of 1.70 A from a 110-V line.
28. Review. An office worker uses an immersion heater to warm
250 g of water in a light, covered, insulated cup from 20.08C
to 1008C in 4.00 min. The heater is a Nichrome resistance
29. A toaster is rated at 600 W when connected to a 120-V
source. What current does the toaster carry, and what is its
resistance?
30. Review. A rechargeable battery of mass 15.0 g delivers an
average current of 18.0 mA to a portable DVD player at
1.60 V for 2.40 h before the battery must be recharged. The
recharger maintains a potential difference of 2.30 V across
the battery and delivers a charging current of 13.5 mA for
4.20 h. (a) What is the efficiency of the battery as an energy
storage device? (b) How much internal energy is produced
in the battery during one charge–discharge cycle? (c) If the
battery is surrounded by ideal thermal insulation and has an
effective specific heat of 975 J/kg ? 8C, by how much will its
temperature increase during the cycle?
31.
An all-electric car (not a hybrid) is designed to run
from a bank of 12.0-V batteries with total energy storage of
2.00 3 107 J. If the electric motor draws 8.00 kW as the car
moves at a steady speed of 20.0 m/s, (a) what is the current
delivered to the motor? (b) How far can the car travel before it is “out of juice”?
32.
Review. A well-insulated electric water heater warms
109 kg of water from 20.08C to 49.08C in 25.0 min. Find the
resistance of its heating element, which is connected across
a 240-V potential difference.
Section 21.6 Sources of emf
33.
A battery has an emf of 15.0 V. The terminal voltage of
the battery is 11.6 V when it is delivering 20.0 W of power
to an external load resistor R. (a) What is the value of R ?
(b) What is the internal resistance of the battery?
34. Two 1.50-V batteries—with their positive terminals in the
same direction—are inserted in series into a flashlight. One
battery has an internal resistance of 0.255 V, and the other
has an internal resistance of 0.153 V. When the switch is
closed, the bulb carries a current of 600 mA. (a) What is the
bulb’s resistance? (b) What fraction of the chemical energy
transformed appears as internal energy in the batteries?
35.
An automobile battery has an emf of 12.6 V and an internal
resistance of 0.080 0 V. The headlights together have an equivalent resistance of 5.00 V (assumed constant). What is the potential difference across the headlight bulbs (a) when they are
the only load on the battery and (b) when the starter motor is
operated, requiring an additional 35.0 A from the battery?
Section 21.7 Resistors in Series and Parallel
36.
For the purpose of
measuring the electric resistance of shoes through the
body of the wearer standing on a metal ground plate,
the
American
National
Standards Institute (ANSI)
specifies the circuit shown in
Figure P21.36. The potential
Figure P21.36
| Problems
maximum power that can safely be delivered to any one resistor is 25.0 W. (a) What is the maximum potential difference
that can be applied to the terminals a and b ? (b) For the voltage determined in part (a), what is the power delivered to
each resistor? (c) What is the total power delivered to the
combination of resistors?
difference ΔV across the 1.00-MV resistor is measured with an
ideal voltmeter. (a) Show that the resistance of the footwear is
R shoes 5
50.0 V 2 DV
DV
(b) In a medical test, a current through the human body
should not exceed 150 A. Can the current delivered by the
ANSI-specified circuit exceed 150 A? To decide, consider
a person standing barefoot on the ground plate.
37. (a) Find the equivalent resistance between points a and
b in Figure P21.37. (b) A potential difference of 34.0 V
is applied between points a
and b. Calculate the current
in each resistor.
Figure P21.37
38. Why is the following situation
impossible? A technician is testing a circuit that contains
a resistance R. He realizes that a better design for the
circuit would include a resistance 73R rather than R . He
has three additional resistors, each with resistance R . By
combining these additional resistors in a certain combination that is then placed in series with the original
resistor, he achieves the desired resistance.
39.
735
42. A young man owns a canister vacuum cleaner marked
“535 W [at] 120 V” and a Volkswagen Beetle, which he
wishes to clean. He parks the car in his apartment parking
lot and uses an inexpensive extension cord 15.0 m long to
plug in the vacuum cleaner. You may assume the cleaner has
constant resistance. (a) If the resistance of each of the two
conductors in the extension cord is 0.900 V, what is the
actual power delivered to the cleaner? (b) If instead
the power is to be at least 525 W, what must be the diameter
of each of two identical copper conductors in the cord he
buys? (c) Repeat part (b) assuming the power is to be at
least 532 W.
43.
Calculate the power delivered to each resistor in the circuit
shown in Figure P21.43.
Consider the circuit shown in Figure P21.39. Find
(a) the current in the 20.0-V resistor and (b) the potential
difference between points a and b.
Figure P21.43
44.
A lightbulb marked “75 W @ 120 V” is screwed into a
socket at one end of a long extension cord, in which each
of the two conductors has resistance 0.800 V. The other
end of the extension cord is plugged into a 120-V outlet.
(a) Explain why the actual power delivered to the lightbulb
cannot be 75 W in this situation. (b) Draw a circuit diagram.
(c) Find the actual power delivered to the lightbulb in this
circuit.
Figure P21.39
40. Four resistors are connected
to a battery as shown in Figure P21.40. The current in
the battery is I, the battery
e
emf is ␧, and the resistor
values are R1 5 R, R2 5
2R, R3 5 4R, and R4 5 3R.
(a) Rank the resistors acFigure P21.40
cording to the potential difference across them, from largest to smallest. Note any cases of
equal potential differences. (b) Determine the potential difference across each resistor in terms of e. (c) Rank the resistors
according to the current in them, from largest to smallest. Note
any cases of equal currents. (d) Determine the current in each
resistor in terms of I. (e) If R 3 is increased, what happens to the
current in each of the resistors? (f) In the limit that R 3 : ∞,
what are the new values of
the current in each resistor
in terms of I, the original
current in the battery?
41.
Three 100-V resistors
are connected as shown
in Figure P21.41. The
Section 21.8 Kirchhoff ’s Rules
Note: The currents are not necessarily in the direction shown
for some circuits.
45.
The ammeter shown in Figure P21.45 reads 2.00 A. Find
I1, I2, and ␧.
e
Figure P21.45
46. The following equations describe an electric circuit:
2I1 (220 V) 1 5.80 V 2 I2 (370 V) 5 0
1I2 (370 V) 1 I3 (150 V) 2 3.10 V 5 0
Figure P21.41
I1 1 I3 2 I2 5 0
736
CHAPTER 21 | Current and Direct Current Circuits
(a) Draw a diagram of the circuit. (b) Calculate the unknowns and identify the physical meaning of each unknown.
47.
The circuit
shown in Figure P21.47
is connected for 2.00
min.
(a) Determine
the current in each
branch of the circuit.
(b) Find the energy delivered by each battery.
Figure P21.47 Problems 47 and 48.
(c) Find the energy delivered to each resistor. (d) Identify the type of energy storage transformation that occurs in the operation of the circuit.
(e) Find the total amount of energy transformed into internal
energy in the resistors.
48. In Figure P21.47, show how to add just enough ammeters
to measure every different current. Show how to add just
enough voltmeters to measure the potential difference
across each resistor and across each battery.
49. Taking R 5 1.00 kV and ␧ 5 250 V in Figure P21.49,
determine the direction and magnitude of the current in
the horizontal wire between a and e.
Figure P21.49
51.
For the
circuit shown in Figure P21.50, we wish to
find the currents I1, I2,
and I3. Use Kirchhoff’s
rules to obtain equations for (a) the upper
loop, (b) the lower
loop, and (c) the junction on the left side.
In each case, suppress
units for clarity and
Figure P21.50
simplify,
combining
the terms. (d) Solve the junction equation for I3. (e) Using
the equation found in part (d), eliminate I3 from the equation found in part (b). (f) Solve the equations found in parts
(a) and (e) simultaneously for the two unknowns I1 and I2.
(g) Substitute the answers found in part (f) into the junction
equation found in part (d), solving for I3. (h) What is the significance of the negative answer for I2?
In the circuit of Figure P21.51, determine (a) the current in each resistor and (b) the potential difference across
the 200-V resistor.
53.
Consider a series RC
circuit as in Figure P21.53
for which R 5 1.00 MΩ,
C 5 5.00 F, and ␧ 5 30.0 V.
Find (a) the time constant
of the circuit and (b) the
maximum charge on the
e
capacitor after the switch is
thrown closed. (c) Find the Figure P21.53 Problems 53,
current in the resistor 10.0 s 67, and 68.
after the switch is closed.
54. In places such as hospital operating rooms or factories for electronic circuit boards, electric sparks must be avoided. A person
standing on a grounded floor and touching nothing else can
typically have a body capacitance of 150 pF, in parallel with a
foot capacitance of 80.0 pF produced by the dielectric soles of
his or her shoes. The person acquires static electric charge from
interactions with his or her surroundings. The static charge
flows to ground through the equivalent resistance of the two
shoe soles in parallel with each other. A pair of rubber-soled
street shoes can present an equivalent resistance of 5.00 3 103
MV. A pair of shoes with special static-dissipative soles can have
an equivalent resistance of 1.00 MV. Consider the person’s
body and shoes as forming an RC circuit with the ground.
(a) How long does it take the rubber-soled shoes to reduce a
person’s potential from 3.00 3 103 V to 100 V? (b) How long
does it take the static-dissipative shoes to do the same thing?
55.
A 2.00-nF capacitor with an initial charge of 5.10 C
is discharged through a 1.30-kV resistor. (a) Calculate the
current in the resistor 9.00 s after the resistor is connected
across the terminals of the capacitor. (b) What charge remains on the capacitor after 8.00 s? (c) What is the maximum current in the resistor?
56. A 10.0-F capacitor is charged by a 10.0-V battery through a
resistance R. The capacitor reaches a potential difference of
4.00 V in a time interval of 3.00 s after charging begins. Find R.
57.
Figure P21.51
Jumper
cables are connected from a
fresh battery in
one car to charge
a dead battery
in another car.
Figure
P21.52
shows the circuit
diagram for this
situation. While
the cables are
connected, the
Figure P21.52
ignition switch
of the car with the dead battery is closed and the starter
is activated to start the engine. Determine the current in
(a) the starter and (b) the dead battery. (c) Is the dead battery being charged while the starter is operating?
Section 21.9 RC Circuits
e
e
50.
52.
In the circuit of Figure P21.57, the switch S has been open
for a long time. It is then suddenly closed. Take ␧ 5 10.0 V,
R1 5 50.0 kΩ, R2 5 100 kΩ, and C 5 10.0 μ F. Determine the
time constant (a) before the switch is closed and (b) after the
switch is closed. (c) Let the switch be closed at t 5 0. Determine the current in the switch as a function of time.
| Problems
at x 5 0.500 m. Find (a) the magnitude and direction of the
electric field in the wire, (b) the resistance of the wire, (c) the
magnitude and direction of the electric current in the wire,
and (d) the current density in the wire. (e) Show that E 5 J.
e
64.
Figure P21.57 Problems 57 and 58.
58.
In the circuit of Figure P21.57, the switch S has been open
for a long time. It is then suddenly closed. Determine the
time constant (a) before the switch is closed and (b) after the
switch is closed. (c) Let the switch be closed at t 5 0. Determine the current in the switch as a function of time.
59.
The circuit in Figure P21.59 has been connected for a
long time. (a) What is the potential difference across the capacitor? (b) If the battery is disconnected from the circuit,
over what time interval does the capacitor discharge to onetenth its initial voltage?
66.
An oceanographer is studying how the
ion concentration in seawater depends on
depth. She makes a measurement by lowering into the water a pair of concentric metallic cylinders (Fig. P21.66) at the end of a
cable and taking data to determine the resistance between these electrodes as a function
of depth. The water between the two cylinders forms a cylindrical shell of inner radius
ra , outer radius rb , and length L much larger Figure P21.66
than rb . The scientist applies a potential difference DV between the inner and outer surfaces, producing
an outward radial current I. Let represent the resistivity of the
water. (a) Find the resistance of the water between the cylinders
in terms of L, , ra , an rb . (b) Express the resistivity of the water
in terms of the measured quantities L, ra , rb , DV, and I.
67.
The values of the components in a simple series RC
circuit containing a switch (Fig. P21.53) are C 5 1.00 F,
R 5 2.00 3 106 V, and ␧ 5 10.0 V. At the instant 10.0 s
after the switch is closed, calculate (a) the charge on the
capacitor, (b) the current in the resistor, (c) the rate at
which energy is being stored in the capacitor, and (d) the
rate at which energy is being delivered by the battery.
68.
A battery is used to charge a capacitor through a resistor as shown in Figure P21.53. Show that half the energy
supplied by the battery appears as internal energy in the
resistor and half is stored in the capacitor.
Figure P21.59
Section 21.10 Context Connection: The Atmosphere
as a Conductor
60. Assume that global lightning on the Earth constitutes a
constant current of 1.00 kA between the ground and an
atmospheric layer at potential 300 kV. (a) Find the power
of terrestrial lightning. (b) For comparison, find the
power of sunlight falling on the Earth. Sunlight has an intensity of 1 370 W/m2 above the atmosphere. Sunlight falls
perpendicularly on the circular projected area that the
Earth presents to the Sun.
61. A current density of 6.00 3 10213 A/m2 exists in the atmosphere at a location where the electric field is 100 V/m. Calculate the electrical conductivity of the Earth’s atmosphere
in this region.
Additional Problems
Lightbulb A is marked “25 W 120 V,” and lightbulb B is
marked “100 W 120 V.” These labels mean that each lightbulb has its respective power delivered to it when it is connected to a constant 120-V source. (a) Find the resistance of
each lightbulb. (b) During what time interval does 1.00 C
pass into lightbulb A? (c) Is this charge different upon its exit
versus its entry into the lightbulb? Explain. (d) In what time interval does 1.00 J pass into lightbulb A? (e) By what mechanisms
does this energy enter and exit the lightbulb? Explain. (f) Find
the cost of running lightbulb A continuously for 30.0 days, assuming the electric company sells its product at $0.110 per kWh.
63. A straight, cylindrical wire lying along the x axis has a
length of 0.500 m and a diameter of 0.200 mm. It is made
of a material described by Ohm’s law with a resistivity of 5
4.00 3 1028 V ? m. Assume a potential of 4.00 V is maintained at the left end of the wire at x 5 0. Also assume V 5 0
A straight, cylindrical wire lying along the x axis has a
length L and a diameter d. It is made of a material described
by Ohm’s law with a resistivity . Assume potential V is maintained at the left end of the wire at x 5 0. Also assume the
potential is zero at x 5 L. In terms of L, d, V, , and physical constants, derive expressions for (a) the magnitude and
direction of the electric field in the wire, (b) the resistance
of the wire, (c) the magnitude and direction of the electric
current in the wire, and (d) the current density in the wire.
(e) Show that E 5 J.
65. Four 1.50-V AA batteries in series are used to power a small
radio. If the batteries can move a charge of 240 C, how long
will they last if the radio has a resistance of 200 V?
m
62.
737
69. Switch S shown in Figure
P21.69 has been closed
for a long time, and the
electric circuit carries a
constant current. Take
C1 5 3.00 F, C 2 5 6.00 F,
R1 5 4.00 kV, and R2 5
7.00 kV. The power deFigure P21.69
livered to R2 is 2.40 W.
(a) Find the charge on C1. (b) Now the switch is opened.
After many milliseconds, by how much has the charge on
C2 changed?
70. Why is the following situation impossible? A battery has an emf
of ␧ 5 9.20 V and an internal resistance of r 5 1.20 V. A resistance R is connected across the battery and extracts from
it a power of P 5 21.2 W.
738
CHAPTER 21 | Current and Direct Current Circuits
potential difference as plotted in Figure P21.74b. What is
the period T of the waveform in terms of R 1, R 2, and C ?
71. The student engineer of a campus radio station wishes to verify
the effectiveness of the lightning rod on the antenna mast
(Fig. P21.71). The unknown resistance R x is between points C
and E. Point E is a true ground,
but it is inaccessible for direct
measurement because this straFigure P21.71
tum is several meters below the
Earth’s surface. Two identical rods are driven into the ground
at A and B, introducing an unknown resistance R y. The procedure is as follows. Measure resistance R 1 between points A and
B, then connect A and B with a heavy conducting wire and
measure resistance R 2 between points A and C. (a) Derive an
equation for R x in terms of the observable resistances, R 1 and
R 2. (b) A satisfactory ground resistance would be R x , 2.00 V.
Is the grounding of the station adequate if measurements give
R 1 5 13.0 V and R 2 5 6.00 V? Explain.
72. The circuit shown in Figure P21.72 is set up in the
laboratory to measure an
unknown capacitance C in
series with a resistance R 5
10.0 MV powered by a battery whose emf is 6.19 V.
The data given in the table
e
are the measured voltages
Figure P21.72
across the capacitor as a
function of time, where
t 5 0 represents the instant at which the switch is thrown to position b. (a) Construct a graph of ln (␧/ΔV) versus t and perform a linear least-squares fit to the data. (b) From the slope
of your graph, obtain a value for the time constant of the circuit and a value for the capacitance.
73.
74.
ΔV (V)
t (s)
6.19
0
5.55
4.87
4.93
11.1
4.34
19.4
3.72
Figure P21.74
75.
An electric heater is rated at 1.50 3 103 W, a toaster
at 750 W, and an electric grill at 1.00 3 103 W. The three appliances are connected to a common 120-V household circuit.
(a) How much current does each draw? (b) If the circuit is
protected with a 25.0-A circuit breaker, will the circuit breaker
be tripped in this situation? Explain your answer.
76.
An experiment is conducted to measure the electrical
resistivity of Nichrome in the form of wires with different
lengths and cross-sectional areas. For one set of measurements, a student uses 30-gauge wire, which has a cross-sectional area of 7.30 3 1028 m2. The student measures the
potential difference across the wire and the current in the
wire with a voltmeter and an ammeter, respectively. For each
set of measurements given in the table taken on wires of
three different lengths, calculate the resistance of the wires
and the corresponding values of the resistivity. (b) What
is the average value of the resistivity? (c) Explain how this
value compares with the value given in Table 21.1.
ln (␧/ΔV )
L (m)
ΔV (V)
I (A)
30.8
0.540
5.22
0.72
3.09
46.6
1.028
5.82
0.414
2.47
67.3
1.543
5.94
0.281
1.83
102.2
A battery has an emf ␧ and internal resistance r. A variable load resistor R is connected across the terminals of the
battery. (a) Determine the value of R such that the potential difference across the terminals is a maximum. (b) Determine the value of R so that the current in the circuit is a
maximum. (c) Determine the value of R so that the power
delivered to the load resistor is a maximum. Choosing the
load resistance for maximum power transfer is a case of
what is called impedance matching in general. Impedance
matching is important in shifting gears on a bicycle, in connecting a loudspeaker to an audio amplifier, in connecting
a battery charger to a bank of solar photoelectric cells, and
in many other applications.
The switch in Figure P21.74a closes when DVc . 23 DV
and opens when DVc , 13 DV . The ideal voltmeter reads a
77.
R (V)
(V ? m)
Four resistors are connected in parallel across a 9.20-V
battery. They carry currents of 150 mA, 45.0 mA, 14.0 mA,
and 4.00 mA. If the resistor with the largest resistance is
replaced with one having twice the resistance, (a) what is
the ratio of the new current in the battery to the original
current? (b) What If? If instead the resistor with the smallest
resistance is replaced with one having twice the resistance,
what is the ratio of the new total current to the original current? (c) On a February night, energy leaves a house by several energy leaks, including 1.50 3 103 W by conduction
through the ceiling, 450 W by infiltration (air-flow) around
the windows, 140 W by conduction through the basement
wall above the foundation sill, and 40.0 W by conduction
through the plywood door to the attic. To produce the
biggest saving in heating bills, which one of these energy
transfers should be reduced first? Explain how you decide.
Clifford Swartz suggested the idea for this problem.
Context
6
CONC LU S I O N
Determining the Number
of Lightning Strikes
N
ow that we have investigated the principles of electricity, let us respond to our
central question for the Lightning Context:
How can we determine the number of lightning strikes on the Earth in a typical day?
We must combine several ideas from our knowledge of electricity to perform this calculation. In
Chapter 20, the atmosphere was modeled as a capacitor. Such modeling was first done by Lord Kelvin,
who modeled the ionosphere as the positive plate
several tens of kilometers above the Earth’s surface.
More sophisticated models have shown the effective
height of the positive plate to be the 5 km that we
used in our earlier calculation.
The Atmospheric Capacitor Model
The plates of the atmospheric capacitor are separated
by a layer of air containing a large number of free
ions that can carry current. Air is a good insulator;
measurements show that the resistivity of air is about
3 3 1013 V ? m. Let us calculate the resistance of the
air between our capacitor plates. The shape of the re- Figure 1 (a) The atmosphere can be modeled as a capacitor, with
sistor is that of a spherical shell between the plates conductive air between the plates. (b) We can imagine an equivalent RC
of the atmospheric capacitor (Fig. 1a). The length of circuit for the atmosphere, with the natural discharge of the capacitor in
balance with the charging of the capacitor by lightning.
5 km, however, is very short compared with the radius
of 6 400 km. Therefore, we can ignore the spherical shape and approximate the resistor as a 5-km slab of flat material whose area is the surface area of the Earth. Using
Equation 21.8,
R5
,
5 3 103 m
5 (3 3 1013 V ? m)
< 3 3 102 V
A
4(6.4 3 106 m)2
The charge on the atmospheric capacitor can pass from the upper plate to the
ground by electric current in the air between the plates. Therefore, we can model
the atmosphere as an RC circuit (Fig. 1b), using the capacitance found in Chapter
20, and the resistance connecting the plates calculated above. The time constant for
this RC circuit is
5 RC 5 (0.9 F)(3 3 102 V) < 3 3 102 s 5 5 min
Therefore, the charge on the atmospheric capacitor should fall to e21 5 37% of
its original value after only 5 min! After 30 min, less than 0.3% of the charge would
739
740
CONTEXT 6 CONCLUSION | Determining the Number of Lightning Strikes
remain! Why doesn’t that happen? What keeps the atmospheric capacitor charged?
The answer is lightning. The processes occurring in cloud charging result in lightning strikes that deliver negative charge to the ground to replace that neutralized by
the flow of charge through the air. On the average, a net charge on the atmospheric
capacitor results from a balance between these two processes.
Now, let’s use this balance to numerically answer our central question. We first
address the charge on the atmospheric capacitor. In Chapter 19, we mentioned a
charge of 5 3 105 C that is spread over the surface of the Earth, which is the charge
on the atmospheric capacitor.
A typical lightning strike delivers about 25 C of negative charge to the ground
in the process of charging the capacitor. Dividing the charge on the capacitor by
the charge per lightning strike tells us the number of lightning strikes required to
charge the capacitor:
Number of lightning strikes 5
5
total charge
charge per lightning strike
5 3 105 C
< 2 3 104 lightning strikes
25 C per strike
According to our calculation for the RC circuit, the atmospheric capacitor almost
completely discharges through the air in about 30 min. Therefore, 2 3 104 lightning
strikes must occur every 30 min, or 4 3 104/h, to keep the charging and discharging
processes in balance. Multiplying by the number of hours in a day gives us
1241 dh2
Number of lightning strikes per day 5 (4 3 104 strikes/h)
< 1 3 106 strokes/day
Despite the simplifications that we have adopted in our calculations, this number is
on the right order of magnitude for the actual number of lightning strikes on the
Earth in a typical day: 1 million!
Problems
1. Consider the atmospheric capacitor described in the text, with the ground as
one plate and positive charges in the atmosphere as the other. On one particular day, the capacitance of the atmospheric capacitor is 0.800 F. The effective
plate separation distance is 4.00 km, and the resistivity of the air between the
plates is 2.00 3 1013 V ? m. If no lightning events occur, the capacitor will discharge through the air. If a charge of 4.00 3 104 C is on the atmospheric capacitor at time t 5 0, at what later time is the charge reduced (a) to 2.00 3 104 C,
(b) to 5.00 3 103 C, and (c) to zero?
2. Consider this alternative line of reasoning to estimate the number of lightning
strikes on the Earth in one day. Using the charge on the Earth of 5.00 3 105 C
and the atmospheric capacitance of 0.9 F, we find that the potential difference
across the capacitor is DV 5 Q/C 5 5.00 3 105 C/0.9 F < 6 3 105 V. The leakage current in the air is I 5 DV/R 5 6 3 105 V/300 V < 2 kA. To keep the capacitor charged, lightning should deliver the same net current in the opposite
direction. (a) If each lightning strike delivers 25 C of charge to the ground,
what is the average time interval between lightning strikes so that the average
current due to lightning is 2 kA? (b) Using this average time interval between
lightning strikes, calculate the number of lightning strikes in one day.
3. Consider again the atmospheric capacitor discussed in the text. (a) Assume that
atmospheric conditions are such that, for one complete day, the lower 2.50 km
of the air between the capacitor plates has resistivity 2.00 3 1013 V ? m and the
upper 2.50 km has resistivity 0.500 3 1013 V ? m. How many lightning strikes
occur on this day? (b) Assume that atmospheric conditions are such that, for
one complete day, resistivity of the air between the plates in the southern hemisphere is 2.00 3 1013 V ? m and the resistivity between the plates in the northern
hemisphere is 0.200 3 1013 V ? m. How many lightning strikes occur on this day?
Context
7
Magnetism in Medicine
N
© Lyroky/Alamy
ow that we have studied electricity, we turn our attention to the closely related topic of magnetism.
Magnetism is prevalent in our everyday life. Magnets
are essential for the operation of motors. Magnets in
generators provide electricity to homes and businesses.
Loudspeaker systems use magnets to convert electrical
signals to sound waves. Magnets are also critical in keeping important data securely fixed to refrigerator doors.
Magnetism has also entered the field of medicine
with a number of applications that can improve health
and save lives. Various medical tests or procedures involve magnets. We will explore some of these important
applications in this Context. We begin, however, by exploring some questionable applications of magnetism in
medicine between the 18th century and the present day.
You may have heard advertisements or even own a
magnetic bracelet, such as those shown in Figure 1.
Such a bracelet is just one example of devices that provide purported magnetic therapy. Additional such devices
include other magnetic jewelry, magnetic straps for various body parts, magnetic shoe inserts, magnetic blankets
and mattresses, and magnetic creams. Despite having
sales of a billion dollars a year, magnetic therapy has
not been shown in any scientific studies to be effective.
Figure 1 Magnetic bracelets are sold to consumers to promote good
health and pain relief. Do you think devices such as these bracelets work?
The United States Food and Drug Administration prohibits marketing any magnetic therapy device as having
proven medical advantages.
Let us now move back in time and investigate some
earlier applications of magnetism in medicine. Some
doctors who employed these applications actually believed that their magnetic instruments would help their
patients. Other so-called quack doctors knew that the instruments would not work, but used them anyway.
Franz Anton Mesmer, from Vienna, was one of the
earliest individuals to develop a theory of medicine involving magnetism. In his doctoral thesis (The Influence
of the Planets on the Human Body, 1767), Mesmer suggested that a universal fluid that he called “animal gravitation” was responsible for all health and illnesses. In
1773, Mesmer began to use magnets to heal diseases. He
claimed that he could “cure” some diseases with a combination of stroking the patient with magnets, various
forms of wailing, and listening to music from the glass
armonica, recently invented by Benjamin Franklin.
By 1776, Mesmer announced that the magnets were
not necessary for his treatment—they were only serving
as conductors of the universal fluid, which by now had
become a magnetic fluid, which he called animal magnetism. Mesmer was very careful about the selection of
diseases he attempted to cure. For organic diseases, he
would refer the patient to a traditional doctor. He used
magnets to treat only nervous or hysterical diseases. The
startling aspect of Mesmer’s practice is that he supposedly restored sight in a blind pianist and relieved many
patients from chronic convulsions. Today we realize that
his magnetic treatments were not really curing patients.
Mesmer was actually hypnotizing patients, using stares,
stroking, glass armonica music, power of suggestion, etc.
In fact, his name is the root of the word mesmerize.
Figure 2 (page 742) shows an example of the Davis
and Kidder Magneto-Electric Machine for Nervous Disorders,
which was used from the 1850s into the latter part of
the 19th century. It is simply an electromagnetic generator, developed shortly after the magnetic induction
discoveries of Michael Faraday. A pair of wire coils
was rotated in the vicinity of a permanent magnet.
The patient held on to the two metal cylinders, which
741
Science & Society Picture Library/SSPL/Getty Images
Dirk Soulis Auctions
742 CONTEXT 7 | Magnetism in Medicine
Figure 2 The Davis & Kidder Magneto-Electric Machine for Nervous
Disorders. The patient would hold a brass tube in each hand, while the
caretaker turned the crank. The patient received a shock from the voltage
generated by the rotating coils in the presence of the magnetic field of the
large permanent magnet in the back of the case.
were connected to the generator. The caretaker then
turned the crank, providing a jolt of electricity to the
patient. Providing jolts of electricity to the patient continued as a supposed treatment until well into the 20th
century, and even has its supporters today. The handcranked Davis and Kidder device, however, was replaced
by plug-in devices such as violet ray machines and the various devices of Albert Abrams. (Check out Abrams and
his struggle with the American Medical Association on
the Internet.)
Another magnetic quack device that arose in the 20th
century was originally manufactured under the name of
the IONACO, developed by Gaylord Wilshire. A large
loop of wire, covered in leather, was plugged into an
electrical socket. The goal was to magnetize the blood
by wearing the loop around the body. Figure 3 shows a
Figure 3 The Theronoid magnetic device. The leather-coated loop of
wire is worn around the body to “magnetize the blood.”
later version of this device, the Theronoid, developed
by Philip Ilsey. In its 1933 annual report, the United
States Federal Trade Commission (FTC) states, “It was
claimed by respondents that the use of said device or
appliance . . . was a beneficial therapeutic agent in the
aid, relief, prevention, or cure of . . . asthma, arthritis,
bladder trouble, bronchitis, catarrh, constipation, diabetes, eczema, heart trouble, hemorrhoids, indigestion, insomnia, lumbago, nervous disorders, neuralgia, neuritis,
rheumatism, sciatica, stomach trouble, varicose veins,
and high blood pressure.” The FTC closes this section of
its report by banning advertising of the Theronoid: “the
Commission issued an order to . . . cease and desist from
representing in any manner whatsoever that the said
belt or device or any similar device or appliance . . . has
any physiotherapeutic effect upon such subject, or
that it is calculated or likely to aid in the prevention,
treatment, or cure of any human ailment, sickness, or
disease.”
In this Context, we will look at scientifically supported uses of magnetism in medicine today as opposed
to the unsubstantiated and, in some cases, fraudulent uses discussed here. We will address the central
question:
How has magnetism entered the field of medicine to
diagnose and cure illnesses and save lives?
Chapter
22
Magnetic Forces
and Magnetic Fields
Chapter Outline
22.1
Historical Overview
22.2 The Magnetic Field
22.3 Motion of a Charged Particle in a
Uniform Magnetic Field
22.4 Applications Involving Charged
Particles Moving in a Magnetic Field
22.5 Magnetic Force on a Current-Carrying
Conductor
22.6 Torque on a Current Loop in a Uniform
Magnetic Field
22.7 The Biot–Savart Law
22.8 The Magnetic Force Between Two
Parallel Conductors
22.9 Ampère’s Law
22.10 The Magnetic Field of a Solenoid
22.11 Magnetism in Matter
22.12 Context Connection: Remote Magnetic
Navigation for Cardiac Catheter
Ablation Procedures
CERN
SUMMARY
T
he list of technological applications of magnetism is very long. For instance, large electromagnets are used to pick up
heavy loads in scrap yards. Magnets are used in such devices as meters, motors, and loudspeakers. Magnetic tapes are routinely used in sound and video
recording equipment. Magnetic stripes on the backs of credit cards allow our
purchase to be completed quickly in a store. Intense magnetic fields generated
by superconducting magnets are currently being used as a means of containing
plasmas at temperatures on the order of 108 K used in controlled nuclear fusion
research.
As we investigate magnetism in this chapter, we shall find that the subject
cannot be divorced from electricity. For example, magnetic fields affect moving
An engineer performs a test on the
electronics associated with one of
the superconducting magnets in the
Large Hadron Collider at the European
Laboratory for Particle Physics, operated
by the European Organization for Nuclear
Research (CERN). The magnets are used
to control the motion of charged particles
in the accelerator. We will study the effects
of magnetic fields on moving charged
particles in this chapter.
743
744 CHAPTER 22 | Magnetic Forces and Magnetic Fields
electric charges, and moving charges produce magnetic fields. This close association
between electricity and magnetism will justify their union into electromagnetism that we
explore in this chapter and the next.
North Wind Picture Archives
22.1 | Historical Overview
Hans Christian Oersted
Danish Physicist and Chemist
(1777–1851)
Oersted is best known for observing that a
compass needle deflects when placed near
a wire carrying a current. This important
discovery was the first evidence of the
connection between electric and magnetic
phenomena. Oersted was also the first to
prepare pure aluminum.
Many historians of science believe that the compass, which uses a magnetic needle,
was used in China as early as the 13th century b.c., its invention being of Arab or
Indian origin. The phenomenon of magnetism was known to the Greeks as early as
about 800 b.c. They discovered that certain stones, made of a material now called
magnetite (Fe3O4), attracted pieces of iron.
In 1269, Pierre de Maricourt (c. 1220–?) mapped out the directions taken by a
magnetized needle when it was placed at various points on the surface of a spherical
natural magnet. He found that the directions formed lines that encircled the sphere
and passed through two points diametrically opposite each other, which he called
the poles of the magnet. Subsequent experiments have shown that every magnet,
regardless of its shape, has two poles, called north (N) and south (S), that exhibit
forces on each other in a manner analogous to electric charges. That is, similar poles
(N–N or S–S) repel each other and dissimilar poles (N–S) attract each other. The
poles received their names because of the behavior of a magnet in the presence of
the Earth’s magnetic field. If a bar magnet is suspended from its midpoint by a piece
of string so that it can swing freely in a horizontal plane, it rotates until its “north”
pole points to the north geographic pole of the Earth (which is a south magnetic
pole) and its “south” pole points to the Earth’s south geographic pole. (The same
idea is used to construct a simple compass.)
In 1600, William Gilbert (1544–1603) extended these experiments to a variety
of materials. Using the fact that a compass needle orients in preferred directions,
Gilbert suggested that magnets are attracted to land masses. In 1750, John Michell
(1724–1793) used a torsion balance to show that magnetic poles exert attractive or
repulsive forces on each other and that these forces vary as the inverse square of
their separation. Although the force between two magnetic poles is similar to the
force between two electric charges, an important difference exists. Electric charges
can be isolated (witness the electron and proton), whereas magnetic poles cannot
be isolated. That is, magnetic poles are always found in pairs. No matter how many
times a permanent magnet is cut, each piece always has a north pole and a south
pole. (Some theories speculate that magnetic monopoles—isolated north or south
poles—may exist in nature, and attempts to detect them currently make up an active
experimental field of investigation. None of these attempts has yet proven successful, however.)
The relationship between magnetism and electricity was discovered in 1819
when, while preparing for a lecture demonstration, Danish scientist Hans
Christian Oersted found that an electric current in a wire deflected a nearby
compass needle. Shortly thereafter, André-Marie Ampère (1775–1836) deduced
quantitative laws of magnetic force between current-carrying conductors. He also
suggested that electric current loops of molecular size are responsible for all magnetic phenomena.
In the 1820s, Faraday and, independently, Joseph Henry (1797–1878) identified
further connections between electricity and magnetism. They showed that an electric current could be produced in a circuit either by moving a magnet near the circuit or by changing the current in a nearby circuit. Their observations demonstrated
that a changing magnetic field produces an electric field. Years later, theoretical
work by James Clerk Maxwell showed that the reverse is also true: a changing electric
field gives rise to a magnetic field.
In this chapter, we shall investigate the effects of constant magnetic fields on
charges and currents, and study the sources of magnetic fields. In the next chapter,
we shall explore the effects of magnetic fields that vary in time.
22.2 | The Magnetic Field 745
22.2 | The Magnetic Field
In earlier chapters, we described the interaction between charged objects in terms of
electric fields. Recall that an electric field surrounds any stationary electric charge.
The region of space surrounding a moving charge includes a magnetic field in addition to the electric field. A magnetic field also surrounds any material with permanent
magnetism. We find that the magnetic field is a vector field, as is the electric field.
To describe any type of vector field, we must define its magnitude and its direc:
tion. The direction of the magnetic field vector B at any location is the direction in
which the north pole of a compass needle points at that location. Active Figure 22.1
shows how the magnetic field of a bar magnet can be traced with the aid of a compass, defining a magnetic field line, similar in many ways to the electric field lines
we studied in Chapter 19. Several magnetic field lines of a bar magnet traced out
in this manner are shown in the two-dimensional pictorial representation in Active
Figure 22.1. Magnetic field patterns can be displayed by small iron filings placed in
the vicinity of a magnet, as in Figure 22.2.
:
We can quantify the magnetic field B by using our model of a particle in a field, like the
model discussed for gravity in Chapter 11 and for electricity in Chapter 19. The existence
of a magnetic field at some point in space can be determined by measuring the magnetic
:
force FB exerted on an appropriate test particle placed at that point. This process is the
same one we followed in defining the electric field in Chapter 19. Our test particle will
be an electrically charged particle such as a proton. If we perform such an experiment,
we find the following results that are similar to those for experiments on electric forces:
Active Figure 22.1 Compass
needles can be used to trace the magnetic field lines in the region outside
a bar magnet.
• The magnetic force is proportional to the charge q of the particle.
• The magnetic force on a negative charge is directed opposite to the force on a
positive charge moving in the same direction.
:
• The magnetic force is proportional to the magnitude of the magnetic field vector B.
We also find the following results, which are totally different from those for experiments on electric forces:
• The magnetic force is proportional to the speed v of the particle.
• If the velocity vector makes an angle with the magnetic field, the magnitude of
the magnetic force is proportional to sin .
• When a charged particle moves parallel to the magnetic field vector, the magnetic force on the charge is zero.
• When a charged particle moves in a direction not parallel to the magnetic field
:
vector, the magnetic force acts in a direction perpendicular to both :
v and B ;
:
that is, the magnetic force is perpendicular to the plane formed by :
v and B .
These results show that the magnetic force on a particle is more complicated
than the electric force. The magnetic force is distinctive because it depends on the
Figure 22.2 Magnetic field
Courtesy of Henry Leap and Jim Lehman
Courtesy of Henry Leap and Jim Lehman
Courtesy of Henry Leap and Jim Lehman
patterns can be displayed with iron
filings sprinkled on paper near
magnets.
746 CHAPTER 22 | Magnetic Forces and Magnetic Fields
Figure 22.3 (a) The direction
:
of the magnetic force FB acting
on a charged particle moving with
a velocity :
v in the presence of a
:
magnetic field B. (b) Magnetic forces
on positive and negative charges. The
dashed lines show the paths of the
particles, which are investigated in
Section 22.3.
S
S
v
B
S S
v F
S
F
S
v
S
B
S
B
u
S
F
S
v
:
velocity of the particle and because its direction is perpendicular to both :
v and B.
Figure 22.3 shows the details of the direction of the magnetic force on a charged
particle. Despite this complicated behavior, these observations can be summarized
in a compact way by writing the magnetic force in the form
:
FB 5 q:
v
c Vector expression for the
magnetic force on a charged
particle moving in a magnetic field
:
22.1b
B
:
where the direction of the magnetic force is that of :
v
B, which, by definition of
:
the cross product, is perpendicular to both :
v and B. Equation 22.1 is analogous to
:
:
Equation 19.4, F e 5 qE , but is clearly more complicated. We can regard Equation 22.1 as an operational definition of the magnetic field at a point in space. The
SI unit of magnetic field is the tesla (T), where
1 T 5 1 N ? s/C ? m
Figure 22.4 reviews two right-hand rules for determining the direction of the cross
:
:
product :
v
B and determining the direction of FB. The rule in Figure 22.4a depends
on our right-hand rule for the cross product in Figure 10.13. You point the four fin:
gers of your right hand along the direction of :
v with the palm facing B and curl them
:
toward B. The extended thumb, which is at a right angle to the fingers, points in the
:
:
: :
direction of :
v
B. Because FB 5 q:
v
B, FB is in the direction of your thumb if q is
positive and opposite the direction of your thumb if q is negative.
A second rule is shown in Figure 22.4b. Here the thumb points in the direction
:
:
of :
v and the extended fingers in the direction of B. Now, the force FB on a positive
charge extends outward from your palm. The advantage of this rule is that the force
on the charge is in the direction that you would push on something with your hand,
outward from your palm. The force on a negative charge is in the opposite direction.
Feel free to use either of these two right-hand rules.
Figure 22.4 Two right-hand rules
for determining the direction of
:
:
the magnetic force F B 5 q:
v
B
acting on a particle with positive
charge q moving with a velocity :
v in
:
a magnetic field B. (a) In this rule,
the magnetic force is in the direction
in which your thumb points. (b) In
this rule, the magnetic force is in the
direction of your palm, as if you are
pushing the particle with your hand.
S
B
S
v
S
S
v
F
S
S
B
B
S
v
S
v
S
B
S
F
22.2 | The Magnetic Field 747
The magnitude of the magnetic force is
FB 5 u q uvB sin :
22.2b
:
where is the angle between v and B. From this expression, we see that FB is zero
:
when :
v is either parallel or antiparallel to B ( 5 0 or 1808). Furthermore, the force
:
:
has its maximum value FB 5 u q uvB when v is perpendicular to B ( 5 908).
Let’s summarize the important differences between electric and magnetic forces
on charged particles:
c Magnitude of the magnetic force
on a charged particle moving in a
magnetic field
• The electric force vector is along the direction of the electric field, whereas the
magnetic force vector is perpendicular to the magnetic field.
• The electric force acts on a charged particle regardless of whether the particle
is moving, whereas the magnetic force acts on a charged particle only when the
particle is in motion.
• The electric force does work in displacing a charged particle, whereas the magnetic force associated with a steady magnetic field does no work when a particle
is displaced.
This last statement is true because when a charge moves in a constant magnetic
field, the magnetic force is always perpendicular to the displacement of its point of
application. That is, for a small displacement d :
s of a particle, the work done by the
:
:
magnetic force on the particle is dW 5 F B d :
s 5 (FB :
v )dt 5 0 because the mag:
netic force is a vector perpendicular to v . From this property and the work–kinetic
energy theorem, we conclude that the kinetic energy of a charged particle cannot
be altered by a constant magnetic field alone. In other words, when a charge moves
with a velocity of :
v , an applied magnetic field can alter the direction of the velocity
vector, but it cannot change the speed of the particle.
In Figures 22.3 and 22.4, we used green arrows to represent magnetic field vectors, which will be the convention in this book. In Active Figure 22.1, we represented
the magnetic field of a bar magnet with green field lines. Studying magnetic fields
presents a complication that we avoided in electric fields. In our study of electric
fields, we drew all electric field vectors in the plane of the page or used perspective
to represent them directed at an angle to the page. The cross product in Equation
22.1 requires us to think in three dimensions for problems in magnetism. Therefore,
in addition to drawing vectors pointing left or right and up or down, we will need a
method of drawing vectors into or out of the page. These methods of representing
the vectors are illustrated in Figure 22.5. A vector coming out of the page is represented by a dot, which we can think of as the tip of the arrowhead representing the
vector coming through the paper toward us (Fig. 22.5a). A vector going into the
page is represented by a cross, which we can think of as the tail feathers of an arrow
going into the page (Fig. 22.5b). This depiction can be used for any type of vector
we will encounter: magnetic field, velocity, force, and so on.
QUICK QUIZ 22.1 An electron moves in the plane of this paper toward the top of the
page. A magnetic field is also in the plane of the page and directed toward the right.
What is the direction of the magnetic force on the electron? (a) toward the top of the
page (b) toward the bottom of the page (c) toward the left edge of the page (d) toward
the right edge of the page (e) upward out of the page (f) downward into the page
S
B
S
B
Figure 22.5 Representations of
magnetic field lines perpendicular to
the page.
THINKING PHYSICS 22.1
On a business trip to Australia, you take along your
U.S.-made compass that you used in your Boy Scout
days. Does this compass work correctly in Australia?
Reasoning Using the compass in Australia presents no
problem. The north pole of the magnet in the compass will be attracted to the south magnetic pole near
the north geographic pole, just as it was in the United
States. The only difference in the magnetic field lines
is that they have an upward component in Australia,
whereas they have a downward component in the
United States. When you hold the compass in a horizontal plane, it cannot detect the vertical component of
the field, however; it only displays the direction of the
horizontal component of the magnetic field. b
748 CHAPTER 22 | Magnetic Forces and Magnetic Fields
Example 22.1 | An Electron Moving in a Magnetic Field
An electron in an old-style television picture tube moves toward the front of the tube
with a speed of 8.0 3 106 m/s along the x axis (Fig. 22.6). Surrounding the neck of the
tube are coils of wire that create a magnetic field of magnitude 0.025 T, directed at an
angle of 608 to the x axis and lying in the xy plane. Calculate the magnetic force on the
electron.
SOLUTION
S
B
S
v
Conceptualize Recall that the magnetic force on a charged particle is perpendicular to the
plane formed by the velocity and magnetic field vectors. Use one of the right-hand rules in
Figure 22.4 to convince yourself that the direction of the force on the electron is downward
in Figure 22.6.
Categorize We evaluate the magnetic force using an equation developed in this section, so
we categorize this example as a substitution problem.
Use Equation 22.2 to find the magnitude of the magnetic
force:
FB 5 |q|vB sin S
F
Figure 22.6 (Example
:
22.1) The magnetic force F B
acting on the electron is in
the negative z direction when
:
:
v and B lie in the xy plane.
5 (1.6 3 10219 C)(8.0 3 106 m/s)(0.025 T)(sin 608)
5 2.8 3 10214 N
For practice using the vector product, evaluate this force in vector notation using Equation 22.1.
22.3 | Motion of a Charged Particle in a Uniform
Magnetic Field
S
F
S
B
S
v
S
F
S
F
S
v
S
F
S
v
In Section 22.2, we found that the magnetic force acting on a charged particle moving in a magnetic field is perpendicular to the particle’s velocity and consequently
the work done by the magnetic force on the particle is zero. Now consider the special case of a positively charged particle moving in a uniform magnetic field with
the initial velocity vector of the particle perpendicular to the field. Let’s assume
the direction of the magnetic field is into the page as in Active Figure 22.7. As the
particle changes the direction of its velocity in response to the magnetic force, the
magnetic force remains perpendicular to the velocity. As we found in Section 5.2, if
the force is always perpendicular to the velocity, the path of the particle is a circle!
Active Figure 22.7 shows the particle moving in a circle in a plane perpendicular to
the magnetic field. Although magnetism and magnetic forces may be new and unfamiliar to you now, we see a magnetic effect that results in something with which we
are familiar: the particle in uniform circular motion!
:
The particle moves in a circle because the magnetic force FB is perpendicular
:
to :
v and B and has a constant magnitude qvB. As Active Figure 22.7 illustrates, the
rotation is counterclockwise for a positive charge in a magnetic field directed into
the page. If q were negative, the rotation would be clockwise. We use the particle
under a net force model to write Newton’s second law for the particle:
S F 5 FB 5 ma
Because the particle moves in a circle, we also model it as a particle in uniform circular motion and we replace the acceleration with centripetal acceleration:
FB 5 qvB 5
Active Figure 22.7 When the velocity of a charged particle is perpendicular to a uniform magnetic field,
the particle moves in a circular path
:
in a plane perpendicular to B.
mv 2
r
This expression leads to the following equation for the radius of the circular path:
r5
mv
qB
22.3b
22.3 | Motion of a Charged Particle in a Uniform Magnetic Field 749
That is, the radius of the path is proportional to the linear momentum mv of the particle and inversely proportional to the magnitude of the charge on the particle and to the
magnitude of the magnetic field. The angular speed of the particle is (from Eq. 10.10)
5
qB
v
5
r
m
22.4b
The period of the motion (the time interval required for the particle to complete
one revolution) is equal to the circumference of the circular path divided by the
speed of the particle:
T5
2r
2
2m
5
5
v
qB
22.5b
These results show that the angular speed of the particle and the period of the circular
motion do not depend on the translational speed of the particle or the radius of the
orbit for a given particle in a given uniform magnetic field. The angular speed is
often referred to as the cyclotron frequency because charged particles circulate at this
angular speed in one type of accelerator called a cyclotron, discussed in Section 22.4.
If a charged particle moves in a uniform magnetic field with its velocity at some
:
arbitrary angle to B, its path is a helix. For example, if the field is in the x direction as in Active Figure 22.8, there is no component of force on the particle in the
x direction. As a result, ax 5 0, and so the x component of velocity of the particle
:
v
B causes the components vy and vz to
remains constant. The magnetic force q:
change in time, however, and the resulting motion of the particle is a helix having
its axis parallel to the magnetic field. The projection of the path onto the yz plane
(viewed along the x axis) is a circle. (The projections of the path onto the xy and xz
planes are sinusoids!) Equations 22.3 to 22.5 still apply provided that v is replaced
by v' 5 "vy2 1 vz2. In the yz plane, the charged particle is modeled as a particle in
uniform circular motion as well as a particle under a net force. In the x direction, the
charged particle is modeled as a particle under constant velocity.
S
B
Active Figure 22.8 A charged
particle having a velocity vector
with a component parallel to a
uniform magnetic field moves in a
helical path.
Q U I CK QUI Z 22.2 A charged particle is moving perpendicular to a magnetic field
in a circle with a radius r. (i) An identical particle enters the field, with :
v perpendic:
ular to B , but with a higher speed than the first particle. Compared with the radius
of the circle for the first particle, is the radius of the circular path for the second particle (a) smaller, (b) larger, or (c) equal in size? (ii) The magnitude of the magnetic
field is increased. From the same choices, compare the radius of the new circular
path of the first particle with the radius of its initial path.
THINKING PHYSICS 22.2
Suppose a uniform magnetic field exists in a finite region of space as in Figure 22.9.
Can you inject a charged particle into this region and have it stay trapped in the
region by the magnetic force?
S
v
Reasoning Consider separately the components of the particle velocity parallel
and perpendicular to the field lines in the region. For the component parallel
to the field lines, no force is exerted on the particle and it continues to move
with the parallel component until it leaves the region of the magnetic field.
Now consider the component perpendicular to the field lines. This component
results in a magnetic force that is perpendicular to both the field lines and the
velocity component. As discussed earlier, if the force acting on a charged particle is always perpendicular to its velocity, the particle moves in a circular path.
Therefore, the particle follows half of a circular arc and exits the field on the
other side of the circle, as shown in Figure 22.9. Therefore, a particle injected
into a uniform magnetic field cannot stay trapped in the field region. b
S
F
Figure 22.9 (Thinking
Physics 22.2) A positively charged
particle enters a region of magnetic
field directed out of the page.
750 CHAPTER 22 | Magnetic Forces and Magnetic Fields
Example 22.2 | A Proton Moving Perpendicular to a Uniform Magnetic Field
A proton is moving in a circular orbit of radius 14 cm in a uniform 0.35-T magnetic field perpendicular to the velocity of
the proton. Find the speed of the proton.
SOLUTION
Conceptualize From our discussion in this section, we know the proton follows a circular path when moving perpendicular
to a uniform magnetic field.
Categorize We evaluate the speed of the proton using an equation developed in this section, so we categorize this example
as a substitution problem.
qBr
v5
Solve Equation 22.3 for the speed of the particle:
mp
Substitute numerical values:
v5
(1.60 3 10219 C)(0.35 T)(0.14 m)
1.67 3 10227 kg
5 4.7 3 106 m/s
What If? What if an electron, rather than a proton, moves in a direction perpendicular to the same magnetic field with
this same speed? Will the radius of its orbit be different?
Answer An electron has a much smaller mass than a proton, so the magnetic force should be able to change its velocity
much more easily than that for the proton. Therefore, we expect the radius to be smaller. Equation 22.3 shows that r is
proportional to m with q, B, and v the same for the electron as for the proton. Consequently, the radius will be smaller by
the same factor as the ratio of masses me /mp .
In an experiment designed to measure the magnitude of a uniform
magnetic field, electrons are accelerated from rest through a potential
difference of 350 V and then enter a uniform magnetic field that is perpendicular to the velocity vector of the electrons. The electrons travel
along a curved path because of the magnetic force exerted on them,
and the radius of the path is measured to be 7.5 cm. (Such a curved
beam of electrons is shown in Fig. 22.10.)
(A) What is the magnitude of the magnetic field?
Figure 22.10
(Example 22.3)
The bending
of an electron
beam in a
magnetic field.
SOLUTION
Conceptualize This example involves electrons accelerating from rest due to an electric force and then moving in a circular
path due to a magnetic force. With the help of Active Figure 22.7 and Figure 22.10, visualize the circular motion of the electrons.
Categorize Equation 22.3 shows that we need the speed v of the electron to find the magnetic field magnitude, and v is
not given. Consequently, we must find the speed of the electron based on the potential difference through which it is accelerated. To do so, we categorize the first part of the problem by modeling an electron and the electric field as an isolated
system. Once the electron enters the magnetic field, we categorize the second part of the problem as one similar to those
we have studied in this section.
Analyze Write the appropriate reduction of the conservation of energy equation, Equation 7.2, for the
electron–electric field system:
DK 1 DU 5 0
Substitute the appropriate initial and final energies:
(12m ev 2 2 0) 1 (q DV ) 5 0
Solve for the speed of the electron:
v5
22q DV
B me
Henry Leap and Jim Lehman
Example 22.3 | Bending an Electron Beam
22.4 | Applications Involving Charged Particles Moving in a Magnetic Field 751
22.3 cont.
Substitute numerical values:
Now imagine the electron entering the magnetic field
with this speed. Solve Equation 22.3 for the magnitude
of the magnetic field:
Substitute numerical values:
v5
22(21.60 3 10219 C)(350 V)
5 1.11 3 107 m/s
9.11 3 10231 kg
B
B5
me v
er
B5
(9.11 3 10 231 kg)(1.11 3 107 m/s)
(1.60 3 10 219 C)(0.075 m)
5 8.4 3 1024 T
(B) What is the angular speed of the electrons?
SOLUTION
5
Use Equation 10.10:
1.11 3 107 m/s
v
5
5 1.5 3 108 rad/s
r
0.075 m
Finalize The angular speed can be represented as 5 (1.5 3 108 rad/s)(1 rev/2 rad) 5 2.4 3 107 rev/s. The electrons
travel around the circle 24 million times per second! This answer is consistent with the very high speed found in part (A).
What If? What if a sudden voltage surge causes the accelerating voltage to increase to 400 V? How does that affect
the angular speed of the electrons, assuming the magnetic
field remains constant?
Answer The increase in accelerating voltage DV causes the
electrons to enter the magnetic field with a higher speed
v. This higher speed causes them to travel in a circle with a
larger radius r. The angular speed is the ratio of v to r. Both
v and r increase by the same factor, so the effects cancel and
the angular speed remains the same. Equation 22.4 is an
expression for the cyclotron frequency, which is the same
as the angular speed of the electrons. The cyclotron frequency depends only on the charge q , the magnetic field
B, and the mass me , none of which have changed. Therefore, the voltage surge has no effect on the angular speed.
(In reality, however, the voltage surge may also increase the
magnetic field if the magnetic field is powered by the same
source as the accelerating voltage. In that case, the angular
speed increases according to Eq. 22.4.)
22.4 | Applications Involving Charged Particles
Moving in a Magnetic Field
:
A charge moving with velocity :
v in the presence of an electric field E and a magnetic
:
:
:
field B experiences both an electric force qE and a magnetic force q:
v
B. The total
force, called the Lorentz force, acting on the charge is therefore the vector sum,
:
:
:
F 5 qE 1 q v
:
B
S
22.6b
In this section, we look at three applications involving particles experiencing the
Lorentz force.
S
B
S
F
E
S
v
S
F
Velocity Selector
In many experiments involving moving charged particles, it is important that all particles move with essentially the same velocity, which can be achieved by applying a
combination of an electric field and a magnetic field oriented as shown in Active Figure 22.11. A uniform electric field is directed to the right (in the plane of the page in
Active Fig. 22.11), and a uniform magnetic field is applied in the direction perpendicular to the electric field (into the page in Active Fig. 22.11). If q is positive and the
:
:
velocity :
v is upward, the magnetic force q:
v
B is to the left and the electric force qE
is to the right. When the magnitudes of the two fields are chosen so that qE 5 qvB, the
charged particle is modeled as a particle in equilibrium and moves in a straight vertical line through the region of the fields. From the expression qE 5 qvB, we find that
v5
E
B
22.7b
Active Figure 22.11 A velocity selector. When a positively charged particle is moving with velocity :
v in the
presence of a magnetic field directed
into the page and an electric field
directed to the right, it experiences
:
an electric force q E to the right and a
:
:
magnetic force q v
B to the left.
752 CHAPTER 22 | Magnetic Forces and Magnetic Fields
Only those particles having this speed pass undeflected through the mutually perpendicular electric and magnetic fields. The magnetic force exerted on particles moving
at speeds greater than that is stronger than the electric force, and the particles are
deflected to the left. Those moving at slower speeds are deflected to the right.
The Mass Spectrometer
S
A mass spectrometer separates ions according to their mass-to-charge ratio. In one
version of this device, known as the Bainbridge mass spectrometer, a beam of ions first
:
passes through a velocity selector and then enters a second uniform magnetic field B0
that has the same direction as the magnetic field in the selector (Active Fig. 22.12).
Upon entering the second magnetic field, the ions move in a semicircle of radius r before striking a detector array at P. If the ions are positively charged, the beam deflects
to the left as Active Figure 22.12 shows. If the ions are negatively charged, the beam
deflects to the right. From Equation 22.3, we can express the ratio m/q as
B
B
S
rB0
m
5
q
v
v
S
Using Equation 22.7 gives
rB0B
m
5
q
E
S
E
Active Figure 22.12 A mass spectrometer. Positively charged particles
are sent first through a velocity selector and then into a region where the
:
magnetic field B0 causes the particles
to move in a semicircular path and
strike a detector array at P.
22.8b
Therefore, we can determine m/q by measuring the radius of curvature and knowing
the field magnitudes B, B0, and E. In practice, one usually measures the masses of
various isotopes of a given ion, with the ions all carrying the same charge q. In this
way, the mass ratios can be determined even if q is unknown.
A variation of this technique was used by J. J. Thomson (1856–1940) in 1897 to
measure the ratio e/me for electrons. Figure 22.13a shows the basic apparatus he used.
Electrons are accelerated from the cathode and pass through two slits. They then drift
into a region of perpendicular electric and magnetic fields. The magnitudes of the two
fields are first adjusted to produce an undeflected beam. When the magnetic field is
turned off, the electric field produces a measurable beam deflection that is recorded
on the fluorescent screen. From the size of the deflection and the measured values of
E and B, the charge-to-mass ratio can be determined. The results of this crucial experiment represent the discovery of the electron as a fundamental particle of nature.
Figure 22.13 (a) Thomson’s apparatus for measuring e/me . (b) J. J. Thomson (left) in the Cavendish
Laboratory, University of Cambridge. The man on the right, Frank Baldwin Jewett, is a distant relative of John
W. Jewett Jr., coauthor of this text.
Reprinted with the permission of Alcatel-Lucent USA Inc.
22.4 | Applications Involving Charged Particles Moving in a Magnetic Field 753
The Cyclotron
A cyclotron can accelerate charged particles to very high speeds. Both electric and
magnetic forces play a key role in its operation. The energetic particles produced are
used to bombard atomic nuclei and thereby produce nuclear reactions of interest
to researchers. A number of hospitals use cyclotron facilities to produce radioactive
substances for diagnosis and treatment as well as beams of high-energy particles for
treating cancer. At the time of this printing, there are 37 proton therapy centers
around the world. These centers use either cyclotrons or another particle accelerator, called a synchrotron, to accelerate protons to high speeds to be used in external
beam radiotherapy for cancer. (See the cover photo.) Conditions that have been
treated with proton therapy include prostate cancer, retinoblastoma (a cancer of the
eye), head and neck cancers, ocular melanoma, and acoustic neuroma.
A schematic drawing of a cyclotron is shown in Figure 22.14a. The charges move
inside two hollow metal semicircular containers, D1 and D2, referred to as dees because
they are shaped like the letter D. A high-frequency alternating potential difference is
applied to the dees, and a uniform magnetic field is directed perpendicular to them.
A positive ion released at P near the center of the magnet moves in a semicircular path
in one dee (indicated by the dashed black line in the drawing) and arrives back at the
gap in a time interval T/2, where T is the time interval needed to make one complete
trip around the two dees, given by Equation 22.5. The frequency of the applied potential difference is chosen so that the polarity of the dees is reversed during the time
interval in which the ion travels around one dee. If the applied potential difference is
adjusted such that D2 is at a lower electric potential than D1 by an amount DV, the ion
accelerates across the gap to D2 and its kinetic energy increases by an amount q DV. It
then moves around D2 in a semicircular path of larger radius (because its speed has
increased). After a time interval T/2, it again arrives at the gap between the dees. By
this time, the polarity across the dees has reversed again and the ion is given another
“kick” across the gap. The motion continues so that for each half-circle trip, the ion
gains additional kinetic energy equal to q DV. When the radius of its path is nearly that
of the dees, the energetic ion leaves the system through the exit slit. It is important to
note that the operation of the cyclotron is based on T being independent of the speed
of the ion and the radius of its circular path (Eq. 22.5).
We can obtain an expression for the kinetic energy of the ion when it exits from
the cyclotron in terms of the radius R of the dees. From Equation 22.3 we know that
v 5 qBR/m. Hence, the kinetic energy is
K 5 12mv 2 5
q 2B 2R 2
2m
Use of cyclotrons in medicine
Pitfall Prevention | 22.1
The Cyclotron Is Not State-of-theArt Technology
The cyclotron is important
historically because it was the first
particle accelerator to produce
particles with very high speeds.
Cyclotrons are still in use in medical
applications, but most accelerators
currently in research use are not
cyclotrons. Research accelerators
work on a different principle and are
generally called synchrotrons.
22.9b
S
Lawrence Berkeley National Lab
B
Figure 22.14 (a) A cyclotron consists of an ion source at P, two dees, D1 and D2, across which an alternating potential difference
is applied, and a uniform magnetic field. (b) The first cyclotron, invented by E. O. Lawrence and M. S. Livingston in 1934.
754 CHAPTER 22 | Magnetic Forces and Magnetic Fields
When the energy of the ions in a cyclotron exceeds about 20 MeV, relativistic effects come into play. For this reason, the moving ions do not remain in phase with
the applied potential difference. Some accelerators solve this problem by modifying
the frequency of the applied potential difference so that it remains in phase with the
moving ions.
22.5 | Magnetic Force on a Current-Carrying
Conductor
Because a magnetic force is exerted on a single charged particle when it moves through
an external magnetic field, it should not surprise you to find that a current-carrying
wire also experiences a magnetic force when placed in an external magnetic field because the current represents a collection of many charged
particles in motion. Hence, the resultant magnetic force
on the wire is due to the sum of the individual magnetic
forces on the charged particles. The force on the particles
is transmitted to the “bulk” of the wire through collisions
with the atoms making up the wire.
The magnetic force on a current-carrying conductor can be demonstrated by hanging a wire between the
poles of a magnet as in Figure 22.15, where the magnetic field is directed into the page. The wire deflects
to the left or right when a current is passed through it.
Let us quantify this discussion by considering a
straight segment of wire of length L and cross-sectional
S
S
S
B
B
B
area A, carrying a current I in a uniform external mag:
netic field B as in Figure 22.16. As a simplification
model, we shall ignore the high-speed zigzag motion of
the charges in the wire (which is valid because the net
velocity associated with this motion is zero) and assume
that the charges simply move with the drift velocity :
vd .
The magnetic force on a charge q moving with drift ve:
locity :
vd is q:
v d B. To find the total magnetic force
on the wire segment, we multiply the magnetic force
Figure 22.15 (a) A wire suspended vertically between the poles of a
on one charge by the number of charges in the segmagnet. (b) through (d) The setup shown in (a) as seen looking at the
south pole of the magnet so that the magnetic field (green crosses) is
ment. Because the volume of the segment is AL, the
directed into the page.
number of charges in the segment is nAL, where n is
the number of charges per unit volume. Hence, the total magnetic force on the wire
of length L is
S
:
FB 5 (q:
vd
S
v ⴛB
B)nAL
This equation can be written in a more convenient form by noting that, from
:
Equation 21.4, the current in the wire is I 5 nqvdA. Therefore, FB can be expressed as
S
F
:
:
FB 5 I L
S
:
22.10b
B
:
B
:
where L is a vector in the direction of the current I; the magnitude of L equals the
length of the segment. Note that this expression applies only to a straight segment
of wire in a uniform external magnetic field.
Now consider an arbitrarily shaped wire of uniform cross-section in an external magnetic field as in Figure 22.17. It follows from Equation 22.10 that the magnetic force
:
on a very small segment of the wire of length ds in the presence of an external field B is
S
:
v
:
d FB 5 Id :
s
S
S
LⴛB
Figure 22.16 A segment of a current:
carrying wire in a magnetic field B.
:
:
B
22.11b
where d s is a vector representing the length segment, with its direction the same as
:
that of the current, and d FB is directed out of the page for the directions assumed
:
in Figure 22.17. We can consider Equation 22.11 as an alternative definition of B to
:
Equation 22.1. That is, the field B can be defined in terms of a measurable force on
22.5 | Magnetic Force on a Current-Carrying Conductor 755
:
a current element, where the force is a maximum when B is perpendicular to the
:
element and zero when B is parallel to the element.
:
To obtain the total magnetic force FB on a length of the wire between arbitrary points
a and b, we integrate Equation 22.11 over the length of the wire between these points:
:
FB 5 I
E
b
:
d:
s
S
22.12b
B
S
S
sⴛB
s
S
a
B
S
s
When this integration is carried out, the magnitude of the magnetic field and the
direction of the field relative to the vector d :
s may vary from point to point.
QUICK QUIZ 22.3 A wire carries current in the plane of this paper toward the top
of the page. The wire experiences a magnetic force toward the right edge of the page.
Is the direction of the magnetic field causing this force (a) in the plane of the page
and toward the left edge, (b) in the plane of the page and toward the bottom edge,
(c) upward out of the page, or (d) downward into the page?
Figure 22.17 A wire segment of
arbitrary shape carrying a current I
:
in a magnetic field B experiences a
magnetic force.
THINKING PHYSICS 22.3
current. Therefore, the vector d :
s is upward, and the
magnetic field vector has a northward component. According to the cross product of the length element and
magnetic field vectors (Eq. 22.11), the lightning stroke
would be deflected to the west. b
In a lightning stroke, negative charge rapidly moves
from a cloud to the ground. In what direction is a lightning stroke deflected by the Earth’s magnetic field?
Reasoning The downward flow of negative charge in
a lightning stroke is equivalent to an upward-moving
Exampl e 22.4 | Force on a Semicircular Conductor
S
B
A wire bent into a semicircle of radius R forms a closed circuit and carries a current
I. The wire lies in the xy plane, and a uniform magnetic field is directed along the
positive y axis as in Figure 22.18. Find the magnitude and direction of the magnetic
force acting on the straight portion of the wire and on the curved portion.
u
S
s
u
SOLUTION
u
:
Conceptualize Using the right-hand rule for cross products, we see that the force F 1
:
on the straight portion of the wire is out of the page and the force F 2 on the curved
:
:
portion is into the page. Is F 2 larger in magnitude than F 1 because the length of the
curved portion is longer than that of the straight portion?
Categorize Because we are dealing with a current-carrying wire in a magnetic field
rather than a single charged particle, we must use Equation 22.12 to find the total
force on each portion of the wire.
:
Analyze Notice that d :
s is perpendicular to B
everywhere on the straight portion of the wire. Use
Equation 22.12 to find the force on this portion:
:
F1 5 I
E
b
d:
s
:
From the geometry in Figure 22.18, write an
expression for ds :
(2) ds 5 R du
E
F2 5 2
E
magnetic force on the straight portion
of the loop is directed out of the page,
and the magnetic force on the curved
portion is directed into the page.
R
B dx k̂ 5 2IRB k̂
2R
s
(1) dF 2 5 I d :
:
B5I
a
To find the magnetic force on the curved part,
first write an expression for the magnetic force
:
dF 2 on the element d :
s in Figure 22.18:
Substitute Equation (2) into Equation (1) and
integrate over the angle from 0 to :
:
Figure 22.18 (Example 22.4) The
0
:
B 5 2IB sin u ds kˆ
ˆ 5 2IRB
IRB sin d k
E
0
sin d k̂ 5 2IRB[2cos ]0 k̂
5 IRB(cos 2 cos 0)k̂ 5 IRB(21 2 1)k̂ 5 22IRB k̂
continued
756 CHAPTER 22 | Magnetic Forces and Magnetic Fields
22.4 cont.
Finalize Two very important general statements follow from this example. First, the force on the curved portion is the same
in magnitude as the force on the straight wire between the same two points. In general, the magnetic force on a curved
current-carrying wire in a uniform magnetic field is equal to that on a straight wire connecting the endpoints and carrying
:
:
the same current. Furthermore, F 1 1 F 2 5 0 is also a general result: the net magnetic force acting on any closed current
loop in a uniform magnetic field is zero.
쩸
22.6 | Torque on a Current Loop in a Uniform
쩺
S
B
S
In the preceding section, we showed how a magnetic force is exerted on a currentcarrying conductor when the conductor is placed in an external magnetic field.
Starting at this point, we shall show that a torque is exerted on a current loop placed
in a magnetic field. The results of this analysis are of great practical value in the design of motors and generators.
Consider a rectangular loop carrying a current I in the presence of a uniform external magnetic field in the plane of the loop as in Figure 22.19a. The magnetic forces on
sides ➀ and ➂, of length b, are zero because these wires are parallel to the field; hence,
:
d:
s
B 5 0 for these sides. Nonzero magnetic forces act on sides ➁ and ➃, however, because these sides are oriented perpendicular to the field. The magnitude of these forces is
쩸
B
쩹
쩻
쩺
쩹
Magnetic Field
F2 5 F4 5 IaB
쩻
:
S
S
F
F
쩹
쩻
We see that the net force on the loop is zero. The direction of F 2, the magnetic force
:
on side ➁, is out of the paper, and that of F 4, the magnetic force on side ➃, is into
the paper. If we look across the plane of the loop from side ➂ as in Figure 22.19b, we
see the forces on ➁ and ➃ directed as shown. If we assume that the loop is pivoted so
that it can rotate about an axis perpendicular to the page and passing through point
O, we see that these two magnetic forces produce a net torque about this axis that
rotates the loop clockwise. The magnitude of the torque, which we will call max , is
max 5 F2
b
b
b
b
1 F4 5 (IaB) 1 (IaB) 5 IabB
2
2
2
2
where the moment arm about this axis is b/2 for each force. Because the area of the
loop is A 5 ab, the magnitude of the torque can be expressed as
S
F
쩹
쩻
S
B
S
F
Figure 22.19 (a) Overhead view
of a rectangular current loop in a
uniform magnetic field. (b) Edge
view of the loop looking from side
➂. The purple dot in the left circle
represents current in wire ➁ coming
toward you; the purple 3 in the right
circle represents current in wire ➃
moving away from you.
max 5 IAB
22.13b
:
Remember that this torque occurs only when the field B is parallel to the plane of
the loop. The sense of the rotation is clockwise when the loop is viewed as in Figure 22.19b. If the current were reversed, the magnetic forces would reverse their
directions and the rotational tendency would be counterclockwise.
Now suppose the loop is rotated so that a line perpendicular to the plane of the loop
makes an angle with the uniform magnetic field as in Active Figure 22.20. Notice that
:
B is still perpendicular to sides ➁ and ➃. In this case, the magnetic forces on sides ➀
and ➂ cancel each other and produce no torque because they have the same line of
:
:
action. The magnetic forces F 2 and F 4 acting on sides ➁ and ➃, however, both produce
a torque about an axis through the center of the loop. Referring to Active Figure 22.20,
:
we note that the moment arm of F 2 about this axis is (b/2) sin . Likewise, the moment
:
arm of F 4 is also (b/2) sin . Because F2 5 F4 5 IaB, the net torque has the magnitude
b
b
sin 1 F4 sin 2
2
b
b
5 (IaB) sin 1 (IaB) sin 5 IabB sin 2
2
5 F2
1
5 IAB sin 2
1
2
22.6 | Torque on a Current Loop in a Uniform Magnetic Field 757
where A 5 ab is the area of the loop. This result shows that the torque has its
maximum value IAB (Eq. 22.13) when the field is parallel to the plane of the loop
( 5 908) and is zero when the field is perpendicular to the plane of the loop
( 5 0). As we see in Active Figure 22.20, the loop tends to rotate in the direction
of decreasing values of (i.e., so that the normal to the plane of the loop rotates
toward the direction of the magnetic field). A convenient vector expression for
the torque is
:
:
:
5 IA
S
F
쩹
S
A
u
u
u
쩻
S
22.14b
B
S
B
F
:
where A, a vector perpendicular to the plane of the loop (Active Fig. 22.20), has a
:
magnitude equal to the area of the loop. The sense of A is determined by the righthand rule illustrated in Figure 22.21. When the four fingers of the right hand are
curled in the direction of the current in the loop, the thumb points in the direction
:
:
of A. The product I A is defined to be the magnetic dipole moment : (often simply
called the “magnetic moment”) of the loop:
:
:
22.15b
5 IA
u
u
Active Figure 22.20 An end view
of the loop in Figure 22.19 with the
normal to the loop at an angle with
respect to the magnetic field.
The SI unit of magnetic dipole moment is the ampere ? meter2 (A ? m2). Using this
definition, the torque can be expressed as
:
5:
:
22.16b
B
c Torque on a magnetic moment
in a magnetic field
:
Although the torque was obtained for a particular orientation of B with respect
to the loop, Equation 22.16 is valid for any orientation. Furthermore, although
the torque expression was derived for a rectangular loop, the result is valid for a
loop of any shape. Once the torque is determined, the loop can be modeled as
a rigid object under a net torque, which we studied in Chapter 10.
If a coil consists of multiple loops such that there are N turns of
S
m
wire, each carrying the same current and each having the same area,
the total magnetic moment of the coil is the product of the number
:
S
of turns and the magnetic moment for one turn, : 5 NIA. ThereA
S
fore, the torque on an N-turn coil is N times greater than that on a
m
one-turn loop.
A common electric motor consists of a coil of wire mounted so
S
that it can rotate in the field of a permanent magnet. The torque
A
on the current-carrying coil is used to rotate a shaft that drives a mechanical device such as the power windows in your car, your household fan, or your electric hedge trimmer.
Imagine the loop in Active Figure 22.20 released from rest.
:
The magnetic moment vector (parallel to A ) will begin to rotate
:
clockwise to line up with the magnetic field vector B . Once : is
:
rule for determining the
aligned with B , which is the equilibrium configuration, the angu- Figure 22.21 Right-hand
:
direction of the vector A . The direction of the magnetic
lar momentum of the loop will carry it past this configuration and moment : is the same as the direction of :
A.
it will slow down due to the restoring torque. The result will be an
oscillation around the equilibrium configuration. Let’s ask a couple of questions
about this situation. From where did the energy come that is associated with the
oscillation of the loop-field system? It came from the work done by an external
agent in initially rotating : away from the equilibrium position. Now, in what
form is the energy in the system before the loop is released? It is in the form of
potential energy, just like when a block on a spring is moved away from the equilibrium configuration. The potential energy of a system of a magnetic dipole in
a magnetic field depends on the orientation of the dipole in the magnetic field
and is given by
U 5 2:
:
B
22.17b
This expression shows that the system has its lowest energy Umin 5 2B when :
:
points in the same direction as B. The system has its highest energy Umax 5 1B
:
when : points in the direction opposite B.
c Potential energy of a system of
a magnetic moment in a magnetic
field
758 CHAPTER 22 | Magnetic Forces and Magnetic Fields
Example 22.5 | The Magnetic Dipole Moment of a Coil
A rectangular coil of dimensions 5.40 cm 3 8.50 cm consists of 25 turns of wire and carries a current of 15.0 mA. A 0.350-T
magnetic field is applied parallel to the plane of the coil.
(A) Calculate the magnitude of the magnetic dipole moment of the coil.
SOLUTION
Conceptualize The magnetic moment of the coil is independent of any magnetic field in which the loop resides, so it depends only on the geometry of the loop and the current it carries.
Categorize We evaluate quantities based on equations developed in this section, so we categorize this example as a substitution problem.
Use Equation 22.15 to calculate the magnetic moment
associated with a coil consisting of N turns:
coil 5 NIA 5 (25)(15.0 3 1023 A)(0.054 0 m)(0.085 0 m)
5 1.72 3 1023 A ? m2
(B) What is the magnitude of the torque acting on the loop?
SOLUTION
:
Use Equation 22.16, noting that B is perpendicular to :coil:
5 coilB 5 (1.72 3 1023 A ? m2)(0.350 T)
5 6.02 3 1024 N ? m
22.7 | The Biot–Savart Law
S
B
r u
S
s
r
S
B
Figure 22.22 The magnetic field
:
d B at a point P due to a current I
through a length element d :
s is given
by the Biot–Savart law.
Pitfall Prevention | 22.2
The Biot–Savart Law
The magnetic field described by the
Biot–Savart law is the field due to a
given current-carrying conductor. Do
not confuse this field with any external
field that may be applied to the
conductor from some other source.
In the previous sections, we investigated the result of placing an object in an existing
magnetic field. When a moving charge is placed in the field, it experiences a magnetic force. A current-carrying wire placed in the field also experiences a magnetic
force; a current loop in the field experiences a torque.
Now we shift our thinking and investigate the source of the magnetic field. Oersted’s 1819 discovery (Section 22.1) that an electric current in a wire deflects a nearby
compass needle indicates that a current acts as a source of a magnetic field. From
their investigations on the force between a current-carrying conductor and a magnet
in the early 19th century, Jean-Baptiste Biot and Félix Savart arrived at an expression
for the magnetic field at a point in space in terms of the current that produces the
field. No point currents exist comparable to point charges (because we must have
a complete circuit for a current to exist). Hence, we must investigate the magnetic
field due to an infinitesimally small element of current that is part of a larger current
distribution. Suppose the current distribution is a wire carrying a steady current I as
:
in Figure 22.22. Experimental results show that the magnetic field d B at point P created by an element of infinitesimal length ds of the wire has the following properties:
:
s (which is in the direction of the cur• The vector d B is perpendicular both to d :
rent) and to the unit vector r̂ directed from the element toward P.
:
• The magnitude of d B is inversely proportional to r 2, where r is the distance from
the element to P.
:
• The magnitude of d B is proportional to the current I and to the length ds of the
element.
:
• The magnitude of d B is proportional to sin , where is the angle between d :
s
and r̂.
The Biot–Savart law describes these results and can be summarized in the following
compact form:
:
d B 5 km
I d:
s
r̂
r2
22.18b
22.7 | The Biot–Savart Law 759
where km is a constant that in SI units is exactly 1027 T ? m/A. The constant km is
usually written 0/4, where 0 is another constant, called the permeability of
free space:
0
4
5 km 5 1027 T ? m/A
0 5 4km 5 4 3 1027 T ? m/A
22.19b
c Permeability of free space
22.20b
c Biot–Savart law
Hence, the Biot–Savart law, Equation 22.18, can also be written
:
dB 5
0 I d :
s
r̂
r2
4
It is important to note that the Biot–Savart law gives the magnetic field at a point
only for a small length element of the conductor. We identify the product I d :
s as a
:
current element. To find the total magnetic field B at some point due to a conductor of finite size, we must sum contributions from all current elements making up
:
the conductor. That is, we evaluate B by integrating Equation 22.20 over the entire
conductor.
There are two similarities between the Biot–Savart law of magnetism and Equation 19.7 for the electric field of a charge distribution, and there are two important
differences. The current element I d :
s produces a magnetic field, and the charge
element dq produces an electric field. Furthermore, the magnitude of the magnetic field varies as the inverse square of the distance from the current element,
as does the electric field due to a charge element. The directions of the two fields
are quite different, however. The electric field due to a charge element is radial;
:
in the case of a positive point charge, E is directed away from the charge. The
magnetic field due to a current element is perpendicular to both the current element and the radius vector. Hence, if the conductor lies in the plane of the page,
:
as in Figure 22.22, d B points out of the page at the point P and into the page at
P 9. Another important difference is that an electric field can be a result either of a
single charge or a distribution of charges, but a magnetic field can only be a result
of a current distribution.
Figure 22.23 shows a convenient right-hand rule for determining the direction of
the magnetic field due to a current. Note that the field lines generally encircle the
current. In the case of current in a long, straight wire, the field lines form circles
that are concentric with the wire and are in a plane perpendicular to the wire. If the
wire is grasped in the right hand with the thumb in the direction of the current, the
:
fingers will curl in the direction of B.
Although the magnetic field due to an infinitely long, current-carrying wire can
be calculated using the Biot–Savart law (Problem 70), in Section 22.9 we use a
different method to show that the magnitude of this field at a distance r from the
wire is
B5
0I
2r
22.21b
Q U I CK QUI Z 22.4 Consider the magnetic field due to the current in the wire shown in
Figure 22.24. Rank the points A, B, and C in terms of magnitude of the magnetic field
that is due to the current in just the length element d :
s shown, from greatest to least.
S
s
Figure 22.24 (Quick Quiz 22.4)
Where is the magnetic field the
greatest?
S
B
Figure 22.23 The right-hand
rule for determining the direction
of the magnetic field surrounding a
long, straight wire carrying a current.
Note that the magnetic field lines
form circles around the wire. The
magnitude of the magnetic field at
a distance r from the wire is given by
Equation 22.21.
c Magnetic field due to a long,
straight wire
760 CHAPTER 22 | Magnetic Forces and Magnetic Fields
Example 22.6 | Magnetic Field on the Axis of a Circular Current Loop
Consider a circular wire loop of radius a located in the yz
plane and carrying a steady current I as in Figure 22.25. Calculate the magnetic field at an axial point P a distance x from
the center of the loop.
S
s
r
u
SOLUTION
S
芯
Conceptualize Figure 22.25 shows the magnetic field contribu-
B
:
tion d B at P due to a single current element at the top of the
ring. This field vector can be resolved into components dBx parallel to the axis of the ring and dB' perpendicular to the axis.
u
Think about the magnetic field contributions from a current element at the bottom of the loop. Because of the symmetry of the
Figure 22.25 (Example 22.6) Geometry for calculating the
situation, the perpendicular components of the field due to elemagnetic field at a point P lying on the axis of a current loop.
ments at the top and bottom of the ring cancel. This cancellation
:
By symmetry, the total field B is along this axis.
occurs for all pairs of segments around the ring, so we can ignore
the perpendicular component of the field and focus solely on the parallel components, which simply add.
Categorize We are asked to find the magnetic field due to a simple current distribution, so this example is a typical problem for which the Biot–Savart law is appropriate.
Analyze In this situation, every length element d :
s is perpendicular to the vector r̂ at the location of the element. There-
fore, for any element, u d :
s
r̂ u 5 (ds)(1) sin 908 5 ds. Furthermore, all length elements around the loop are at the same
distance r from P, where r 2 5 a2 1 x 2.
0 I u d :
0I
s
r̂ u
ds
:
dB 5
5
22.22b
Use Equation 22.20 to find the magnitude of d B due to
2
2 1 x 2)
4
r
4
(a
:
the current in any length element d s :
Find the x component of the field element:
dBx 5
Integrate over the entire loop:
Bx 5
0I
ds
cos u
4 (a 2 1 x 2)
R
dBx 5
0I
R
4
a
cos u 5 2
(a 1 x 2)1/2
From the geometry, evaluate cos :
Substitute this expression for cos into the integral and
note that x and a are both constant:
Bx 5
Integrate around the loop:
Bx 5
0I
4
R
ds cos u
a2 1 x2
0I
R
0Ia 2
2
2(a 1 x 2)3/2
22.23b
ds
a
a2 1 x 2
(a 2 1 x 2)1/2
0I
a
4
(a 2 1 x 2)3/2
(2a) 5
5
a
4 (a 2 1 x 2)3/2
ds
Finalize To find the magnetic field at the center of the loop, set x 5 0 in Equation 22.23. At this special point,
B5
0I
2a
(at x 5 0)
22.24b
The pattern of magnetic field lines for a circular current loop is shown in Figure 22.26a. For clarity, the lines are drawn for only the plane that contains the axis of the
loop. The field-line pattern is axially symmetric and looks like the pattern around a bar
magnet, which is shown in Figure 22.26b.
What If? What if we consider points on the x axis very far from the loop? How does the
magnetic field behave at these distant points?
Answer In this case, in which x .. a, we can neglect the term a2 in the denominator of
Equation 22.23 and obtain
B<
0Ia 2
2x 3
(for x .. a)
22.25b
Figure 22.26 (Example 22.6)
(a) Magnetic field lines surrounding
a current loop. (b) Magnetic field
lines surrounding a bar magnet.
Notice the similarity between this line
pattern and that of a current loop.
22.8 | The Magnetic Force Between Two Parallel Conductors 761
22.6 cont.
The magnitude of the magnetic moment : of the loop is defined as the product of current and loop area (see Eq. 22.15):
5 I(a2) for our circular loop. We can express Equation 22.25 as
B<
0 2 x 3
22.26b
This result is similar in form to the expression for the electric field due to an electric dipole, E 5 ke(p/y 3) (see Example 19.4), where p 5 2aq is the electric dipole moment.
22.8 | The Magnetic Force Between Two Parallel
S
B
Conductors
ᐉ
In Section 22.5, we described the magnetic force that acts on a current-carrying
conductor when the conductor is placed in an external magnetic field. Because a
current in a conductor sets up its own magnetic field, it is easy to understand that
two current-carrying conductors exert magnetic forces on each other. As we shall
see, such forces can be used as the basis for defining the ampere and the coulomb.
Consider two long, straight, parallel wires separated by the distance a and
carrying currents I1 and I2 in the same direction as in Active Figure 22.27. We shall
adopt a simplification model in which the radii of the wires are much smaller than
a so that the radius plays no role in the calculation. We can determine the force on
one wire due to the magnetic field set up by the other wire. Wire 2, which carries
:
current I2, sets up a magnetic field B2 at the position of wire 1. The direction of
:
B2 is perpendicular to the wire as shown in Active Figure 22.27. According
:
:
:
to Equation
22.10, the magnetic force on a length / of wire 1 is F1 5 I 1 ᐍ
B 2.
:
:
:
Because ᐍ is perpendicular to B2, the magnitude of F1 is F1 5 I1,B 2. Because the field
due to wire 2 is given by Equation 22.21, we see that
0I2
0I1I2
F1 5 I1,B2 5 I1,
5
,
2a
2a
ᐉ
S
B
S
F
Active Figure 22.27 Two parallel
wires that each carry a steady current exert a force on each other. The
force is attractive if the currents are
parallel (as shown) and repulsive if
the currents are antiparallel.
1 2
We can rewrite this expression in terms of the force per unit length as
0I1I2
F1
5
,
2a
:
:
:
The direction of F 1 is downward, toward wire 2, because ᐍ
B is downward. If one
: 2
considers the field set up at wire 2 due to wire 1, the force F 2 on wire 2 is found to
:
be equal in magnitude and opposite in direction to F 1. That is what one would expect because Newton’s third law must be obeyed. Therefore, we can drop the force
subscript so that the magnetic force per unit length exerted by each long currentcarrying wire on the other is
0I1I2
F
5
22.27b
,
2a
This equation also applies if one of the wires is of finite length. In the discussion
above, we used the equation for the magnetic field of an infinite wire carrying current I2, but did not require that wire 1 be of infinite length.
When the currents are in opposite directions, the magnetic forces are reversed
and the wires repel each other. Hence, we find that parallel conductors carrying currents in the same direction attract each other, whereas parallel conductors carrying
currents in opposite directions repel each other.
The magnetic force between two parallel wires, each carrying a current, is used to
define the ampere:
If two long, parallel wires 1 m apart carry the same current and the force per
unit length on each wire is 2 3 1027 N/m, the current is defined to be 1 A.
c Magnetic force per unit length
between parallel current-carrying
wires
762 CHAPTER 22 | Magnetic Forces and Magnetic Fields
The numerical value of 2 3 1027 N/m is obtained from Equation 22.27, with
I 1 5 I 2 5 1 A and a 5 1 m.
The SI unit of charge, the coulomb, can now be defined in terms of the ampere: If
a conductor carries a steady current of 1 A, the quantity of charge that flows through
a cross section of the conductor in 1 s is 1 C.
QUI C K QU IZ 22.5 A loose spiral spring carrying no current is hung from a ceiling.
When a switch is thrown so that a current exists in the spring, do the coils (a) move
closer together, (b) move farther apart, or (c) not move at all?
Example 22.7 | Suspending a Wire
S
S
F
Two infinitely long, parallel wires are lying on the ground
a distance a 5 1.00 cm apart as shown in Figure 22.28a. A
third wire, of length L 5 10.0 m and mass 400 g, carries a
current of I1 5 100 A and is levitated above the first two
wires, at a horizontal position midway between them. The
infinitely long wires carry equal currents I2 in the same direction, but in the direction opposite that in the levitated
wire. What current must the infinitely long wires carry so
that the three wires form an equilateral triangle?
F
u
S
F
Figure 22.28 (Example 22.7) (a) Two current-carrying wires
lie on the ground and suspend a third wire in the air by magnetic
forces. (b) End view. In the situation described in the example,
the three wires form an equilateral triangle. The two magnetic
:
forces on the levitated wire are F B,L , the force due to the left-hand
:
wire on the ground, and F B,R , the force due to the right-hand
:
wire. The gravitational force F g on the levitated wire is also shown.
SOLUTION
Conceptualize Because the current in the short wire is opposite those in the long wires, the short wire is repelled from
both of the others. Imagine that the currents in the long wires
in Figure 22.28a are increased. The repulsive force becomes
stronger, and the levitated wire rises to the point at which the wire is once again levitated in equilibrium at a higher position.
Figure 22.28b shows the desired situation with the three wires forming an equilateral triangle.
Categorize Because the levitated wire is subject to forces but does not accelerate, it is modeled as a particle in equilibrium.
Analyze The horizontal components of the magnetic forces on the levitated wire cancel. The vertical components are both
positive and add together. Choose the z axis to be upward through the top wire in Figure 22.28b and in the plane of the page.
Find the total magnetic force in the upward direction on
the levitated wire:
Find the gravitational force on the levitated wire:
:
0I1I2
0I1I2
1 2a ,2 cos u kˆ 5 a , cos u kˆ
FB 5 2
:
F g 5 2mg kˆ
:
:
:
0I1I2
Apply the particle in equilibrium model by adding the
forces and setting the net force equal to zero:
o F 5 F B 1 F g 5 a , cos u kˆ 2 mg kˆ 5 0
Solve for the current in the wires on the ground:
I2 5
Substitute numerical values:
I2 5
mg a
0I1, cos u
(0.400 kg)(9.80 m/s2)(0.010 0 m)
(4 3 1027 T?m/A)(100 A)(10.0 m) cos 30.08
5 113 A
Finalize The currents in all wires are on the order of 102 A. Such large currents would require specialized equipment.
Therefore, this situation would be difficult to establish in practice. Is the equilibrium of wire 1 stable or unstable?
22.9 | Ampère’s Law
A simple experiment first carried out by Oersted in 1820 clearly demonstrates that
a current-carrying conductor produces a magnetic field. In this experiment, several compass needles are placed in a horizontal plane near a long vertical wire as
© Richard Megna, Fundamental Photographs
22.9 | Ampère’s Law 763
S
B
S
s
Active Figure 22.29 (a) and
(b) Compasses show the effects of the
current in a nearby wire. (c) Circular
magnetic field lines surrounding a
current-carrying conductor, displayed
with iron filings.
in Active Figure 22.29a. When the wire carries no current, all needles point in the
same direction (that of the Earth’s magnetic field), as one would expect. When the
wire carries a strong, steady current, however, the needles all deflect in a direction
tangent to the circle as in Active Figure 22.29b. These observations show that the
:
direction of B is consistent with the right-hand rule described in Section 22.7. When
the current is reversed, the needles in Active Figure 22.29b also reverse.
:
:
Because the needles point in the direction of B, we conclude that the lines of B
form circles about the wire, as discussed in Section 22.7. By symmetry, the magnitude
:
of B is the same everywhere on a circular path that is centered on the wire and lies
in a plane perpendicular to the wire. By varying the current and distance from the
:
wire, one finds that B is proportional to the current and inversely proportional to
the distance from the wire.
In Chapter 19, we investigated Gauss’s law, which is a relationship between an
electric charge and the electric field it produces. Gauss’s law can be used to determine the electric field in highly symmetric situations. We now consider an analogous
relationship in magnetism between a current and the magnetic field it produces.
This relationship can be used to determine the magnetic field created by a highly
symmetric current distribution.
:
Let us evaluate the product B d :
s for a small length element d :
s on the circular
1
path centered on the wire in Active Figure 22.29b. Along this path, the vectors d :
s
:
:
:
and B are parallel at each point, so B d :
s 5 B ds. Furthermore, by symmetry, B is
constant in magnitude on this circle and is given by Equation 22.21. Therefore, the
sum of the products B ds over the closed path, which is equivalent to the line integral
:
of B d :
s , is
0I
:
B d:
22.28b
s 5 B ds 5
(2r) 5 0I
2r
R
R
where r ds 5 2r is the circumference of the circle.
This result was calculated for the special case of a circular path surrounding a
wire. It can, however, also be applied in the general case in which a steady current
passes through the area surrounded by an arbitrary closed path. The general result
is Ampère's law:
:
The line integral of B d :
s around any closed path equals 0I, where I is the
total steady current passing through any surface bounded by the closed path:
R
:
s 5 0I
B d:
22.29b
1You may wonder why we would choose to do this evaluation. The origin of Ampère’s law is in 19th-century science, in
which a “magnetic charge” (the supposed analog to an isolated electric charge) was imagined to be moved around a
:
circular field line. The work done on the charge was related to B d :
s , just like the work done moving an electric charge
:
s . Therefore, Ampère’s law, a valid and useful principle, arose from an erroneous
in an electric field is related to E d :
and abandoned work calculation!
c Ampère’s law
764 CHAPTER 22 | Magnetic Forces and Magnetic Fields
:
QUI C K QU IZ 22.6 Rank the magnitudes of r B d:
s for the closed paths a through d
in Figure 22.30, from greatest to least.
:
© iStockphoto.com/HultonArchive
s for the closed paths a through d
QUI C K QU IZ 22.7 Rank the magnitudes of r B d:
in Figure 22.31, from greatest to least.
Andre-Marie Ampère
French Physicist (1775–1836)
Ampère is credited with the discovery
of electromagnetism, the relationship
between electric currents and magnetic
fields. Ampère’s genius, particularly in
mathematics, became evident by the time
he was 12 years old; his personal life,
however, was filled with tragedy. His father,
a wealthy city official, was guillotined during
the French Revolution, and his wife died
young, in 1803. Ampère died at the age of
61 of pneumonia. His judgment of his life
is clear from the epitaph he chose for his
gravestone: Tandem Felix (Happy at Last).
Figure 22.30 (Quick Quiz 22.6)
Figure 22.31 (Quick Quiz 22.7)
Four closed paths around three
current-carrying wires.
Four closed paths near a single
current-carrying wire.
Ampère’s law is valid only for steady currents. Furthermore, even though Ampère’s
law is true for all current configurations, it is only useful for calculating the magnetic
fields of configurations with high degrees of symmetry.
In Section 19.10, we provided some conditions to be sought when defining a
gaussian surface. Similarly, to apply Equation 22.29 to calculate a magnetic field, we
must determine a path of integration (sometimes called an amperian loop) such that
each portion of the path satisfies one or more of the following conditions:
1. The value of the magnetic field can be argued by symmetry to be constant over
the portion of the path.
2. The dot product in Equation 22.29 can be expressed as a simple algebraic
:
product B ds because B and d :
s are parallel.
:
3. The dot product in Equation 22.29 is zero because B and d :
s are
perpendicular.
4. The magnetic field can be argued to be zero at all points on the portion of the
path.
The following examples illustrate some symmetric current configurations for which
Ampère’s law is useful.
Example 22.8 | The Magnetic Field Created by a Long Current-Carrying Wire
A long, straight wire of radius R carries a steady current I that is uniformly distributed through the cross section of the wire (Fig. 22.32). Calculate the magnetic field a distance r from the center of the wire in the regions r $ R and
r , R.
SOLUTION
Conceptualize Study Figure 22.32 to understand the structure of the wire and the
current in the wire. The current creates magnetic fields everywhere, both inside and
outside the wire.
Categorize Because the wire has a high degree of symmetry, we categorize this ex-
ample as an Ampère’s law problem. For the r $ R case, we should arrive at the same
result as shown in Equation 22.21.
Analyze For the magnetic field exterior to the wire, let us choose for our path of in:
tegration circle 1 in Figure 22.32. From symmetry, B must be constant in magnitude
and parallel to d :
s at every point on this circle, satisfying conditions 1 and 2 above.
S
s
Figure 22.32 (Example 22.8)
A long, straight wire of radius R
carrying a steady current I uniformly
distributed across the cross section
of the wire. The magnetic field at
any point can be calculated from
Ampère’s law using a circular path of
radius r, concentric with the wire.
22.9 | Ampère’s Law 765
22.8 cont.
Note that the total current passing through the plane of
the circle is I and apply Ampère’s law:
Solve for B:
R
:
B d:
s 5B
B5
0I
R
ds 5 B(2r) 5 0I
(for r $ R)
2r
Now consider the interior of the wire, where r , R. Here the current I9 passing through the plane of circle 2 is less than
the total current I.
Set the ratio of the current I9 enclosed by circle 2 to the
entire current I equal to the ratio of the area r 2 enclosed
by circle 2 to the cross-sectional area R 2 of the wire:
r 2
I9
5
I
R 2
Solve for I9:
I9 5
Apply Ampère’s law to circle 2:
R
Solve for B:
B5
r2
I
R2
2
1Rr I2
:
B d:
s 5 B(2r) 5 0I 9 5 0
0I
12R 2r
2
2
22.30b
(for r , R)
Finalize The magnetic field exterior to the wire is identical in
form to Equation 22.21. As is often the case in highly symmetric
Figure 22.33 (Example 22.8)
situations, it is much easier to use Ampère’s law than the Biot–
Magnitude of the magnetic
field versus r for the wire shown
Savart law to find this result. The magnetic field interior to the
in
Figure
22.32.
The
fi
eld
is
wire is similar in form to the expression for the electric field inproportional to r inside the wire
side a uniformly charged sphere (see Example 19.10). The magniand varies as 1/r outside the wire.
tude of the magnetic field versus r for this configuration is plotted
in Figure 22.33. Inside the wire, B : 0 as r : 0. Furthermore, the results for r . R and r , R give the same value of the
magnetic field at r 5 R, demonstrating that the magnetic field is continuous at the surface of the wire.
Exampl e 22.9 | The Magnetic Field Created by a Toroid
A device called a toroid (Fig. 22.34) is often used to create an almost
uniform magnetic field in some enclosed area. The device consists of
a conducting wire wrapped around a ring (a torus) made of a nonconducting material. For a toroid having N closely spaced turns of wire, calculate the magnetic field in the region occupied by the torus, a distance
r from the center.
S
s
S
B
SOLUTION
Conceptualize Study Figure 22.34 carefully to understand how the wire is
wrapped around the torus. The torus could be a solid material or it could
be air, with a stiff wire wrapped into the shape shown in Figure 22.34 to
form an empty toroid.
Categorize Because the toroid has a high degree of symmetry, we catego-
rize this example as an Ampère’s law problem.
Analyze Consider the circular amperian loop (loop 1) of radius r in
the plane of Figure 22.34. By symmetry, the magnitude of the field is con:
stant on this circle and tangent to it, so B d:
s 5 B ds. Furthermore, the
wire passes through the loop N times, so the total current through the
loop is NI.
Figure 22.34 (Example 22.9) A toroid consisting
of many turns of wire. If the turns are closely spaced,
the magnetic field in the interior of the toroid is
tangent to the dashed circle (loop 1) and varies as
1/r. The dimension a is the cross-sectional radius of
the torus. The field outside the toroid is very small
and can be described by using the amperian loop
(loop 2) at the right side, perpendicular to the page.
continued
766 CHAPTER 22 | Magnetic Forces and Magnetic Fields
22.9 cont.
Apply Ampère’s law to loop 1:
R
Solve for B:
B5
Finalize This result shows that B varies as 1/r and hence is
nonuniform in the region occupied by the torus. If, however,
r is very large compared with the cross-sectional radius a
of the torus, the field is approximately uniform inside the
torus.
For an ideal toroid, in which the turns are closely
spaced, the external magnetic field is close to zero, but it
is not exactly zero. In Figure 22.34, imagine the radius r of
amperian loop 1 to be either smaller than b or larger than c.
In either case, the loop encloses zero net current, so
:
r B d:
s 5 0. You might think this result proves that
:
B 5 0, but it does not. Consider the amperian loop (loop 2)
on the right side of the toroid in Figure 22.34. The plane
of this loop is perpendicular to the page, and the toroid
passes through the loop. As charges enter the toroid as
indicated by the current directions in Figure 22.34, they
work their way counterclockwise around the toroid. Therefore, a current passes through the perpendicular amperian
loop! This current is small, but not zero. As a result, the
toroid acts as a current loop and produces a weak external field of the form shown in Figure 22.26. The reason
:
r B d:
s 5 0 for amperian loop 1 of radius r , b or
r . c is that the field lines are perpendicular to d :
s , not
:
because B 5 0.
:
B d:
s 5B
R
ds 5 B(2r) 5 0NI
0NI
2r
22.31b
Henry Leap and Jim Lehman
22.10 | The Magnetic Field of a Solenoid
Figure 22.35 (a) Magnetic field
lines for a tightly wound solenoid
of finite length, carrying a steady
current. The field in the interior
space is strong and nearly uniform.
(b) The magnetic field pattern of a
bar magnet, displayed with small iron
filings on a sheet of paper.
A solenoid is a long wire wound in the form of a helix. If the turns are closely spaced,
this configuration can produce a reasonably uniform magnetic field throughout the
volume enclosed by the solenoid, except close to its ends. Each of the turns can be
modeled as a circular loop, and the net magnetic field is the vector sum of the fields
due to all the turns.
If the turns are closely spaced and the solenoid is of finite length, the field lines
are as shown in Figure 22.35a. In this case, the field lines diverge from one end and
converge at the opposite end. An inspection of this field distribution exterior to the
solenoid shows a similarity to the field of a bar magnet (Fig. 22.35b). Hence, one
end of the solenoid behaves like the north pole of a magnet and the opposite end
behaves like the south pole. As the length of the solenoid increases, the field within it
becomes more and more uniform. When the solenoid’s turns are closely spaced and
its length is large compared with its radius, it approaches the case of an ideal solenoid.
For an ideal solenoid, the field outside the solenoid is negligible and the field inside
is uniform. We will use the ideal solenoid as a simplification model for a real solenoid.
If we consider the amperian loop (loop 1) perpendicular to the page in Figure 22.36, surrounding the ideal solenoid, we see that it it encloses a small current as the charges in the wire move turn by turn along the length of the solenoid.
Therefore, there is a nonzero magnetic field outside the solenoid. It is a weak field,
with circular field lines, like those due to a line of current as in Figure 22.23. For an
ideal solenoid, it is the only field external to the solenoid. We could eliminate this
field in Figure 22.36 by adding a second layer of turns of wire outside the first layer.
If the first layer of turns is wrapped so that the turns progress from the bottom of
Figure 22.36 to the top and the second layer has turns progressing from the top to
the bottom, the net current along the axis is zero.
We can use Ampère’s law to obtain an expression for the magnetic field inside
an ideal solenoid with a single layer of wire. A longitudinal cross-section of part of
:
our ideal solenoid (Fig. 22.36) carries current I. Here, B inside the ideal solenoid
is uniform and parallel to the axis. Consider a rectangular path (loop 2) of length
/ and width w as shown in Figure 22.36. We can apply Ampère’s law to this path by
:
evaluating the integral of B d :
s over each of the four sides of the rectangle. The
contribution along side 3 is zero because the magnetic field lines are perpendicular
to the path in this region, which matches condition 3 in Section 22.9. The contri:
butions from sides 2 and 4 are both zero because B is perpendicular to d :
s along
these paths, both inside and outside the solenoid. Side 1, whose length is /, gives a
22.11 | Magnetism in Matter 767
:
contribution to the integral because B along this portion of the path is constant in
magnitude and parallel to d :
s , which matches conditions 1 and 2. The integral over
the closed rectangular path therefore has the value
R
:
B d:
s 5
E
:
B d:
s 5B
side 1
E
side 1
ds 5 B ,
S
B
The right side of Ampère’s law involves the total current that passes through the
surface bounded by the path of integration. In our case, the total current through
the rectangular path equals the current through each turn of the solenoid multiplied by the number of turns enclosed by the path of integration. If N is the number
of turns in the length /, the total current through the rectangle equals NI. Ampère’s
law applied to this path therefore gives
R
ᐉ
:
B d:
s 5 B, 5 0NI
N
I 5 0nI
22.32b
,
where n 5 N// is the number of turns per unit length (not to be confused with N, the
number of turns).
We also could obtain this result in a simpler manner by reconsidering the magnetic field of a toroidal coil (Example 22.9). If the radius r of the toroidal coil containing N turns is large compared with its cross-sectional radius a, a short section
of the toroidal coil approximates a short section of a solenoid, with n 5 N/2r.
In this limit, we see that Equation 22.31 derived for the toroidal coil agrees with
Equation 22.32.
Equation 22.32 is valid only for points near the center of a very long solenoid. As
you might expect, the field near each end is smaller than the value given by Equation 22.32. At the very end of a long solenoid, the magnitude of the field is about
one-half that of the field at the center (see Problem 56 at the end of this chapter).
B 5 0
Q U I CK QUI Z 22.8 Consider a solenoid that is very long compared with its radius.
Of the following choices, what is the most effective way to increase the magnetic field
in the interior of the solenoid? (a) double its length, keeping the number of turns
per unit length constant (b) reduce its radius by half, keeping the number of turns
per unit length constant (c) overwrap the entire solenoid with an additional layer of
current-carrying wire
Figure 22.36 Cross-sectional
view of an ideal solenoid, where the
interior magnetic field is uniform and
the exterior field is close to zero.
22.11 | Magnetism in Matter
The magnetic field produced by a current in a loop of wire gives a hint about what
causes certain materials to exhibit strong magnetic properties. To understand why
some materials are magnetic, it is instructive to begin this discussion with the Bohr
structural model of the atom that we discussed in Chapter 11. In this model, electrons are assumed to move in circular orbits about the much more massive nucleus.
Figure 22.37 shows the angular momentum associated with the electron. In the Bohr
model, each electron, with its charge of magnitude 1.6 3 10219 C, circles the atom
once in about 10216 s. If we divide the electronic charge by this time interval, we find
that the orbiting electron is equivalent to a current of 1.6 3 1023 A. Each orbiting
electron is therefore viewed as a tiny current loop with a corresponding magnetic
moment. Because the charge of the electron is negative, the magnetic moment is
directed opposite to the angular momentum as shown in Figure 22.37.
In most substances, the magnetic moment of one electron in an atom is canceled
by that of another electron in the atom, orbiting in the opposite direction. The net
result is that the magnetic effect produced by the orbital motion of the electrons is
either zero or very small for most materials.
In addition to its orbital angular momentum, an electron has an intrinsic angular
momentum, called spin, which also contributes to its magnetic moment. The spin of
S
L
S
m
S
L
S
m
Figure 22.37 An electron moving
in the direction of the gray arrow in a
circular orbit of radius r. Because the
electron carries a negative charge,
the direction of the current due to its
motion about the nucleus is opposite
the direction of that motion.
768 CHAPTER 22 | Magnetic Forces and Magnetic Fields
Pitfall Prevention | 22.3
The Electron Does Not Spin
The electron is not physically
spinning. It has an intrinsic angular
momentum as if it were spinning, but
the notion of rotation for a point
particle is meaningless. Rotation
applies only to a rigid object, with an
extent in space, as in Chapter 10.
Spin angular momentum is actually a
relativistic effect.
an electron is an angular momentum separate from its orbital angular momentum,
just as the spin of the Earth is separate from its orbital motion about the Sun. Even
if the electron is at rest, it still has an angular momentum associated with spin. We
shall investigate spin more deeply in Chapter 29.
In atoms or ions containing multiple electrons, many electrons are paired up with
their spins in opposite directions, an arrangement that results in a cancellation of
the spin magnetic moments. An atom with an odd number of electrons, however,
must have at least one “unpaired” electron and a corresponding spin magnetic moment. The net magnetic moment of the atom leads to various types of magnetic
behavior. The magnetic moments of several atoms and ions are listed in Table 22.1.
Ferromagnetic Materials
TABLE 22.1 |
Magnetic Moments of Some
Atoms and Ions
Atom
or Ion
Magnet Moment
per Atom
or Ion (10224 J/T)
H
9.27
He
0
Ne
0
Ce31
19.8
Yb31
37.1
Iron, cobalt, nickel, gadolinium, and dysprosium are strongly magnetic materials
and are said to be ferromagnetic. Ferromagnetic substances, used to fabricate permanent magnets, contain atoms with spin magnetic moments that tend to align parallel to each other even in a weak external magnetic field. Once the moments are
aligned, the substance remains magnetized after the external field is removed. This
permanent alignment is due to strong coupling between neighboring atoms, which
can only be understood using quantum physics.
All ferromagnetic materials contain microscopic regions called domains, within
which all magnetic moments are aligned. The domains range from about 10212 to 1028
m3 in volume and contain 1017 to 1021 atoms. The boundaries between domains having
different orientations are called domain walls. In an unmagnetized sample, the domains
are randomly oriented so that the net magnetic moment is zero as in Figure 22.38a.
When the sample is placed in an external magnetic field, domains with magnetic moment vectors initially oriented along the external field grow in size at the expense of
other domains, which results in a magnetized sample, as in Figures 22.38b and 22.38c.
When the external field is removed, the sample may retain most of its magnetism.
The extent to which a ferromagnetic substance retains its magnetism is described
by its classification as being magnetically hard or soft. Soft magnetic materials, such
as iron, are easily magnetized but also tend to lose their magnetism easily. When a
soft magnetic material is magnetized and the external magnetic field is removed,
thermal agitation produces domain motion and the material quickly returns to an
unmagnetized state. In contrast, hard magnetic materials, such as cobalt and nickel,
are difficult to magnetize but tend to retain their magnetism, and domain alignment
persists in them after the external magnetic field is removed. Such hard magnetic
materials are referred to as permanent magnets. Rare-earth permanent magnets,
such as samarium–cobalt, are now regularly used in industry.
S
B
S
B
S
B
S
A
S
B
S
B
Figure 22.38 Orientation of magnetic dipoles before and after a magnetic field is applied to a ferromagnetic substance.
22.12 | Context Connection: Remote Magnetic Navigation for Cardiac Catheter Ablation Procedures 769
22.12 | Context Connection: Remote Magnetic
In the Heart Attacks Context, we studied the role of fluid flow in blood vessels and the
dangerous effect of plaque buildup on delivery of blood to the heart. In Section 21.2, we
looked at the heart again as we investigated the details of cardiac catheter ablation for a
patient who suffers from atrial fibrillation. In this Context Connection, we return to atrial
fibrillation in the heart, but consider a newer development in the ablation procedure.
There are a number of risks with a traditional cardiac catheter ablation procedure. One possible outcome is a perforation of the heart wall with one of the catheters. Because the esophagus passes right behind the heart, it is possible to burn
through too much tissue during a particular ablation and create an esophageal fistula. Other risks come from exposure to x-rays. In order to observe the positions of
the catheters, the electrophysiologist must use x-rays and a fluoroscope to make the
heart and the catheters visible. As a result, the patient receives a relatively high dose
of radiation during the procedure. In addition, despite the use of lead aprons, the
electrophysiologist receives radiation from each ablation procedure during his or
her entire career. In addition to the effects of this prolonged radiation exposure,
studies have shown that a large percentage of electrophysiologists have been treated
for back and neck pain due to the long hours of wearing lead aprons.
One possibility for reducing risks to both patient and doctor is the use of remote magnetic navigation in catheter ablation procedures. This procedure uses softer and more
flexible catheters than the traditional approach, reducing the risk of perforation and
allowing catheters to reach areas of the heart unavailable to the stiffer traditional catheters. The tips of the catheters are guided magnetically with the aid of a computer. The
electrophysiologist can sit comfortably at a computer in another room and guide the
catheters with a joystick, avoiding exposure to radiation. Figure 22.39 shows a typical
computer display that helps the electrophysiologist guide the catheters.
During a catheter ablation procedure using remote magnetic navigation, the patient is
located between two strong magnets as shown in Figure 22.40 (page 770). The magnets
can be moved over a wide range of positions and orientations relative to the patient. The
magnetic field from these magnets is strong, but only about 10% as strong as that used
in magnetic resonance imaging (to be discussed in the Context Conclusion). The tip of
the catheter includes ferromagnetic material so that its orientation can be precisely controlled by the positions of the external magnets. Once the tip is
correctly oriented, it can be advanced mechanically as with the
traditional approach.
In addition to the increased safety of the softer catheter and
the precise magnetic orientation of its tip, the computer control
of the procedure provides additional advantages. For example,
locations of ablations can be “memorized” by the computer.
The catheter tip can be quickly returned to this exact location
for a repeated ablation by calling up the memorized location.
While there are many advantages to remote magnetic
navigation, clinical evidence shows one disadvantage. The
total procedure time with remote navigation has been measured to be significantly longer than that with the traditional
approach.2 Reasons for this longer time interval include the
learning curve for the procedure, “interruption time” because
Figure 22.39 In remote magnetic navigation procedures for cardiac
the electrophysiologist is available to other staff in a room
catheter ablations, the electrophysiologist views a computer model of the
separate from the patient, and increased time for the more
heart such as the front and back images shown here. The yellow dots are
lesions around the pulmonary veins made by the ablation process.
complicated mapping procedures.
2A. Arya, R. Zaker-Shahrak, P. Sommer, A. Bollmann, U. Wetzel, T. Gaspar, S. Richter, D. Husser, C. Piorkowski, and
G. Hindricks, “Catheter Ablation of Atrial Fibrillation Using Remote Magnetic Catheter Navigation: A Case–Control
Study,” Europace, 13, pp. 45–50 (2011).
Needell M.D./Custom Medical Stock Photo
Navigation for Cardiac Catheter Ablation
Procedures
770 CHAPTER 22 | Magnetic Forces and Magnetic Fields
© Courtesy of Stereotaxis Inc.
Figure 22.40 A cardiac
catheterization laboratory using
remote magnetic navigation stands
ready to receive a patient suffering
from atrial fibrillation. The large
white objects on either side of
the operating table are housings
for strong magnets that place the
patient in a magnetic field. The
electrophysiologist performing a
catheter ablation procedure sits at
a computer in the room to the left.
With guidance from the magnetic
field, he or she uses a joystick
and other controls to thread the
magnetically sensitive tip of a cardiac
catheter through blood vessels and
into the chambers of the heart.
As more electrophysiologists are trained in remote magnetic navigation and the
mapping techniques are streamlined, perhaps the procedure times can be shortened. In that case, the magnetic technique will have a clear advantage over the traditional mechanical approach.
SUMMARY |
The magnetic force that acts on a charge q moving with ve:
locity :
v in an external magnetic field B is
:
v
FB 5 q:
:
22.1b
B
This force is in a direction perpendicular both to the velocity of the particle and to the magnetic field and given by
the right-hand rules shown in Figure 22.4. The magnitude of
the magnetic force is
FB 5 u q uvB sin :
22.2b
:
where is the angle between v and B.
A particle with mass m and charge q moving with velocity
:
:
v perpendicular to a uniform magnetic field B follows a circular path of radius
mv
r5
qB
22.3b
If a straight conductor of length L carries current I, the
magnetic force on that conductor when placed in a uniform
:
external magnetic field B is
:
:
:
FB 5 I L
:
22.10b
B
:
where L is in the direction of the current and u L u 5 L, the
length of the conductor.
If an arbitrarily shaped wire carrying current I is placed
in an external magnetic field, the magnetic force on a very
small length element d :
s is
:
s
d FB 5 I d :
:
B
22.11b
To determine the total magnetic force on the wire, one must
integrate Equation 22.11 over the wire.
The magnetic dipole moment : of a loop carrying current
I is
:
:
22.15b
5 IA
:
:
where A is perpendicular to the plane of the loop and u A u
is equal to the area of the loop. The SI unit of magnetic
moment : is the ampere ? meter squared, or A ? m2.
The torque : on a current loop when the loop is placed
:
in a uniform external magnetic field B is
:
5:
:
22.16b
B
The potential energy of the system of a magnetic dipole
in a magnetic field is
:
U52: B
22.17b
:
The Biot–Savart law says that the magnetic field d B at a point
P due to a wire element d :
s carrying a steady current I is
:
dB 5
0 I d :
s
4
r2
r̂
22.20b
where 0 5 4 3 1027 T ? m/A is the permeability of free
space and r is the distance from the element to the point
P. To find the total field at P due to a current distribution,
one must integrate this vector expression over the entire
distribution.
| Objective Questions 771
The magnitude of the magnetic field at a distance r from
a long, straight wire carrying current I is
B5
0I
2r
22.21b
The field lines are circles concentric with the wire.
The magnetic force per unit length between two
parallel wires (at least one of which is long) separated
by a distance a and carrying currents I1 and I2 has the
magnitude
0I1I2
F
5
,
2a
22.27b
The force is attractive if the currents are in the same direction and repulsive if they are in opposite directions.
:
Ampère’s law says that the line integral of B d:
s around
any closed path equals 0I, where I is the total steady current
passing through any surface bounded by the closed path:
R
:
B d:
s 5 0I
Using Ampère’s law, one finds that the magnitudes of the
magnetic fields inside a toroidal coil and solenoid are
B5
0NI
(toroid)
2r
N
B 5 0 I 5 0nI (solenoid)
,
2. What creates a magnetic field? More than one answer may
be correct. (a) a stationary object with electric charge (b) a
moving object with electric charge (c) a stationary conductor carrying electric current (d) a difference in electric potential (e) a charged capacitor disconnected from a battery
and at rest. Note: In Chapter 24, we will see that a changing
electric field also creates a magnetic field.
3. A charged particle is traveling through a uniform magnetic
field. Which of the following statements are true of the magnetic field? There may be more than one correct statement.
(a) It exerts a force on the particle parallel to the field.
(b) It exerts a force on the particle along the direction of its
motion. (c) It increases the kinetic energy of the particle.
(d) It exerts a force that is perpendicular to the direction
of motion. (e) It does not change the magnitude of the momentum of the particle.
4. A proton moving horizontally enters a region where a
uniform magnetic field is directed perpendicular to the S
v
proton’s velocity as shown
in Figure OQ22.4. After the
proton enters the field, does
it (a) deflect downward, with
Figure OQ22.4
its speed remaining constant; (b) deflect upward,
moving in a semicircular path with constant speed, and exit
the field moving to the left; (c) continue to move in the
horizontal direction with constant velocity; (d) move in a
circular orbit and become trapped by the field; or (e) deflect out of the plane of the paper?
22.31b
22.32b
where N is the total number of turns and n is the number of
turns per unit length.
denotes answer available in Student
Solutions Manual/Study Guide
OBJECTIVE QUESTIONS |
1. A spatially uniform magnetic field cannot exert a magnetic
force on a particle in which of the following circumstances?
There may be more than one correct statement. (a) The
particle is charged. (b) The particle moves perpendicular
to the magnetic field. (c) The particle moves parallel to the
magnetic field. (d) The magnitude of the magnetic field
changes with time. (e) The particle is at rest.
22.29b
5. Two long, parallel wires
each carry the same current I in the same direction
(Fig. OQ22.5). Is the total
magnetic field at the point
P midway between the wires
(a) zero, (b) directed into
the page, (c) directed out
of the page, (d) directed to
the left, or (e) directed to
the right?
Figure OQ22.5
6. Two long, straight wires
cross each other at a
right angle, and each carries the same current I
(Fig. OQ22.6). Which of
the following statements
is true regarding the total
magnetic field due to the
Figure OQ22.6
two wires at the various
points in the figure? More
than one statement may be correct. (a) The field is strongest at points B and D. (b) The field is strongest at points A
and C. (c) The field is out of the page at point B and into
the page at point D. (d) The field is out of the page at point
C and out of the page at point D. (e) The field has the same
magnitude at all four points.
7. Answer each question yes or no. Assume the motions and currents mentioned are along the x axis and fields are in the y
direction. (a) Does an electric field exert a force on a stationary charged object? (b) Does a magnetic field do so? (c) Does
an electric field exert a force on a moving charged object?
(d) Does a magnetic field do so? (e) Does an electric field exert
a force on a straight current-carrying wire? (f) Does a magnetic
field do so? (g) Does an electric field exert a force on a beam of
moving electrons? (h) Does a magnetic field do so?
8. At a certain instant, a proton is moving in the positive x direction through a magnetic field in the negative z direction. What
772 CHAPTER 22 | Magnetic Forces and Magnetic Fields
is the direction of the magnetic force exerted on the proton?
(a) positive z direction (b) negative z direction (c) positive y direction (d) negative y direction (e) The force is zero.
9. Answer each question yes or no. (a) Is it possible for each
of three stationary charged particles to exert a force of attraction on the other two? (b) Is it possible for each of three
stationary charged particles to repel both of the other particles? (c) Is it possible for each of three current-carrying
metal wires to attract the other two wires? (d) Is it possible
for each of three current-carrying metal wires to repel the
other two wires? André-Marie Ampère’s experiments on
electromagnetism are models of logical precision and included observation of the phenomena referred to in this
question.
10. A long, straight wire carries
a current I (Fig. OQ22.10).
Which of the following
statements is true regarding the magnetic field due
Figure OQ22.10
to the wire? More than one
statement may be correct.
(a) The magnitude is proportional to I/r, and the direction
is out of the page at P. (b) The magnitude is proportional
to I/r2, and the direction is out of the page at P. (c) The
magnitude is proportional to I/r, and the direction is into
the page at P. (d) The magnitude is proportional to I/r2,
and the direction is into the page at P. (e) The magnitude is
proportional to I, but does not depend on r.
11. A thin copper rod 1.00 m long has a mass of 50.0 g. What
is the minimum current in the rod that would allow it to
levitate above the ground in a magnetic field of magnitude 0.100 T? (a) 1.20 A (b) 2.40 A (c) 4.90 A (d) 9.80 A
(e) none of those answers
12. A magnetic field exerts a torque on each of the current
carrying single loops of wire shown in Figure OQ22.12.
The loops lie in the xy plane, each carrying the same magnitude current, and the uniform magnetic field points in
the positive x direction. Rank the loops by the magnitude
of the torque exerted on them by the field from largest to
smallest
CONCEPTUAL QUESTIONS |
1. Two charged particles are projected in the same direction
into a magnetic field perpendicular to their velocities. If the
particles are deflected in opposite directions, what can you
say about them?
S
B
Figure OQ22.12
13. Two long, parallel wires
carry currents of 20.0 A
and 10.0 A in opposite directions (Fig. OQ22.13).
Which of the following
statements is true? More
than one statement may
be correct. (a) In region Figure OQ22.13 Objective
Questions 13 and 14.
I, the magnetic field is
into the page and is never zero. (b) In region II, the field is
into the page and can be zero. (c) In region III, it is possible
for the field to be zero. (d) In region I, the magnetic field
is out of the page and is never zero. (e) There are no points
where the field is zero.
14. Consider the two parallel wires carrying currents in opposite directions in Figure OQ22.13. Due to the magnetic interaction between the wires, does the lower wire experience
a magnetic force that is (a) upward, (b) downward, (c) to
the left, (d) to the right, or (e) into the paper?
15. A long solenoid with closely spaced turns carries electric
current. Does each turn of wire exert (a) an attractive force
on the next adjacent turn, (b) a repulsive force on the next
adjacent turn, (c) zero force on the next adjacent turn, or
(d) either an attractive or a repulsive force on the next turn,
depending on the direction of current in the solenoid?
16. Solenoid A has length L and N turns, solenoid B has length
2L and N turns, and solenoid C has length L /2 and 2N
turns. If each solenoid carries the same current, rank the
magnitudes of the magnetic fields in the centers of the solenoids from largest to smallest.
denotes answer available in Student
Solutions Manual/Study Guide
5. Is it possible to orient a current loop in a uniform magnetic
field such that the loop does not tend to rotate? Explain.
6. Is Ampère’s law valid for all closed paths surrounding a con:
ductor? Why is it not useful for calculating B for all such paths?
2. One pole of a magnet attracts a nail. Will the other pole
of the magnet attract the nail? Explain. Also explain how a
magnet sticks to a refrigerator door.
7. A hollow copper tube carries a current along its length. Why
is B 5 0 inside the tube? Is B nonzero outside the tube?
3. A magnet attracts a piece of iron. The iron can then attract
another piece of iron. On the basis of domain alignment,
explain what happens in each piece of iron.
8. Imagine you have a compass whose needle can rotate
vertically as well as horizontally. Which way would the compass needle point if you were at the Earth’s north magnetic
pole?
4. How can the motion of a moving charged particle be used
to distinguish between a magnetic field and an electric
field? Give a specific example to justify your argument.
9. How can a current loop be used to determine the presence
of a magnetic field in a given region of space?
| Problems 773
Cengage Learning Charles D. Winters
10. Can a constant magnetic field set into
motion an electron initially at rest?
Explain your answer.
11. Explain why two parallel wires carrying currents in opposite directions
repel each other.
12. Figure CQ22.12 shows four permanent magnets, each having a hole
through its center. Notice that the
blue and yellow magnets are levitated above the red ones. (a) How
does this levitation occur? (b) What purpose do the rods
serve? (c) What can you say about the poles of the magnets
from this observation? (d) If the blue magnet were inverted,
what do you suppose would happen?
13. Is the magnetic field created by a current loop uniform?
Explain.
14. Consider a magnetic field that is uniform in direction
throughout a certain volume. (a) Can the field be uniform
in magnitude? (b) Must it be uniform in magnitude? Give
evidence for your answers.
Figure CQ22.12
PROBLEMS |
The problems found in this chapter may be assigned
online in Enhanced WebAssign.
denotes guided problem
denotes Master It tutorial available in Enhanced
WebAssign
denotes asking for quantitative and conceptual reasoning
denotes symbolic reasoning problem
denotes “paired problems” that develop reasoning with
symbols and numerical values
denotes Watch It video solution available in Enhanced
WebAssign
1. denotes straightforward problem; 2. denotes intermediate
problem; 3. denotes challenging problem
1. denotes full solution available in the Student Solutions Manual/
Study Guide
1. denotes problems most often assigned in Enhanced
WebAssign.
denotes biomedical problem
Section 22.2 The Magnetic Field
4.
1. A proton travels with a speed of 3.00 3 106 m/s at an angle
of 37.08 with the direction of a magnetic field of 0.300 T in
the 1y direction. What are (a) the magnitude of the magnetic force on the proton and (b) its acceleration?
2.
Determine the initial direction of the deflection of
charged particles as they enter the magnetic fields as shown
in Figure P22.2.
S
S
B
B
S
B
S
5. Consider an electron near the Earth’s equator. In which direction does it tend to deflect if its velocity is (a) directed
downward? (b) Directed northward? (c) Directed westward? (d) Directed southeastward?
6. At the equator, near the surface of the Earth, the magnetic
field is approximately 50.0 T northward and the electric
field is about 100 N/C downward in fair weather. Find the
gravitational, electric, and magnetic forces on an electron
in this environment, assuming that the electron has an instantaneous velocity of 6.00 3 106 m/s directed to the east.
7.
A proton moves perpendicular to a uniform magnetic
:
field B at a speed of 1.00 3 107 m/s and experiences an
acceleration of 2.00 3 1013 m/s2 in the positive x direction
when its velocity is in the positive z direction. Determine the
magnitude and direction of the field.
8.
A proton moves with a velocity of :
v 5 (2î 2 4 ĵ 1 k̂) m/s
:
in a region in which the magnetic field is B 5 (î 2 2 ĵ 2 k̂) T.
What is the magnitude of the magnetic force this particle
experiences?
B
Figure P22.2
3.
An electron is accelerated through 2.40 3 103 V from
rest and then enters a uniform 1.70-T magnetic field. What
are (a) the maximum and (b) the minimum values of the
magnetic force this particle experiences?
A proton moving at 4.00 3 106 m/s through a magnetic
field of magnitude 1.70 T experiences a magnetic force of
magnitude 8.20 3 10213 N. What is the angle between the
proton’s velocity and the field?
Section 22.3 Motion of a Charged Particle in a Uniform
Magnetic Field
9. Review. An electron moves in a circular path perpendicular to a constant magnetic field of magnitude 1.00 mT. The
angular momentum of the electron about the center of the
774 CHAPTER 22 | Magnetic Forces and Magnetic Fields
circle is 4.00 3 10225 kg ? m2/s. Determine (a) the radius of
the circular path and (b) the speed of the electron.
10.
A cosmic-ray proton in interstellar space has an energy
of 10.0 MeV and executes a circular orbit having a radius equal
to that of Mercury’s orbit around the Sun (5.80 3 1010 m).
What is the magnetic field in that region of space?
the vertical at the ring’s
location, what are (a) the
magnitude and (b) the
direction of the resultant
magnetic force on the ring?
19.
11. Review. One electron collides elastically with a second electron initially at rest. After the collision, the radii of their
trajectories are 1.00 cm and 2.40 cm. The trajectories are perpendicular to a uniform magnetic field of magnitude 0.044 0
T. Determine the energy (in keV) of the incident electron.
12.
Review. One electron collides elastically with a second
electron initially at rest. After the collision, the radii of their
trajectories are r1 and r2. The trajectories are perpendicular
to a uniform magnetic field of magnitude B. Determine the
energy of the incident electron.
A velocity selector consists of electric and magnetic
:
:
fields described by the expressions E 5 E k̂ and B 5 B ĵ
with B 5 15.0 mT. Find the value of E such that a 750-eV
electron moving in the negative x direction is undeflected.
14.
A cyclotron designed to accelerate protons has a magnetic field of magnitude 0.450 T over a region of radius
1.20 m. What are (a) the cyclotron frequency and (b) the
maximum speed acquired by the protons?
15.
Consider the mass spectrometer shown schematically in
Active Figure 22.12. The magnitude of the electric field between the plates of the velocity selector is 2.50 3 103 V/m,
and the magnetic field in both the velocity selector and the
deflection chamber has a magnitude of 0.035 0 T. Calculate
the radius of the path for a singly charged ion having a mass
m 5 2.18 3 10226 kg.
16. A cyclotron (Fig. 22.14) designed to accelerate protons has an
outer radius of 0.350 m. The protons are emitted nearly at rest
from a source at the center and are accelerated through 600 V
each time they cross the gap between the dees. The dees are between the poles of an electromagnet where the field is 0.800 T.
(a) Find the cyclotron frequency for the protons in this cyclotron. Find (b) the speed at which protons exit the cyclotron
and (c) their maximum kinetic energy. (d) How many revolutions does a proton make in the cyclotron? (e) For what time
interval does the proton accelerate?
17. The picture tube in an old black-and-white television uses magnetic deflection coils rather than electric deflection plates.
Suppose an electron beam is accelerated through a 50.0-kV
potential difference and then through a region of uniform
magnetic field 1.00 cm wide. The screen is located 10.0 cm
from the center of the coils and is 50.0 cm wide. When the
field is turned off, the electron beam hits the center of the
screen. Ignoring relativistic corrections, what field magnitude
is necessary to deflect the beam to the side of the screen?
Section 22.5 Magnetic Force on a Current-Carrying
Conductor
18.
A strong magnet is placed under a horizontal conducting ring of radius r that carries current I as shown in Fig:
ure P22.18. If the magnetic field B makes an angle with
A wire carries a steady
current of 2.40 A. A straight
section of the wire is 0.750 m
long and lies along the x axis
within uniform magnetic
:
field, B 5 160 k̂ T. If the current is in the positive x direction, what is the magnetic
force on the section of wire?
u
S
B
Figure P22.18
20.
In Figure P22.20, the cube is 40.0 cm on each edge.
Four straight segments of wire—ab, bc, cd, and da—form a
closed loop that carries a current I 5 5.00 A in the direction shown. A uniform magnetic
S
field of magnitude B 5 0.020 0 T
B
is in the positive y direction. Determine the magnetic force vector on (a) ab, (b) bc, (c) cd, and
(d) da. (e) Explain how you could
find the force exerted on the
fourth of these segments from the
forces on the other three, without
Figure P22.20
further calculation involving the
magnetic field.
21.
A wire having a mass per unit length of 0.500 g/cm carries a 2.00-A current horizontally to the south. What are
(a) the direction and (b) the magnitude of the minimum
magnetic field needed to lift this wire vertically upward?
Section 22.4 Applications Involving Charged Particles
Moving in a Magnetic Field
13.
u
22. Why is the following situation impossible? Imagine a copper wire
with radius 1.00 mm encircling the Earth at its magnetic
equator, where the field direction is horizontal. A power
supply delivers 100 MW to the wire to maintain a current
in it, in a direction such that the magnetic force from the
Earth’s magnetic field is upward. Due to this force, the wire
is levitated immediately above the ground.
23.
A wire 2.80 m in length carries a current of 5.00 A in a
region where a uniform magnetic field has a magnitude of
0.390 T. Calculate the magnitude of the magnetic force on
the wire assuming the angle between the magnetic field and
the current is (a) 60.08, (b) 90.08, and (c) 1208.
Section 22.6 Torque on a Current Loop in a Uniform
Magnetic Field
24.
A current loop with magnetic dipole moment : is placed
:
in a uniform magnetic field B, with its moment making angle
with the field. With the arbitrary
S
B
choice of U 5 0 for 5 908, prove
that the potential energy of the di:
pole-field system is U 5 2: B.
25.
A rectangular coil consists of
N 5 100 closely wrapped turns
and has dimensions a 5 0.400 m
and b 5 0.300 m. The coil is
hinged along the y axis, and its
plane makes an angle 5 30.08
with the x axis (Fig. P22.25).
u
Figure P22.25
| Problems 775
(a) What is the magnitude of the torque exerted on the
coil by a uniform magnetic field B 5 0.800 T directed in
the positive x direction when the current is I 5 1.20 A in
the direction shown? (b) What is the expected direction
of rotation of the coil?
26.
27.
A rectangular loop of
wire has dimensions 0.500 m
by 0.300 m. The loop is pivoted at the x axis and lies in
S
B
the xy plane as shown in Fig°
ure P22.26. A uniform magnetic field of magnitude 1.50 T
is directed at an angle of
40.08 with respect to the y axis
Figure P22.26
with field lines parallel to the
yz plane. The loop carries a current of 0.900 A in the direction shown. (Ignore gravitation.) We wish to evaluate
the torque on the current loop. (a) What is the direction of the magnetic force exerted on wire segment ab?
(b) What is the direction of the torque associated with
this force about an axis through the origin? (c) What
is the direction of the magnetic force exerted on segment cd? (d) What is the direction of the torque associated with this force about an axis through the origin?
(e) Can the forces examined in parts (a) and (c) combine to cause the loop to rotate around the x axis?
(f) Can they affect the motion of the loop in any way?
Explain. (g) What is the direction of the magnetic force
exerted on segment bc? (h) What is the direction of the
torque associated with this force about an axis through
the origin? (i) What is the torque on segment ad about
an axis through the origin? (j) From the point of view of
Figure P22.26, once the loop is released from rest at the
position shown, will it rotate clockwise or counterclockwise around the x axis? (k) Compute the magnitude of
the magnetic moment of the loop. (l) What is the angle
between the magnetic moment vector and the magnetic
field? (m) Compute the torque on the loop using the
results to parts (k) and (l).
A current of 17.0 mA is maintained in a single circular
loop of 2.00 m circumference. A magnetic field of 0.800 T
is directed parallel to the plane of the loop. (a) Calculate
the magnetic moment of the loop. (b) What is the magnitude of the torque exerted by the magnetic field on the
loop?
28. The rotor in a certain electric motor is a flat, rectangular coil with 80 turns of wire and dimensions 2.50 cm by
4.00 cm. The rotor rotates in a uniform magnetic field
of 0.800 T. When the plane of the rotor is perpendicular
to the direction of the magnetic field, the rotor carries
a current of 10.0 mA. In this orientation, the magnetic
moment of the rotor is directed opposite the magnetic
field. The rotor then turns through one-half revolution.
This process is repeated to cause the rotor to turn steadily
at an angular speed of 3.60 3 103 rev/min. (a) Find the
maximum torque acting on the rotor. (b) Find the peak
power output of the motor. (c) Determine the amount
of work performed by the magnetic field on the rotor in
every full revolution. (d) What is the average power of
the motor?
Section 22.7 The Biot–Savart Law
29.
Calculate the magnitude
of the magnetic field at a point
25.0 cm from a long, thin conductor carrying a current of
2.00 A.
30.
An infinitely long wire
carrying a current I is bent
Figure P22.30
at a right angle as shown in
Figure P22.30. Determine the
magnetic field at point P, located a distance x from the
corner of the wire.
31. A conductor consists of a circular loop of radius R 5 15.0 cm
and two long, straight sections as shown in Figure P22.31.
The wire lies in the plane of the paper and carries a current
I 5 1.00 A. Find the magnetic field at the center of the loop.
Figure P22.31 Problems 31 and 32.
32.
A conductor consists of a circular loop of radius R and
two long, straight sections as shown in Figure P22.31. The
wire lies in the plane of the paper and carries a current I.
(a) What is the direction of the magnetic field at the center
of the loop? (b) Find an expression for the magnitude of
the magnetic field at the center of the loop.
33.
One long wire carries current 30.0 A to the left along the
x axis. A second long wire carries current 50.0 A to the right
along the line (y 5 0.280 m, z 5 0). (a) Where in the plane
of the two wires is the total magnetic field equal to zero?
(b) A particle with a charge of 22.00 C is moving with a
velocity of 150 î Mm/s along the line (y 5 0.100 m, z 5 0).
Calculate the vector magnetic force acting on the particle.
(c) What If? A uniform electric field is applied to allow this
particle to pass through this region undeflected. Calculate
the required vector electric field.
34.
Two long, straight, parallel wires carry currents that
are directed perpendicular to the page as shown in Figure P22.34. Wire 1 carries a current I1 into the page (in the
negative z direction) and passes through the x axis at x 5 1a.
Wire 2 passes through the x axis at x 5 22a and carries an unknown current I2. The total magnetic field at the origin due to
the current-carrying wires has the magnitude 20I1/(2a). The
current I2 can have either of two possible values. (a) Find the
value of I2 with the smaller magnitude, stating it in terms of I1
and giving its direction. (b) Find the other possible value of I2.
Figure P22.34
35. A current path shaped as shown in Figure P22.35 produces a magnetic field
at P, the center of the arc. If the arc
subtends an angle of 5 30.08 and the
u
Figure P22.35
776 CHAPTER 22 | Magnetic Forces and Magnetic Fields
radius of the arc is 0.600 m, what are the magnitude and
direction of the field produced at P if the current is 3.00 A?
36. Consider a flat, circular current loop of radius R carrying a
current I. Choose the x axis to be along the axis of the loop,
with the origin at the loop’s center. Plot a graph of the ratio
of the magnitude of the magnetic field at coordinate x to
that at the origin for x 5 0 to x 5 5R. It may be helpful to
use a programmable calculator or a computer to solve this
problem.
37. Three long, parallel
conductors
each carry a current of I 5 2.00 A.
Figure P22.37 is
an end view of the
conductors,
with
each current coming out of the page.
Taking a 5 1.00 cm,
Figure P22.37
determine the magnitude and direction of the magnetic field at (a) point A, (b) point B, and
(c) point C.
38.
42.
Two long wires hang vertically. Wire 1 carries an upward current of 1.50 A. Wire 2, 20.0 cm to the right of wire
1, carries a downward current of 4.00 A. A third wire, wire 3,
is to be hung vertically and located such that when it carries
a certain current, each wire experiences no net force. (a) Is
this situation possible? Is it possible in more than one way?
Describe (b) the position of wire 3 and (c) the magnitude
and direction of the current in wire 3.
43.
In Figure P22.43, the current in the long, straight wire
is I1 5 5.00 A and the wire lies
in the plane of the rectangular
loop, which carries a current
I2 5 10.0 A. The dimensions
in the figure are c 5 0.100 m,
a 5 0.150 m, and / 5 0.450 m.
Find the magnitude and direction of the net force exerted
on the loop by the magnetic
field created by the wire.
40. A lightning bolt may carry a current of 1.00 3 104 A for
a short time interval. What is the resulting magnetic field
100 m from the bolt? Assume that the bolt extends far above
and below the point of observation.
(a) A conducting loop in the
shape of a square of edge length / 5
0.400 m carries a current I 5 10.0 A as
shown in Figure P22.41. Calculate the
magnitude and direction of the magnetic field at the center of the square.
(b) What If? If this conductor is reshaped to form a circular loop and carries the same current, what is the value
of the magnetic field at the center?
ᐉ
Figure P22.41
ᐉ
Figure P22.43 Problems
44.
In Figure P22.43, the cur- 43 and 44.
rent in the long, straight wire is
I1 and the wire lies in the plane of a rectangular loop, which
carries a current I2. The loop is of length / and width a. Its
left end is a distance c from the wire. Find the magnitude and
direction of the net force exerted on the loop by the magnetic
field created by the wire.
45.
Two long, parallel conductors, separated by 10.0 cm,
carry currents in the same direction. The first wire carries
a current I1 5 5.00 A, and the second carries I2 5 8.00 A.
(a) What is the magnitude of the magnetic field created by
I1 at the location of I2? (b) What is the force per unit length
exerted by I1 on I2? (c) What is the magnitude of the magnetic field created by I2 at the location of I1? (d) What is the
force per length exerted by I2 on I1?
In Niels Bohr’s 1913 model of the hydrogen atom, an
electron circles the proton at a distance of 5.29 3 10211 m
with a speed of 2.19 3 106 m/s. Compute the magnitude of
the magnetic field this motion produces at the location of
the proton.
39. Review. In studies of the
possibility of migrating
birds using the Earth’s
magnetic field for navigation, birds have been
fitted with coils as “caps”
and “collars” as shown
in Figure P22.39. (a) If
the identical coils have
radii of 1.20 cm and are
2.20 cm apart, with 50
turns of wire apiece, what
current should they both
Figure P22.39
carry to produce a magnetic field of 4.50 3 1025 T halfway between them? (b) If
the resistance of each coil is 210 V, what voltage should the
battery supplying each coil have? (c) What power is delivered to each coil?
41.
Section 22.8 The Magnetic Force Between Two Parallel
Conductors
46. Why is the following situation impossible? Two parallel copper conductors each have length / 5 0.500 m and radius
r 5 250 m. They carry currents I 5 10.0 A in opposite
directions and repel each other with a magnetic force
FB 5 1.00 N.
Section 22.9 Ampère’s Law
47.
Figure P22.47 is a crosssectional view of a coaxial
cable. The center conductor is surrounded by a
rubber layer, an outer conductor, and another rubber layer. In a particular
application, the current in
the inner conductor is I1 5
1.00 A out of the page and
the current in the outer
conductor is I2 5 3.00 A
into the page. Assuming
Figure P22.47
the distance d 5 1.00 mm,
determine the magnitude and direction of the magnetic
field at (a) point a and (b) point b.
| Problems 777
48.
49.
A packed bundle of 100 long, straight, insulated wires
forms a cylinder of radius R 5 0.500 cm. If each wire carries
2.00 A, what are (a) the magnitude and (b) the direction of
the magnetic force per unit length acting on a wire located
0.200 cm from the center of the bundle? (c) What If? Would
a wire on the outer edge of the bundle experience a force
greater or smaller than the value calculated in parts (a) and
(b)? Give a qualitative argument for your answer.
The magnetic coils of a tokamak fusion reactor are in
the shape of a toroid having an inner radius of 0.700 m and
an outer radius of 1.30 m. The toroid has 900 turns of large
diameter wire, each of which carries a current of 14.0 kA.
Find the magnitude of the magnetic field inside the toroid
along (a) the inner radius and (b) the outer radius.
carries a clockwise current of 15.0 A. Find (a) the force on each
side of the loop and (b) the torque acting on the loop.
56.
Consider a solenoid of length / and radius a containing N closely spaced turns and carrying a steady current I.
(a) In terms of these parameters, find the magnetic field at
a point along the axis as a function of position x from the
end of the solenoid. (b) Show that as / becomes very long,
B approaches 0NI/2/ at each end of the solenoid.
57.
A long solenoid that has 1 000 turns uniformly distributed over a length of 0.400 m produces a magnetic field
of magnitude 1.00 3 1024 T at its center. What current is
required in the windings for that to occur?
58. A solenoid 10.0 cm in diameter and 75.0 cm long is made from
copper wire of diameter 0.100 cm, with very thin insulation. The
wire is wound onto a cardboard tube in a single layer, with adjacent turns touching each other. What power must be delivered
to the solenoid if it is to produce a field of 8.00 mT at its center?
50. Niobium metal becomes a superconductor when cooled below
9 K. Its superconductivity is destroyed when the surface magnetic field exceeds 0.100 T. In the absence of any external
magnetic field, determine the maximum current a 2.00-mmdiameter niobium wire can carry and remain superconducting.
Section 22.11 Magnetism in Matter
51.
59.
52.
The magnetic field 40.0 cm away from a long, straight wire
carrying current 2.00 A is 1.00 T. (a) At what distance is it
0.100 T? (b) What If? At one instant, the two conductors in a
long household extension cord carry equal 2.00-A currents in
opposite directions. The two wires are 3.00 mm apart. Find the
magnetic field 40.0 cm away from the middle of the straight
cord, in the plane of the two wires. (c) At what distance is it
one-tenth as large? (d) The center wire in a coaxial cable carries current 2.00 A in one direction, and the sheath around
it carries current 2.00 A in the opposite direction. What magnetic field does the cable create at points outside the cable?
The magnetic field created by a large current passing
through plasma (ionized gas) can force current-carrying
particles together. This pinch effect has been used in designing fusion reactors. It can be demonstrated by making an
empty aluminum can carry a large current parallel to its
axis. Let R represent the radius of the can and I the current,
uniformly distributed over the can’s curved wall. Determine
the magnetic field (a) just inside the wall and (b) just outside. (c) Determine the pressure on the wall.
60. In Niels Bohr’s 1913 model of the hydrogen atom, the single
electron is in a circular orbit of radius 5.29 3 10211 m and its
speed is 2.19 3 106 m/s. (a) What is the magnitude of the magnetic moment due to the electron’s motion? (b) If the electron
moves in a horizontal circle, counterclockwise as seen from
above, what is the direction of this magnetic moment vector?
Section 22.12 Context Connection: Remote Magnetic
Navigation for Cardiac Catheter Ablation Procedures
61.
A typical magnitude of the external magnetic field in a
catheter ablation procedure using remote magnetic navigation is B 5 0.080 T. Suppose that the permanent magnet in
the catheter used in the procedure is inside the left atrium
of the heart and subject to this external magnetic field. The
permanent magnet has a magnetic moment of 0.10 A · m2.
The orientation of the permanent magnet is 308 from the direction of the external magnetic field lines. (a) What is the
magnitude of the torque on the tip of the catheter containing
this permanent magnet? (b) What is the potential energy of
the system consisting of the permanent magnet in the catheter and the magnetic field provided by the external magnets?
62.
Review. Energy from the tip of a cardiac catheter is delivered by electromagnetic radiation to the tissue of the heart
during a cardiac ablation procedure at a typical rate of 50.0 W.
A typical target temperature for ablation is 65.08C. Suppose
a typical application of the ablation at one site in the heart
lasts for 15.0 s. (a) How much energy is delivered to the ablation site during this single ablation? (b) If the tissue rises
to the target temperature at the 15.0-s mark, what is the mass
of tissue that has been ablated? Assume the specific heat of
heart tissue is that of water and that the initial temperature
of the tissue is normal body temperature, 37.08C. Also assume that the energy delivered to the tissue stays within the
tissue during the 15.0-s burn time.
53. A long, straight wire lies on a horizontal table and carries
a current of 1.20 A. In a vacuum, a proton moves parallel
to the wire (opposite the current) with a constant speed of
2.30 3 104 m/s at a distance d above the wire. Ignoring the
magnetic field due to the Earth, determine the value of d.
54.
Four long, parallel conductors
carry equal currents of I 5 5.00 A. Figure P22.54 is an end view of the conductors. The current direction is into
the page at points A and B and out of
the page at points C and D. Calculate
(a) the magnitude and (b) the direction of the magnetic field at point P,
located at the center of the square of
edge length / 5 0.200 m.
,
,
Figure P22.54
Section 22.10 The Magnetic Field of a Solenoid
55.
A single-turn square loop of wire, 2.00 cm on each edge,
carries a clockwise current of 0.200 A. The loop is inside a solenoid, with the plane of the loop perpendicular to the magnetic field of the solenoid. The solenoid has 30.0 turns/cm and
The magnetic moment of the Earth is approximately
8.00 3 1022 A · m2. Imagine that the planetary magnetic
field were caused by the complete magnetization of a huge
iron deposit with density 7 900 kg/m3 and approximately
8.50 3 1028 iron atoms/m3. (a) How many unpaired electrons,
each with a magnetic moment of 9.27 3 10224 A · m2, would
participate? (b) At two unpaired electrons per iron atom,
how many kilograms of iron would be present in the deposit?
778 CHAPTER 22 | Magnetic Forces and Magnetic Fields
Additional Problems
63.
64.
A particle with positive charge q 5 3.20 3 10219 C moves
with a velocity :
v 5 (2 î 1 3 ĵ 2 k̂) m/s through a region
where both a uniform magnetic field and a uniform electric field exist. (a) Calculate the total force on the moving
:
particle (in unit-vector notation), taking B 5 (2î 1 4 ĵ 1 k̂) T
:
and E 5 (4 î 2 ĵ 2 2k̂) V/m. (b) What angle does the force
vector make with the positive x axis?
An infinite sheet of current
lying in the yz plane carries a
surface current of linear density
Js . The current is in the positive
z direction, and Js represents the
current per unit length measured
along the y axis. Figure P22.64 is
an edge view of the sheet. Prove
that the magnetic field near the
sheet is parallel to the sheet and
perpendicular to the current direction, with magnitude 0 Js /2.
67. Review. A 0.200-kg metal rod carrying a current of 10.0 A
glides on two horizontal rails 0.500 m apart. If the coefficient of kinetic friction between the rod and rails is 0.100,
what vertical magnetic field is required to keep the rod moving at a constant speed?
68.
Review. A metal rod of mass m carrying a current I glides
on two horizontal rails a distance d apart. If the coefficient
of kinetic friction between the rod and rails is , what vertical magnetic field is required to keep the rod moving at a
constant speed?
69.
Two circular loops
are parallel, coaxial, and
almost in contact, with
their centers 1.00 mm
apart (Fig. P22.69). Each
loop is 10.0 cm in radius.
The top loop carries a
Figure P22.69
clockwise current of I 5
140 A. The bottom loop
carries a counterclockwise current of I 5 140 A. (a) Calculate the magnetic force exerted by the bottom loop on the
top loop. (b) Suppose a student thinks the first step in solving part (a) is to use Equation 22.23 to find the magnetic
field created by one of the loops. How would you argue for or
against this idea? (c) The upper loop has a mass of 0.021 0 kg.
Calculate its acceleration, assuming the only forces acting
on it are the force in part (a) and the gravitational force.
70.
Consider a thin,
straight wire segment carrying a constant current I
and placed along the x axis
as shown in Figure P22.70.
(a) Use the Biot–Savart law
to show that the total magnetic field at the point P,
located a distance a from
the wire, is
Figure P22.64
65. Carbon-14 and carbon-12 ions (each with charge of magnitude e) are accelerated in a cyclotron. If the cyclotron has a
magnetic field of magnitude 2.40 T, what is the difference in
cyclotron frequencies for the two ions?
66. The Hall effect finds important application in the electronics
industry. It is used to find the sign and density of the carriers of
electric current in semiconductor chips. The arrangement is
shown in Figure P22.66. A semiconducting block of thickness
t and width d carries a current I in the x direction. A uniform
magnetic field B is applied in the y direction. If the charge
carriers are positive, the magnetic force deflects them in the
z direction. Positive charge accumulates on the top surface of
the sample and negative charge on the bottom surface, creating a downward electric field. In equilibrium, the downward
electric force on the charge carriers balances the upward magnetic force and the carriers move through the sample without
deflection. The Hall voltage ΔVH 5 Vc 2 Va between the top and
bottom surfaces is measured, and the density of the charge
carriers can be calculated from it. (a) Demonstrate that if the
charge carriers are negative the Hall voltage will be negative.
Hence, the Hall effect reveals the sign of the charge carriers,
so the sample can be classified as p-type (with positive majority
charge carriers) or n-type (with negative). (b) Determine the
number of charge carriers per unit volume n in terms of I, t, B,
ΔVH, and the magnitude q of the carrier charge.
S
B
S
F
S
v
S
B
B5
4a
(cos u1 2 cos u2)
Figure P22.70
(b) Assuming that the wire is infinitely long, show that the
result in part (a) gives a magnetic field that agrees with that
obtained by using Ampère’s law in Example 22.8.
71. Assume the region to the right of
a certain plane contains a uniform
magnetic field of magnitude 1.00 mT
and the field is zero in the region to
S
the left of the plane as shown in Figv
ure P22.71. An electron, originally
traveling perpendicular to the boundary plane, passes into the region of the
field. (a) Determine the time interval
required for the electron to leave the
Figure P22.71
“field-filled” region, noting that the
electron’s path is a semicircle. (b) Assuming the maximum
depth of penetration into the field is 2.00 cm, find the kinetic energy of the electron.
72.
Figure P22.66
0I
θu
θu
Heart–lung machines and artificial kidney
machines employ electromagnetic blood pumps. The blood
is confined to an electrically insulating tube, cylindrical in
practice but represented here for simplicity as a rectangle of
| Problems 779
interior width w and height
h. Figure P22.72 shows a
rectangular section of blood
within the tube. Two elecS
trodes fit into the top and
B
the bottom of the tube.
The potential difference between them establishes an
electric current through the
blood, with current density J
Figure P22.72
over the section of length L
shown in Figure P22.72. A perpendicular magnetic field exists in the same region. (a) Explain why this arrangement
produces on the liquid a force that is directed along the
length of the pipe. (b) Show that the section of liquid in
the magnetic field experiences a pressure increase JLB.
(c) After the blood leaves the pump, is it charged? (d) Is
it carrying current? (e) Is it magnetized? (The same electromagnetic pump can be used for any fluid that conducts
electricity, such as liquid sodium in a nuclear reactor.)
73.
A heart
surgeon
monitors
the flow rate of blood
through an artery
using
an
electromagnetic flowmeter
(Fig. P22.73). Electrodes A and B make
contact with the outer
surface of the blood
Figure P22.73
vessel, which has a diameter of 3.00 mm. (a) For a magnetic field magnitude of
0.040 0 T, an emf of 160 V appears between the electrodes.
Calculate the speed of the blood. (b) Explain why electrode
A has to be positive as shown. (c) Does the sign of the emf
depend on whether the mobile ions in the blood are predominantly positively or negatively charged? Explain.
74. Why is the following situation impossible? The magnitude of
the Earth’s magnetic field at either pole is approximately
7.00 3 1025 T. Suppose the field fades away to zero before
its next reversal. Several scientists propose plans for artificially generating a replacement magnetic field to assist with
devices that depend on the presence of the field. The plan
that is selected is to lay a copper wire around the equator
and supply it with a current that would generate a magnetic
field of magnitude 7.00 3 1025 T at the poles. (Ignore magnetization of any materials inside the Earth.) The plan is
implemented and is highly successful.
75. Protons having a kinetic
energy of 5.00 MeV
(1 eV 5 1.60 3 10219 J)
are moving in the positive x direction and enter
:
a magnetic field B 5
0.050 0k̂ T directed out of
the plane of the page and
Figure P22.75
extending from x 5 0
to x 5 1.00 m as shown in Figure P22.75. (a) Ignoring relativistic effects, find the angle between the initial velocity vector of the proton beam and the velocity vector after the beam
emerges from the field. (b) Calculate the y component of the
protons’ momenta as they leave the magnetic field.
76.
Review. Rail guns have been suggested for launching
projectiles into space without chemical rockets. A tabletop
model rail gun (Fig. P22.76) consists of two long, parallel,
horizontal rails / 5 3.50 cm apart, bridged by a bar of mass
m 5 3.00 g that is free to slide without friction. The rails and
bar have low electric resistance, and the current is limited
to a constant I 5 24.0 A by a power supply that is far to
the left of the figure, so it has no magnetic effect on the
bar. Figure P22.76 shows the bar at rest at the midpoint of
the rails at the moment the current is established. We wish
to find the speed with which the bar leaves the rails after
being released from the midpoint of the rails. (a) Find the
magnitude of the magnetic field at a distance of 1.75 cm
from a single long wire carrying a current of 2.40 A. (b) For
purposes of evaluating the magnetic field, model the rails
as infinitely long. Using the result of part (a), find the magnitude and direction of the magnetic field at the midpoint
of the bar. (c) Argue that this value of the field will be the
same at all positions of the bar to the right of the midpoint
of the rails. At other points along the bar, the field is in the
same direction as at the midpoint, but is larger in magnitude. Assume the average effective magnetic field along the
bar is five times larger than the field at the midpoint. With
this assumption, find (d) the magnitude and (e) the direction of the force on the bar. (f) Is the bar properly modeled
as a particle under constant acceleration? (g) Find the velocity of the bar after it has traveled a distance d 5 130 cm
to the end of the rails.
S
v ,
Figure P22.76
77.
A nonconducting ring of radius 10.0 cm is uniformly
charged with a total positive charge 10.0 C. The ring rotates at a constant angular speed 20.0 rad/s about an axis
through its center, perpendicular to the plane of the ring.
What is the magnitude of the magnetic field on the axis of
the ring 5.00 cm from its center?
78.
A nonconducting ring of radius R is uniformly charged
with a total positive charge q. The ring rotates at a constant
angular speed about an axis through its center, perpendicular to the plane of the ring. What is the magnitude
of the magnetic field on the axis of the ring a distance 21R
from its center?
79. Model the electric motor in a handheld electric mixer as a
single flat, compact, circular coil carrying electric current
in a region where a magnetic field is produced by an external permanent magnet. You need consider only one instant
in the operation of the motor. (We will consider motors
again in Chapter 23.) Make order-of-magnitude estimates
of (a) the magnetic field, (b) the torque on the coil, (c) the
current in the coil, (d) the coil’s area, and (e) the number
of turns in the coil. The input power to the motor is electric,
given by P 5 I ΔV, and the useful output power is mechanical, P 5 .
80. Why is the following situation impossible? Figure P22.80 (page
780) shows an experimental technique for altering the
780 CHAPTER 22 | Magnetic Forces and Magnetic Fields
direction of travel
u
for a charged particle. A particle of
charge q 5 1.00 C
and mass m 5 2.00 3
S
B
10213 kg enters the
bottom of the region
of
uniform
magS
v
netic field at speed
v 5 2.00 3 105 m/s,
with a velocity vecFigure P22.80
tor perpendicular to
the field lines. The magnetic force on the particle causes
its direction of travel to change so that it leaves the region of the magnetic field at the top traveling at an
angle from its original direction. The magnetic field has
magnitude B 5 0.400 T and is directed out of the page.
The length h of the magnetic field region is 0.110 m.
An
experimenter
performs the technique and measures
the angle at which
the particles exit the
top of the field. She
finds that the angles
of deviation are exactly as predicted.
81.
A very long,
thin strip of metal
Figure P22.81
of width w carries a
current I along its length as shown in Figure P22.81. The
current is distributed uniformly across the width of the strip.
Find the magnetic field at point P in the diagram. Point P
is in the plane of the strip at distance b away from its edge.
Chapter
23
Faraday’s Law and Inductance
Chapter Outline
23.1 Faraday’s Law of Induction
23.2 Motional emf
23.3 Lenz’s Law
23.4 Induced emfs and Electric Fields
23.5 Inductance
23.6 RL Circuits
23.7 Energy Stored in a Magnetic Field
23.8 Context Connection: The Use of
Transcranial Magnetic Stimulation
in Depression
Marine Current Turbines TM Ltd.
SUMMARY
O
ur studies in electromagnetism so
far have been concerned with the
electric fields due to stationary charges and
the magnetic fields produced by moving
charges. This chapter introduces a new
type of electric field, one that is due to a
changing magnetic field.
As we learned in Section 19.1, experiments conducted by Michael Faraday
in England in the early 1800s and independently by Joseph Henry in the United
States showed that an electric current can be induced in a circuit by a changing
magnetic field. The results of those experiments led to a very basic and important law of electromagnetism known as Faraday’s law of induction. Faraday’s
law explains how generators, as well as other practical devices, work.
Faraday’s law is also the basis for a new circuit element, the inductor. This
new circuit element combines with resistors and capacitors to allow for a
variety of useful electric circuits.
An artist’s impression of the Skerries
SeaGen Array, a tidal energy generator
under development near the island
of Anglesey, North Wales. When it is
brought on line, possibly in 2015, it will
offer 10.5 MW of power from generators
turned by tidal streams. The image shows
the underwater blades that are driven
by the tidal currents. The second blade
system has been raised from the water for
servicing. We will study generators in this
chapter.
23.1 | Faraday’s Law of Induction
We begin discussing the concepts in this chapter by considering a simple experiment that builds on material presented in Chapter 22. Imagine that a straight
metal conductor resides in a uniform magnetic field directed into the page as in
Figure 23.1 (page 782). Within the conductor, there are free electrons. Suppose
the conductor is now moved with a velocity :
v toward the right. Equation 22.1
781
782
CHAPTER 23 | Faraday’s Law and Inductance
a
Figure 23.1 A straight electrical
conductor moving with a velocity :
v
:
through a uniform magnetic field B
:
directed perpendicular to v .
a
b
c
© iStockphoto.com/Steven Wynn Photography
Active Figure 23.2 A simple experiment showing that a current is induced in a loop when a magnet
is moved toward or away from the loop.
Michael Faraday
British Physicist and Chemist (1791–1867)
Faraday is often regarded as the greatest
experimental scientist of the 1800s. His
many contributions to the study of electricity
include the invention of the electric motor,
the electric generator, and the transformer
as well as the discovery of electromagnetic
induction and the laws of electrolysis.
Greatly influenced by his religious beliefs,
he refused to work on the development of
poison gas for the British military.
tells us that a magnetic force acts on the electrons in the conductor. Using the righthand rule, the force on the electrons is downward in Figure 23.1 (remember that the
electrons carry a negative charge). Because this direction is along the conductor, the
electrons move along the conductor in response to this force. Therefore, a current is
produced in the conductor as it moves through a magnetic field!
Let us consider another simple experiment that demonstrates that an electric
current can be produced by a magnetic field. Consider a loop of wire connected
to a sensitive ammeter, a device that measures current, as illustrated in Active
Figure 23.2. If a magnet is moved toward the loop, the ammeter display shows
the existence of a current as in Active Figure 23.2a. When the magnet is held stationary as in Active Figure 23.2b, the ammeter shows no current. If the magnet is
moved away from the loop as in Active Figure 23.2c, the ammeter display shows
a current in the opposite direction from that caused by the motion of the magnet toward the ammeter. Finally, if the magnet is held stationary and the loop is
moved either toward or away from it, the ammeter shows a current again. From
these observations comes the conclusion that an electric current is set up in the
loop as long as relative motion occurs between the magnet and the loop.
These results are quite remarkable when we consider that a current exists in a
loop of wire even though no batteries are connected to the wire. We call such a current an induced current, and it is produced by an induced emf.
Another experiment, first conducted by Faraday, is illustrated in Active Figure 23.3. Part of the apparatus consists of a coil of insulated wire connected to a
switch and a battery. We shall refer to this coil as the primary coil of wire and to the
corresponding circuit as the primary circuit. The coil is wrapped around an iron
ring to intensify the magnetic field produced by the current through the coil. A
second coil of insulated wire at the right is also wrapped around the iron ring and
is connected to a sensitive ammeter. We shall refer to this coil as the secondary coil
and to the corresponding circuit as the secondary circuit. The secondary circuit
has no battery, and the secondary coil is not electrically connected to the primary
coil. The purpose of this apparatus is to detect any current that might be generated in the secondary circuit by a change in the magnetic field produced by the
primary circuit.
Initially, you might guess that no current would ever be detected in the secondary circuit. Something quite surprising happens, however, when the switch
23.1 | Faraday’s Law of Induction
783
in the primary circuit is opened or thrown closed. At the instant the
switch is thrown closed, the ammeter display briefly shows a current
and then returns to zero. When the switch is opened, the ammeter display shows a current in the opposite direction and then again returns
to zero. Finally, the ammeter reads zero when the primary circuit carries a steady current.
As a result of these observations, Faraday concluded that an electric current can be produced by a time-varying magnetic field. A current cannot be produced by a steady magnetic field. In the experiment
shown in Active Figure 23.2, the changing magnetic field is a result of
the relative motion between the magnet and the loop of wire. As long
as the motion persists, the current is maintained. In the experiment
shown in Active Figure 23.3, the current produced in the secondary circuit occurs for only an instant after the switch is closed while the magnetic field acting on the secondary coil builds from its zero value to its
final value. In effect, the secondary circuit behaves as though a source
of emf were connected to it for an instant. It is customary to say that an
Active Figure 23.3 Faraday’s experiment.
emf is induced in the secondary circuit by the changing magnetic field
produced by the current in the primary circuit.
To quantify such observations, we define a quantity called magnetic flux. The
flux associated with a magnetic field is defined in a similar manner to the electric
flux (Section 19.8) and is proportional to the number of magnetic field lines passing through an area. Consider an element of area dA on an arbitrarily shaped open
:
surface as in Figure 23.4. If the magnetic field at the location of this element is B,
:
:
:
the magnetic flux through the element is B ? dA, where dA is a vector perpendicular
to the surface whose magnitude equals the area dA. Hence, the total magnetic flux
FB through the surface is
FB 5
E
:
:
B ? dA
23.1b
c Magnetic flux
The SI unit of magnetic flux is a tesla ? meter squared, which is named the weber
(Wb); 1 Wb 5 1 T ? m2.
The two experiments illustrated in Figures 23.2 and 23.3 have one thing in
common. In both cases, an emf is induced in a circuit when the magnetic flux
through the surface bounded by the circuit changes with time. A general statement known as Faraday’s law of induction summarizes such experiments involving induced emfs:
The emf induced in a circuit is equal to the time rate of change of magnetic
flux through the circuit:
dF
e 5 2 dt B
23.2b
In Equation 23.2, FB is the magnetic flux through the surface bounded by
the circuit and is given by Equation 23.1. The negative sign in Equation 23.2
will be discussed in Section 23.3. If the circuit is a coil consisting of N identical
and concentric loops and if the field lines pass through all loops, the induced
emf is
e 5 2N
dFB
dt
23.3b
The emf is increased by the factor N because all the loops are in series, so the emfs
in the individual loops add to give the total emf.
u
Figure 23.4 The magnetic flux
:
through an area element dA is given
:
:
by B ? dA 5 B dA cos ␪. Note that
:
vector dA is perpendicular to the
surface.
c Faraday’s law
Pitfall Prevention | 23.1
Induced emf Requires a Change
The existence of a magnetic flux
through an area is not sufficient to
create an induced emf. The magnetic
flux must change to induce an emf.
784
CHAPTER 23 | Faraday’s Law and Inductance
:
Suppose a loop enclosing an area A lies in a uniform magnetic field B as in Figure 23.5. In this case, the magnetic flux through the loop is
u
FB 5
u
E
:
:
B ? dA 5
E
B dA cos ␪ 5 B cos ␪
E
dA 5 BA cos ␪
Hence, the induced emf is
e 5 2 dtd (BA cos ␪)
Figure 23.5 A conducting loop
that encloses an area A in the
presence of a uniform magnetic field
:
:
B. The angle between B and the
normal to the loop is ␪.
© Cengage Learning/Charles D. Winters
a
23.4b
This expression shows that an emf can be induced in a circuit by changing the mag:
netic flux in several ways: (1) the magnitude of B can vary with time, (2) the area
:
A of the circuit can change with time, (3) the angle ␪ between B and the normal
to the plane can change with time, and (4) any combination of these changes can
occur.
An interesting application of Faraday’s law is the production of sound in an electric guitar (Fig. 23.6). The coil in this case, called the pickup coil, is placed near the vibrating guitar string, which is made of a metal that can be magnetized. A permanent
magnet inside the coil magnetizes the portion of the string nearest the coil. When
the string vibrates at some frequency, its magnetized segment produces a changing
magnetic flux through the coil. The changing flux induces an emf in the coil that
is fed to an amplifier. The output of the amplifier is sent to the loudspeakers, which
produce the sound waves we hear.
Q U I C K Q U I Z 23.1 A circular loop of wire is held in a uniform magnetic field,
with the plane of the loop perpendicular to the field lines. Which of the following will not cause a current to be induced in the loop? (a) crushing the loop
(b) rotating the loop about an axis perpendicular to the field lines (c) keeping the
orientation of the loop fixed and moving it along the field lines (d) pulling the
loop out of the field
Q U I C K Q U I Z 23.2 Figure 23.7 shows a graphical representation of the field
magnitude versus time for a magnetic field that passes through a fixed loop and
that is oriented perpendicular to the plane of the loop. The magnitude of the
magnetic field at any time is uniform over the area of the loop. Rank the magnitudes of the emf generated in the loop at the five instants indicated, from largest
to smallest.
b
Figure 23.7 (Quick Quiz 23.2)
The time behavior of a magnetic field
through a loop.
Figure 23.6 (a) In an electric
guitar, a vibrating magnetized string
induces an emf in a pickup coil.
(b) The pickups (the circles beneath
the metallic strings) of this electric
guitar detect the vibrations of the
strings and send this information
through an amplifier and into
speakers. (A switch on the guitar
allows the musician to select which
set of six pickups is used.)
THINKING PHYSICS 23.1
The ground fault circuit interrupter (GFCI) is a safety device that protects users
of electric power against electric shock when they touch appliances. Its essential
parts are shown in Figure 23.8. How does the operation of a GFCI make use of
Faraday’s law?
785
23.1 | Faraday’s Law of Induction
Reasoning Wire 1 leads from the wall outlet to the appliance being protected,
and wire 2 leads from the appliance back to the wall outlet. An iron ring surrounds the two wires. A sensing coil wrapped around part of the iron ring
activates a circuit breaker when changes in magnetic flux occur. Because the currents in the two wires are in opposite directions during normal operation of the
appliance, the net magnetic field through the sensing coil due to the currents is
zero. A change in magnetic flux through the sensing coil can happen, however,
if one of the wires on the appliance loses its insulation and accidentally touches
the metal case of the appliance, providing a direct path to ground. When such a
short to ground occurs, a net magnetic flux occurs through the sensing coil that
alternates in time because household current is alternating. This changing flux
produces an induced voltage in the coil, which in turn triggers a circuit breaker,
stopping the current before it reaches a level that might be harmful to the person using the appliance. b
Figure 23.8 (Thinking Physics
23.1) Essential components of a
ground fault circuit interrupter.
Exampl e 23.1 | Inducing an emf in a Coil
A coil consists of 200 turns of wire. Each turn is a square of side d 5 18 cm, and a uniform magnetic field directed perpendicular to the plane of the coil is turned on. If the field changes linearly from 0 to 0.50 T in 0.80 s, what is the magnitude
of the induced emf in the coil while the field is changing?
SOLUTION
Conceptualize From the description in the problem, imagine magnetic field lines passing through the coil. Because the
magnetic field is changing in magnitude, an emf is induced in the coil.
Categorize We will evaluate the emf using Faraday’s law from this section, so we categorize this example as a substitution
problem.
DFB
Bf 2 Bi
D(BA)
DB
5 NA
5 Nd 2
Dt
Dt
Dt
Evaluate Equation 23.3 for the situation described here,
noting that the magnetic field changes linearly with time:
ueu 5 N
Substitute numerical values:
u e u 5 (200)(0.18 m)2
Dt
5N
(0.50 T 2 0)
5 4.0 V
0.80 s
What If? What if you were asked to find the magnitude of the induced current in the coil while the field is changing? Can
you answer that question?
Answer If the ends of the coil are not connected to a circuit, the answer to this question is easy: the current is zero!
(Charges move within the wire of the coil, but they cannot move into or out of the ends of the coil.) For a steady current to
exist, the ends of the coil must be connected to an external circuit. Let’s assume the coil is connected to a circuit and the
total resistance of the coil and the circuit is 2.0 V. Then, the magnitude of the induced current in the coil is
I5
ueu
4.0 V
5
5 2.0 A
R
2.0 V
Exampl e 23.2 | An Exponentially Decaying Magnetic Field
A loop of wire enclosing an area A is placed in a region where the magnetic field is perpendicular to the plane of the
:
loop. The magnitude of B varies in time according to the expression B 5 Bmaxe2at, where a is some constant. That is, at
t 5 0, the field is Bmax, and for t . 0, the field decreases exponentially as in Figure 23.9 (page 786). Find the induced emf
in the loop as a function of time.
continued
786
CHAPTER 23 | Faraday’s Law and Inductance
23.2 cont.
SOLUTION
Conceptualize The physical situation is similar to that in
Example 23.1 except for two things: there is only one loop,
and the field varies exponentially with time rather than
linearly.
Categorize We will evaluate the emf using Faraday’s law
from this section, so we categorize this example as a substitution problem.
Evaluate Equation 23.2 for the situation
described here:
Figure 23.9 (Example 23.2)
Exponential decrease in the
magnitude of the magnetic
field with time. The induced
emf and induced current vary
with time in the same way.
dF
d
e 5 2 dt B 5 2 dt (ABmax e2at ) 5 2ABmax dtd e2at 5 aABmaxe2at
This expression indicates that the induced emf decays exponentially in time. The maximum emf occurs at t 5 0, where
emax 5 aABmax. The plot of e versus t is similar to the B-versus-t curve shown in Figure 23.9.
23.2 | Motional emf
ᐉ
Examples 23.1 and 23.2 are cases in which an emf is produced in a circuit when the
magnetic field changes with time. In this section, we describe motional emf, in which
an emf is induced in a conductor moving through a magnetic field. This is the situation described in Figure 23.1 at the beginning of Section 23.1.
Consider a straight conductor of length , moving with constant velocity through
a uniform magnetic field directed into the page as in Figure 23.10. For simplicity,
we shall assume that the conductor is moving with a velocity that is perpendicular
to the field. The electrons in the conductor experience a force along the conduc:
:
tor with magnitude u F B u 5 u q:
v 3 B u 5 qvB. According to Newton’s second law,
the electrons accelerate in response to this force and move along the conductor.
Once the electrons move to the lower end of the conductor, they accumulate there,
leaving a net positive charge at the upper end. As a result of this charge separation,
:
an electric field E is produced within the conductor. The charge at the ends of the
conductor builds up until the magnetic force qvB on an electron in the conductor
is balanced by the electric force qE on the electron as shown in Figure 23.10. At this
point, charge stops flowing. In this situation, the zero net force on an electron allows
us to relate the electric field to the magnetic field:
:
Figure 23.10 A straight
electrical conductor of length ,
moving with a velocity :
v through a
:
uniform magnetic field B directed
perpendicular to :
v.
:
:
o F 5 F e 2 F B 5 0 : qE 5 qvB
:
E 5 vB
Because the electric field produced in the conductor is uniform, it is related to
the potential difference across the ends of the conductor according to the relation
DV 5 E , (Section 20.2). Therefore,
DV 5 E , 5 B ,v
where the upper end is at a higher potential than the lower end. Therefore, a potential difference is maintained as long as the conductor is moving through the magnetic field. If the motion is reversed, the polarity of DV is also reversed.
An interesting situation occurs if we now consider what happens when the
moving conductor is part of a closed circuit. Consider a circuit consisting of a
conducting bar of length , sliding along two fixed parallel conducting rails as
in Active Figure 23.11a. For simplicity, we assume that the moving bar has zero
electrical resistance and that the stationary part of the circuit has a resistance R.
:
A uniform and constant magnetic field B is applied perpendicular to the plane
of the circuit.
As the bar is pulled to the right with a velocity :
v under the influence of an applied
:
force F app, free charges in the bar experience a magnetic force along the length of
787
23.2 | Motional emf
the bar. Because the moving bar is part of a complete circuit, a continuous current is established in
the circuit. In this case, the rate of change of magnetic flux through the loop and the accompanying
induced emf across the moving bar are proportional
to the change in loop area as the bar moves through
the magnetic field.
Because the area of the circuit at any instant is ,x,
the magnetic flux through the circuit is
FB 5 B ,x
e
ᐉ
ᐉ
where x is the width of the circuit, a parameter that
changes with time. Using Faraday’s law, we find that
the induced emf is
dF
e 5 2 dt B 5 2 dtd (B ,x) 5 2 B , dx
dt
e 5 2 B,v
b
a
23.5b
Active Figure 23.11 (a) A conducting bar sliding with a velocity :
v along two
Because the resistance of the circuit is R, the magnitude of the induced current is
conducting rails under the action of an applied force F app. (b) The equivalent
circuit diagram for the pictorial representation in (a).
:
I5
u eu
B ,v
5
R
R
23.6b
The equivalent circuit diagram for this example is shown in Active Figure 23.11b.
The moving bar is behaving like a battery in that it is a source of emf as long as the
bar continues to move.
Let us examine this situation using energy considerations in the nonisolated system model, with the system being the entire circuit. Because the circuit has no battery, you might wonder about the origin of the induced current and the energy
:
delivered to the resistor. Note that the external force F app does work on the conductor, thereby moving charges through a magnetic field, which causes the charges
to move along the conductor with some average drift velocity. Hence, a current is
established. From the viewpoint of the conservation of energy equation (Eq. 7.2),
the total work done on the system by the applied force while the bar moves with
constant speed must equal the increase in internal energy in the resistor during this
time interval. (This statement assumes that the energy stays in the resistor; in reality,
energy leaves the resistor by heat and electromagnetic radiation.)
:
As the conductor of length , moves through the uniform magnetic field B, it ex:
periences a magnetic force F B of magnitude I ,B (Eq. 22.10), where I is the current
induced due to its motion. The direction of this force is opposite the motion of the
bar, or to the left in Active Figure 23.11a.
If the bar is to move with a constant velocity, it is modeled as a particle in equilib:
rium, so the applied force F app must be equal in magnitude and opposite in direction
to the magnetic force, or to the right in Active Figure 23.11a. (If the magnetic force
acted in the direction of motion, it would cause the bar to accelerate once it was in
motion, thereby increasing its speed. This state of affairs would represent a violation of the principle of energy conservation.) Using Equation 23.6 and that Fapp 5
FB 5 I ,B, we find that the power delivered by the applied force is
P 5 Fappv 5 (I,B)v 5
B 2,2v 2
B ,v 2
5
R 5 I 2R
R
R
1 2
23.7b
This power is equal to the rate at which energy is delivered to the resistor, as we
expect.
Q U I CK QUI Z 23.3 You wish to move a rectangular loop of wire into a region of uniform magnetic field at a given speed so as to induce an emf in the loop. The plane of
the loop must remain perpendicular to the magnetic field lines. In which orientation
788
CHAPTER 23 | Faraday’s Law and Inductance
should you hold the loop while you move it into the region of magnetic field so as
to generate the largest emf ? (a) with the long dimension of the loop parallel to the
velocity vector (b) with the short dimension of the loop parallel to the velocity vector
(c) either way because the emf is the same regardless of orientation
QUI C K QU IZ 23.4 In Active Figure 23.11, a given applied force of magnitude Fapp
results in a constant speed v and a power input P. Imagine that the force is increased
so that the constant speed of the bar is doubled to 2v. Under these conditions, what
are the new force and the new power input? (a) 2F and 2P (b) 4F and 2P (c) 2F
and 4P (d) 4F and 4P
Example 23.3 | Motional emf Induced in a Rotating Bar
A conducting bar of length , rotates with a constant angular speed ␻ about a pivot
:
at one end. A uniform magnetic field B is directed perpendicular to the plane of
rotation as shown in Figure 23.12. Find the motional emf induced between the ends
of the bar.
ᐉ
SOLUTION
Conceptualize The rotating bar is different in nature from the sliding bar in Active
Figure 23.11. Consider a small segment of the bar, however. It is a short length of conductor moving in a magnetic field and has an emf generated in it like the sliding bar.
By thinking of each small segment as a source of emf, we see that all segments are in
series and the emfs add.
Categorize Based on the conceptualization of the problem, we approach this example by modeling each small segment of the bar as a conductor moving in a magnetic
field, with the added feature that the short segments of the bar are traveling in circular
paths.
Analyze Evaluate the magnitude of the emf induced in a
segment of the bar of length dr having a velocity :
v from
Equation 23.5:
Find the total emf between the ends of the bar by adding
the emfs induced across all segments:
The tangential speed v of an element is related to
the angular speed ␻ through the relationship v 5 r␻
(Eq. 10.10); use that fact and integrate:
Figure 23.12 (Example 23.3) A
conducting bar rotating around a pivot
at one end in a uniform magnetic field
that is perpendicular to the plane of
rotation. A motional emf is induced
between the ends of the bar.
d e 5 Bv dr
␧5
E
␧5B
Bv dr
E
v dr 5 B ␻
E
,
0
r dr 5 12 B␻,2
Finalize In Equation 23.5 for a sliding bar, we can increase e by increasing B, ,, or v. Increasing any one of these variables by
a given factor increases e by the same factor. Therefore, you would choose whichever of these three variables is most convenient to increase. For the rotating rod, however, there is an advantage to increasing the length of the rod to raise the emf because , is squared. Doubling the length gives four times the emf, whereas doubling the angular speed only doubles the emf.
What If? Suppose, after reading through this example, you come up with a brilliant idea. A Ferris wheel has radial metallic spokes between the hub and the circular rim. These spokes move in the magnetic field of the Earth, so each spoke acts
like the bar in Figure 23.12. You plan to use the emf generated by the rotation of the Ferris wheel to power the lightbulbs
on the wheel. Will this idea work?
Answer Let’s estimate the emf that is generated in this situation. The magnitude of the Earth’s magnetic field is about
B 5 0.5 3 1024 T. A typical spoke on a Ferris wheel might have a length on the order of 10 m. Suppose the period of rotation is on the order of 10 s.
2␲
2␲
␻5
5
5 0.63 s21 , 1 s21
Determine the angular speed of the spoke:
T
10 s
23.2 | Motional emf
789
23.3 cont.
Assume the magnetic field lines of the Earth are horizontal at the location of the Ferris wheel and perpendicular
to the spokes. Find the emf generated:
␧ 5 12B␻,2 5 12(0.5 3 1024 T)(1 s21 )(10 m)2
5 2.5 3 1023 V , 1 mV
This value is a tiny emf, far smaller than that required to operate lightbulbs.
An additional difficulty is related to energy. Even assuming you could find lightbulbs that operate using a potential difference on the order of millivolts, a spoke must be part of a circuit to provide a voltage to the lightbulbs. Consequently, the
spoke must carry a current. Because this current-carrying spoke is in a magnetic field, a magnetic force is exerted on the
spoke in the direction opposite its direction of motion. As a result, the motor of the Ferris wheel must supply more energy
to perform work against this magnetic drag force. The motor must ultimately provide the energy that is operating the lightbulbs, and you have not gained anything for free!
Exampl e 23.4 | Magnetic Force Acting on a Sliding Bar
The conducting bar illustrated in Figure 23.13 moves on two frictionless, parallel rails
in the presence of a uniform magnetic field directed into the page. The bar has mass
m, and its length is ,. The bar is given an initial velocity :
v i to the right and is released
at t 5 0.
ᐉ
(A) Using Newton’s laws, find the velocity of the bar as a function of time.
SOLUTION
Conceptualize As the bar slides to the right in Figure 23.13, a counterclockwise current
is established in the circuit consisting of the bar, the rails, and the resistor. The upward
current in the bar results in a magnetic force to the left on the bar as shown in the figure.
Therefore, the bar must slow down, so our mathematical solution should demonstrate that.
Categorize The text already categorizes this problem as one that uses Newton’s laws. We
Figure 23.13 (Example 23.4)
A conducting bar of length , on
two fixed conducting rails is given
an initial velocity :
v i to the right.
model the bar as a particle under a net force.
Analyze From Equation 22.10, the magnetic force is FB 5 2I ,B, where the negative sign indicates that the force is to the
left. The magnetic force is the only horizontal force acting on the bar.
Apply Newton’s second law to the bar in the horizontal
direction:
Fx 5 ma
Substitute I 5 B,v/R from Equation 23.6:
m
Rearrange the equation so that all occurrences of the
variable v are on the left and those of t are on the right:
Integrate this equation using the initial condition that
v 5 vi at t 5 0 and noting that (B 2, 2/mR) is a constant:
:
2I ,B 5 m
dv
dt
B 2, 2
dv
52
v
dt
R
dv
B 2, 2
dt
52
v
mR
1 2
E
v dv
vi
v
52
B 2, 2
mR
E
t
dt
0
1vv 2 5 21BmR, 2t
2 2
ln
i
Define the constant ␶ 5 mR/B 2,2 and solve for the
velocity:
(1) v 5 vie2t/␶
continued
790
CHAPTER 23 | Faraday’s Law and Inductance
23.4 cont.
Finalize This expression for v indicates that the velocity of the bar decreases with time under the action of the magnetic
force as expected from our conceptualization of the problem.
(B) Show that the same result is found by using an energy approach.
SOLUTION
Categorize The text of this part of the problem tells us to use an energy approach for the same situation. We model the
entire circuit in Figure 23.13 as an isolated system.
Analyze Consider the sliding bar as one system component possessing kinetic energy, which decreases because energy is
transferring out of the bar by electrical transmission through the rails. The resistor is another system component possessing
internal energy, which rises because energy is transferring into the resistor. Because energy is not leaving the system, the rate
of energy transfer out of the bar equals the rate of energy transfer into the resistor.
Equate the power entering the resistor to that leaving the
bar:
Substitute for the electrical power delivered to the resistor
and the time rate of change of kinetic energy for the bar:
Use Equation 23.6 for the current and carry out the
derivative:
Rearrange terms:
Presistor 5 2Pbar
d
I 2R 5 2 (12mv 2)
dt
B 2, 2v 2
dv
5 2mv
R
dt
dv
B 2, 2
dt
52
v
mR
1 2
Finalize This result is the same expression to be integrated that we found in part (A).
What If? Suppose you wished to increase the distance through which the bar moves between the time it is initially projected and the time it essentially comes to rest. You can do so by changing one of three variables—vi , R , or B—by a factor of
2 or 12 . Which variable should you change to maximize the distance, and would you double it or halve it?
Answer Increasing vi would make the bar move farther. Increasing R would decrease the current and therefore the magnetic force, making the bar move farther. Decreasing B would decrease the magnetic force and make the bar move farther.
Which method is most effective, though?
dx
v5
5 vi e2t/␶
Use Equation (1) to find the distance the bar moves by
dt
integration:
x5
E
`
0
vi e2t/␶ dt 5 2vi ␶e2t/␶
5 2vi ␶(0 2 1) 5 vi ␶ 5 vi
u
`
0
1BmR, 2
2 2
This expression shows that doubling vi or R will double the distance. Changing B by a factor of 12 , however, causes the
distance to be four times as great!
The Alternating-Current Generator
The alternating-current (AC) generator is a device in which energy is transferred in
by work and out by electrical transmission. A simplified pictorial representation of
an AC generator is shown in Active Figure 23.14a. It consists of a coil of wire rotated
in an external magnetic field by some external agent, which represents the work
input. In commercial power plants, the energy required to rotate the loop can be
derived from a variety of sources. In a hydroelectric plant, for example, falling water
directed against the blades of a turbine produces the rotary motion; in a coal-fired
23.3 | Lenz’s Law
plant, the high temperature produced by burning
the coal is used to convert water to steam and this
steam is directed against turbine blades. As the loop
rotates, the magnetic flux through it changes with
time, inducing an emf and a current in a circuit
connected to the coil.
Suppose the coil has N turns, all of the same area
A, and suppose the coil rotates with a constant angular speed ␻ about an axis perpendicular to the
magnetic field. If ␪ is the angle between the magnetic field and the direction perpendicular to the
plane of the coil, the magnetic flux through the
loop at any time t is given by
791
e
e
FB 5 BA cos ␪ 5 ⟩⟨ cos ␻t
where we have used the relationship between angular
position and a constant angular speed, ␪ 5 ␻t. (See
Eq. 10.7 and set the angular acceleration ␣ equal to
zero.) Hence, the induced emf in the coil is
e 5 2N
dFB
dt
5 2NBA
a
b
Active Figure 23.14 (a) Schematic diagram of an AC generator. (b) A
graphical representation of the alternating emf induced in the loop as a function of time.
d
(cos ␻t) 5 NBA␻ sin ␻t
dt
23.8b
This result shows that the emf varies sinusoidally with time as shown in Active
Figure 23.14b. From Equation 23.8, we see that the maximum emf has the value
emax 5 NBA␻, which occurs when ␻t 5 908 or 2708. In other words, e 5 emax when
the magnetic field is in the plane of the coil, and the time rate of change of flux
is a maximum. In this position, the velocity vector for a wire in the loop is perpendicular to the magnetic field vector. Furthermore, the emf is zero when ␻t 5 0 or
:
1808—that is, when B is perpendicular to the plane of the coil—and the time rate
of change of flux is zero. In this orientation, the velocity vector for a wire in the
loop is parallel to the magnetic field vector.
The sinusoidally varying emf in Equation 23.8 is the source of alternating current
delivered to customers of electrical utility companies. It is called AC voltage as opposed to the DC voltage from a source such as a battery.
23.3 | Lenz’s Law
Let us now address the negative sign in Faraday’s law. When a change occurs in the
magnetic flux, the direction of the induced emf and induced current can be found
from Lenz’s law:
The induced current in a loop is in the direction that creates a magnetic field
that opposes the change in magnetic flux through the area enclosed by the
loop. That is, the induced current tends to keep the original magnetic flux
through the loop from changing.
Notice that no equation is associated with Lenz’s law. The law is in words only and
provides a means for determining the direction of the current in a circuit when a
magnetic change occurs.
THINKING PHYSICS 23.2
A transformer (Fig. 23.15) consists of a pair of coils wrapped around an iron
form. When AC voltage is applied to one coil, the primary, the magnetic field
lines cutting through the other coil, the secondary, induce an emf across a load
resistor RL. (This arrangement is used in Faraday’s experiment shown in Active
Fig. 23.3.) By varying the number of turns of wire on each coil, the AC voltage
Figure 23.15 (Thinking Physics
23.2) An ideal transformer consists of
two coils of wire wound on the same
iron core.
792
CHAPTER 23 | Faraday’s Law and Inductance
in the secondary can be made larger or smaller than that in the primary. Clearly,
this device cannot work with DC voltage. What’s more, if DC voltage is applied,
the primary coil sometimes overheats and burns. Why?
Reasoning When a current exists in the primary coil, the magnetic field lines
from this current pass through the coil itself. Therefore, any change in the current causes a change in the magnetic field that in turn induces a current in the
same coil. According to Lenz’s law, this current is in the direction opposite the
original current. The result is that when an AC voltage is applied, the opposing
emf due to Lenz’s law limits the current in the coil to a low value. If DC voltage
is applied, no opposing emf occurs and the current can rise to a higher value.
This increased current causes the temperature of the coil to rise, to the point at
which the insulation on the wire sometimes burns. b
Pitfall Prevention | 23.2
Induced Current Opposes
the Change
The induced current in a circuit
opposes the change in the magnetic
field, not the field itself. Therefore,
in some cases the magnetic field due
to the induced current is in the same
direction as the changing external
magnetic field. Such is the case if the
external magnetic field is decreasing
in magnitude, for example.
To attain a better understanding of Lenz’s law, let us return to the example of
a bar moving to the right on two parallel rails in the presence of a uniform magnetic field directed into the page (Fig. 23.16a). As the bar moves to the right, the
magnetic flux through the circuit increases with time because the area of the loop
increases. Lenz’s law says that the induced current must be in such a direction
that the magnetic field it produces opposes the change in the magnetic flux of the
external magnetic field. Because the flux is due to an external field into the page
and is increasing, the induced current, if it is to oppose the change, must produce
a magnetic field through the circuit out of the page. Hence, the induced current
must be counterclockwise when the bar moves to the right to give a counteracting field out of the page in the region inside the loop. (Use the right-hand rule
to verify this direction.) If the bar is moving to the left, as in Figure 23.16b, the
magnetic flux through the loop decreases with time. Because the magnetic field
is into the page, the induced current has to be clockwise to produce a magnetic
field into the page inside the loop. In either case, the induced current attempts to
maintain the original flux through the circuit.
Let us examine this situation from the viewpoint of energy considerations.
Suppose the bar is given a slight push to the right. In the preceding analysis, we
found that this motion leads to a counterclockwise current in the loop. What
happens if we incorrectly assume that the current is clockwise? For a clockwise
Figure 23.16 (a) Lenz’s law can be
used to determine the direction of
the induced current. (b) When the
bar moves to the left, the induced
current must be clockwise. Why?
a
b
23.3 | Lenz’s Law
current I, the direction of the magnetic force I <B on the sliding bar would be
to the right. According to Newton’s second law, this force would accelerate the
rod and increase its speed, which in turn would cause the area of the loop to
increase more rapidly. This increase would increase the induced current, which
would increase the force, which would increase the current, and so on. In effect, the system would acquire energy with no additional energy input. This
result is clearly inconsistent with all experience and with the conservation of
energy equation. Therefore, we are forced to conclude that the current must be
counterclockwise.
Q U I CK QUI Z 23.5 In equal-arm balances from the early 20th century (Fig. 23.17),
an aluminum sheet hangs from one of the arms and passes between the poles of a
magnet, which causes the oscillations of the balance to decay rapidly. In the absence
of such magnetic braking, the oscillation might continue for a long time and the
experimenter would have to wait to take a reading. Why do the oscillations decay?
(a) because the aluminum sheet is attracted to the magnet (b) because currents in
the aluminum sheet set up a magnetic field that opposes the oscillations (c) because
aluminum is paramagnetic
Figure 23.17 (Quick
John W. Jewett, Jr.
Quiz 23.5) In an oldfashioned equal-arm
balance, an aluminum sheet
hangs between the poles of
a magnet.
THINKING PHYSICS 23.3
A magnet is placed near a metal loop as shown in Figure 23.18a.
(A) Find the direction of the induced current in the loop when the magnet is pushed toward the loop.
a
b
c
Figure 23.18 (Thinking Physics 23.3) A moving bar magnet induces a current in a conducting loop.
d
793
794
CHAPTER 23 | Faraday’s Law and Inductance
Reasoning As the magnet moves to the right toward the loop, the external magnetic flux through the loop in-
creases with time. To counteract this increase in flux due to a field toward the right, the induced current produces
its own magnetic field to the left as illustrated in Figure 23.18b; hence, the induced current is in the direction
shown. Knowing that like magnetic poles repel each other, we conclude that the left face of the current loop acts
like a north pole and the right face acts like a south pole.
(B) Find the direction of the induced current in the loop when the magnet is pulled away from the loop.
Reasoning If the magnet moves to the left as in Figure 23.18c, its flux through the area enclosed by the loop de-
creases in time. Now the induced current in the loop is in the direction shown in Figure 23.18d because this current
direction produces a magnetic field in the same direction as the external field. In this case, the left face of the loop
is a south pole and the right face is a north pole. b
23.4 | Induced emfs and Electric Fields
We have seen that a changing magnetic flux induces an emf and a current in a conducting loop. We can also interpret this phenomenon from another point of view.
Because the normal flow of charges in a circuit is due to an electric field in the wires
set up by a source such as a battery, we can interpret the changing magnetic field as
creating an induced electric field. This electric field applies a force on the charges to
cause them to move. With this approach, then, we see that an electric field is created
in the conductor as a result of changing magnetic flux. In fact, the law of electromagnetic induction can be interpreted as follows: An electric field is always generated by a changing magnetic flux, even in free space where no charges are present.
This induced electric field, however, has quite different properties from those of the
electrostatic field produced by stationary charges.
Let us illustrate this point by considering a conducting loop of radius r, situated
in a uniform magnetic field that is perpendicular to the plane of the loop as in Figure 23.19. If the magnetic field changes with time, Faraday’s law tells us that an emf
e 5 2d FB/dt is induced in the loop. The:induced current thus produced implies
the presence of an induced electric field E that must be tangent to the loop so as
to provide an electric force on the charges around the loop. The work done by the
electric field on the loop in moving a test charge q once around the loop is equal to
W 5 q e. Because the magnitude of the electric force on the charge is qE , the work
done on the charge by the electric field can also be expressed from Equation 6.8
:
as W 5 # F ? d :
r 5 qE(2␲r), where 2␲r is the circumference of the loop. These two
expressions for the work must be equal; therefore, we see that
q e 5 qE(2␲r)
E5
Figure 23.19 A conducting loop
of radius r in a uniform magnetic
field perpendicular to the plane of
the loop.
e
2␲r
Using this result along with Faraday’s law and that FB 5 BA 5 B␲r 2 for a circular
loop, we find that the induced electric field can be expressed as
E52
1 dFB
1 d
r dB
52
(B␲r 2) 5 2
2␲r dt
2␲r dt
2 dt
This expression can be used to calculate the induced electric field if the time variation
of the magnetic field is specified. The negative sign indicates that the induced electric
:
field E results in a current that opposes the change in the magnetic field. It is important
to understand that this result is also valid in the absence of a conductor or charges. That
is, the same electric field is induced by the changing magnetic field in empty space.
In general, the magnitude of the emf for any closed path can be expressed as the
:
:
line integral of E ? d :
s over that path: e 5 rE ? d :
s (Eq. 20.3). Hence, the general
form of Faraday’s law of induction is
c Faraday’s law in general form
R
:
E ? d:
s 52
dFB
dt
23.9b
23.4 | Induced emfs and Electric Fields
795
:
It is important to recognize that the induced electric field E that appears in Equation 23.9 is a nonconservative field that is generated by a changing magnetic field.
We call it a nonconservative field because the work done in moving a charge around
a closed path (the loop in Fig. 23.19) is not zero. This type of electric field is very
different from an electrostatic field.
Q U I CK QUI Z 23.6 In a region of space, a magnetic field is uniform over space but
increases at a constant rate. This changing magnetic field induces an electric field
that (a) increases in time, (b) is conservative, (c) is in the direction of the magnetic
field, or (d) has a constant magnitude.
THINKING PHYSICS 23.4
In studying electric fields, we noted that electric field lines begin on positive
charges and end on negative charges. Do all electric field lines begin and end
on charges?
Reasoning The statement that electric field lines begin and end on charges is
true only for electrostatic fields, that is, electric fields due to stationary charges.
Electric field lines due to changing magnetic fields form closed loops, with no
beginning and no end, and are independent of the presence of charges. b
Exampl e 23.5 | Electric Field Induced by a Changing Magnetic Field in a Solenoid
A long solenoid of radius R has n turns of wire per unit length and carries a time-varying
current that varies sinusoidally as I 5 Imax cos ␻t, where Imax is the maximum current and
␻ is the angular frequency of the alternating current source (Fig. 23.20).
(A) Determine the magnitude of the induced electric field outside the solenoid at a
distance r . R from its long central axis.
SOLUTION
Conceptualize Figure 23.20 shows the physical situation. As the current in the coil
changes, imagine a changing magnetic field at all points in space as well as an induced
electric field.
Categorize Because the current varies in time, the magnetic field is changing, leading
to an induced electric field as opposed to the electrostatic electric fields due to stationary
electric charges.
Analyze First consider an external point and take the path for the line integral to be a
circle of radius r centered on the solenoid as illustrated in Figure 23.20.
Evaluate the right side of Equation 23.9, noting
:
that the magnetic field B inside the solenoid is
perpendicular to the circle bounded by the path of
integration:
(1) 2
dFB
dt
52
dB
d
(B␲R 2 ) 5 2 ␲R 2
dt
dt
Evaluate the magnetic field in the solenoid from
Equation 22.32:
(2) B 5 ␮0nI 5 ␮0nImax cos ␻t
Substitute Equation (2) into Equation (1):
(3) 2
Evaluate the left side of Equation 23.9, noting that
:
the magnitude of E is constant on the path of inte:
gration and E is tangent to it:
(4)
R
d FB
dt
:
Figure 23.20 (Example 23.5)
A long solenoid carrying a
time-varying current given by
I 5 Imax cos ␻t. An electric field is
induced both inside and outside
the solenoid.
5 2 ␲R 2␮0nImax
E ?d:
s 5 E(2␲r)
d
(cos ␻t) 5 ␲R 2␮0nImax␻ sin ␻t
dt
796
CHAPTER 23 | Faraday’s Law and Inductance
23.5 cont.
Substitute Equations (3) and (4) into Equation 23.9:
E(2␲r) 5 ␲R 2␮0nImax␻ sin ␻t
Solve for the magnitude of the electric field:
E5
␮0nImax␻R 2
2r
sin ␻t
(for r . R)
Finalize This result shows that the amplitude of the electric field outside the solenoid falls off as 1/r and varies sinusoidally
with time. As we will learn in Chapter 24, the time-varying electric field creates an additional contribution to the magnetic
field. The magnetic field can be somewhat stronger than we first stated, both inside and outside the solenoid. The correction to the magnetic field is small if the angular frequency ␻ is small. At high frequencies, however, a new phenomenon can
dominate: The electric and magnetic fields, each re-creating the other, constitute an electromagnetic wave radiated by the
solenoid as we will study in Chapter 24.
(B) What is the magnitude of the induced electric field inside the solenoid, a distance r from its axis?
SOLUTION
Analyze For an interior point (r , R), the magnetic flux through an integration loop is given by FB 5 B␲r 2.
d FB
dB
d
(B␲r 2) 5 2 ␲r 2
dt
dt
Evaluate the right side of Equation 23.9:
(5) 2
Substitute Equation (2) into Equation (5):
(6) 2
Substitute Equations (4) and (6) into Equation 23.9:
E(2␲r) 5 ␲r 2␮0nImax␻ sin ␻t
Solve for the magnitude of the electric field:
E5
dt
d FB
dt
52
5 2 ␲r 2␮0 nImax
␮0nImax␻
2
r sin ␻t
d
(cos ␻t) 5 ␲r 2␮0nImax␻ sin ␻t
dt
(for r , R)
Finalize This result shows that the amplitude of the electric field induced inside the solenoid by the changing magnetic
flux through the solenoid increases linearly with r and varies sinusoidally with time.
23.5 | Inductance
e
Figure 23.21 Self-induction in a
simple circuit
Consider an isolated circuit consisting of a switch, a resistor, and a source of emf
as in Figure 23.21. The circuit diagram is represented in perspective so that we
can see the orientations of some of the magnetic field lines due to the current in
the circuit. When the switch is thrown to its closed position, the current doesn’t
immediately jump from zero to its maximum value e/R; the law of electromagnetic induction (Faraday’s law) describes the actual behavior. As the current increases with time, the magnetic flux through the loop of the circuit itself due to
the current also increases with time. This increasing magnetic flux from the circuit
induces an emf in the circuit (sometimes referred to as a back emf ) that opposes
the change in the net magnetic flux through the loop of the circuit. By Lenz’s law,
the induced electric field in the wires must therefore be opposite the direction
of the current, and the opposing emf results in a gradual increase in the current.
This effect is called self-induction because the changing magnetic flux through the
circuit arises from the circuit itself. The emf set up in this case is called a selfinduced emf.
To obtain a quantitative description of self-induction, we recall from Faraday’s law
that the induced emf is the negative time rate of change of the magnetic flux. The
magnetic flux is proportional to the magnetic field, which in turn is proportional to
the current in the circuit. Therefore, the self-induced emf is always proportional to
the time rate of change of the current. For a closely spaced coil of N turns of fixed
797
23.5 | Inductance
geometry (a toroidal coil or the ideal solenoid), we can express this proportionality
as follows:
dF
eL 5 2 N B 5 2 L dI
23.10b
dt
dt
c Self-induced emf
L5
N FB
I
Brady-Handy Collection, Library of Congress Prints and
Photographs Division [LC-BH83- 997]
where L is a proportionality constant, called the inductance of the coil, that depends
on the geometric features of the coil and other physical characteristics. From this
expression, we see that the inductance of a coil containing N turns is
23.11b
where it is assumed that the same magnetic flux passes through each turn. Later we
shall use this equation to calculate the inductance of some special coil geometries.
From Equation 23.10, we can also write the inductance as the ratio
L52
eL
23.12b
dI/dt
which is usually taken to be the defining equation for the inductance of any coil,
regardless of its shape, size, or material characteristics. If we compare Equation 23.10
with Equation 21.6, R 5 DV/I, we see that resistance is a measure of opposition to
current, whereas inductance is a measure of opposition to the change in current.
The SI unit of inductance is the henry (H), which, from Equation 23.12, is seen to
be equal to 1 volt ? second per ampere:
Joseph Henry
American Physicist (1797–1878)
Henry became the first director of the
Smithsonian Institution and first president
of the Academy of Natural Science. He
improved the design of the electromagnet
and constructed one of the first motors.
He also discovered the phenomenon of
self-induction, but he failed to publish his
findings. The unit of inductance, the henry,
is named in his honor.
1 H 5 1 V ? s/A
As we shall see, the inductance of a coil depends on its geometry. Because inductance calculations can be quite difficult for complicated geometries, the examples
we shall explore involve simple situations for which inductances are easily evaluated.
Exampl e 23.6 | Inductance of a Solenoid
Consider a uniformly wound solenoid having N turns and length ,. Assume , is much longer than the radius of the
windings and the core of the solenoid is air.
(A) Find the inductance of the solenoid.
SOLUTION
Conceptualize The magnetic field lines from each turn of the solenoid pass through all the turns, so an induced emf in
each coil opposes changes in the current.
Categorize Because the solenoid is long, we can use the results for an ideal solenoid obtained in Chapter 22.
Analyze Find the magnetic flux through each turn of
FB 5 BA 5 ␮0nIA 5 ␮0
area A in the solenoid, using the expression for the magnetic field from Equation 22.32:
Substitute this expression into Equation 23.11:
(1) L 5
N FB
I
5 ␮0
N
IA
,
N2
A
,
(B) Calculate the inductance of the solenoid if it contains 300 turns, its length is 25.0 cm, and its cross-sectional area is
4.00 cm2.
SOLUTION
Substitute numerical values into Equation (1):
L 5 (4␲ 3 1027 T . m/A)
3002
(4.00 3 1024 m2)
25.0 3 1022 m
5 1.81 3 1024 T ? m2/A 5 0.181 mH
continued
798
CHAPTER 23 | Faraday’s Law and Inductance
23.6 cont.
(C) Calculate the self-induced emf in the solenoid if the current it carries decreases at the rate of 50.0 A/s.
SOLUTION
Substitute dI/dt 5 250.0 A/s and the answer to part (B)
into Equation 23.10:
␧L 5 2L dI 5 2(1.81 3 1024 H)(250.0 A/s)
dt
5 9.05 mV
Finalize The result for part (A) shows that L depends on geometry and is proportional to the square of the number of
turns. Because N 5 n,, we can also express the result in the form
L 5 ␮0
(n,)2
A 5 ␮0n2A, 5 ␮0n2V
,
where V 5 A, is the interior volume of the solenoid.
23.6 | RL Circuits
e
Active Figure 23.22 An RL
circuit. When switch S2 is in position
a, the battery is in the circuit.
If a circuit contains a coil such as a solenoid, the inductance of the coil prevents
the current in the circuit from increasing or decreasing instantaneously. A circuit
element that has a large inductance is called an inductor and has the circuit symbol
. We always assume the inductance of the remainder of a circuit is negligible
compared with that of the inductor. Keep in mind, however, that even a circuit without a coil has some inductance that can affect the circuit’s behavior.
Because the inductance of an inductor results in a back emf, an inductor in a
circuit opposes changes in the current in that circuit. The inductor attempts to keep
the current the same as it was before the change occurred. If the battery voltage in
the circuit is increased so that the current rises, the inductor opposes this change
and the rise is not instantaneous. If the battery voltage is decreased, the inductor
causes a slow drop in the current rather than an immediate drop. Therefore, the
inductor causes the circuit to be “sluggish” as it reacts to changes in the voltage.
Consider the circuit shown in Active Figure 23.22, which contains a battery of
negligible internal resistance. This circuit is an RL circuit because the elements connected to the battery are a resistor and an inductor. The curved lines on switch S2
suggest this switch can never be open; it is always set to either a or b. (If the switch is
connected to neither a nor b, any current in the circuit suddenly stops.) Suppose S2
is set to a and switch S1 is open for t , 0 and then thrown closed at t 5 0. The current in the circuit begins to increase, and a back emf (Eq. 23.10) that opposes the
increasing current is induced in the inductor.
With this point in mind, let’s apply Kirchhoff’s loop rule to this circuit, traversing
the circuit in the clockwise direction:
e 2 IR 2 L dI
50
dt
23.13b
where IR is the voltage drop across the resistor. (Kirchhoff’s rules were developed
for circuits with steady currents, but they can also be applied to a circuit in which
the current is changing if we imagine them to represent the circuit at one instant of
time.) Now let’s find a solution to this differential equation, which is similar to that
for the RC circuit (see Section 21.9).
A mathematical solution of Equation 23.13 represents the current in the circuit
as a function of time. To find this solution, we change variables for convenience, letting x 5 (e/R) 2 I, so dx 5 2dI. With these substitutions, Equation 23.13 becomes
x1
L dx
50
R dt
23.6 | RL Circuits
799
Rearranging and integrating this last expression gives
E
x dx
52
R
L
E
x0
x
ln
x
R
52 t
x0
L
t
dt
0
where x0 is the value of x at time t 5 0. Taking the antilogarithm of this result gives
x 5 x0e2Rt/L
Because I 5 0 at t 5 0, note from the definition of x that x0 5 e/R. Hence, this last
expression is equivalent to
e 2 I 5 e e2Rt/L
R
R
I5
e (1 2 e2Rt/L)
R
This expression shows how the inductor affects the current. The current does not
increase instantly to its final equilibrium value when the switch is closed, but instead increases according to an exponential function. If the inductance is removed
from the circuit, which corresponds to letting L approach zero, the exponential
term becomes zero and there is no time dependence of the current in this case; the
current increases instantaneously to its final equilibrium value in the absence of the
inductance.
We can also write this expression as
I5
e (1 2 e2t/␶)
R
L
R
23.15b
dI
50
dt
t
Active Figure 23.23 Plot of the
current versus time for the RL circuit
shown in Active Figure 23.22. The
time constant ␶ is the time interval
required for I to reach 63.2% of its
maximum value.
23.16b
This result shows that the time rate of change of the current is a maximum (equal to
e/L) at t 5 0 and falls off exponentially to zero as t approaches infinity (Fig. 23.24).
Now consider the RL circuit in Active Figure 23.22 again. Suppose switch S2 has
been set at position a long enough (and switch S1 remains closed) to allow the current to reach its equilibrium value e/R. In this situation, the circuit is described by
the outer loop in Active Figure 23.22. If S2 is thrown from a to b, the circuit is now
described by only the right-hand loop in Active Figure 23.22. Therefore, the battery
has been eliminated from the circuit. Setting e 5 0 in Equation 23.13 gives
IR 1 L
e
t
Physically, ␶ is the time interval required for the current in the circuit to reach
(1 2 e21) 5 0.632 5 63.2% of its final value e/R. The time constant is a useful parameter for comparing the time responses of various circuits.
Active Figure 23.23 shows a graph of the current versus time in the RL circuit.
Notice that the equilibrium value of the current, which occurs as t approaches infinity, is e/R. That can be seen by setting dI/dt equal to zero in Equation 23.13 and
solving for the current I. (At equilibrium, the change in the current is zero.) Therefore, the current initially increases very rapidly and then gradually approaches the
equilibrium value e/R as t approaches infinity.
Let’s also investigate the time rate of change of the current. Taking the first time
derivative of Equation 23.14 gives
dI
e
5 e 2t/␶
dt
L
e
23.14b
where the constant ␶ is the time constant of the RL circuit:
␶5
e
23.17b
e
Figure 23.24 Plot of dI/dt
versus time for the RL circuit shown
in Active Figure 23.22. The rate
decreases exponentially with time as I
increases toward its maximum value.
800
CHAPTER 23 | Faraday’s Law and Inductance
It is left as a problem (Problem 38) to show that the solution of this differential
equation is
e
I5
e
Active Figure 23.25 Current
versus time for the right-hand loop
of the circuit shown in Active Figure 23.22. For t , 0, switch S2 is at
position a.
e 2t/␶
R
e
5 Ii e2t/␶
23.18b
where e is the emf of the battery and Ii 5 e/R is the initial current at the instant the
switch is thrown to b.
If the circuit did not contain an inductor, the current would immediately decrease
to zero when the battery is removed. When the inductor is present, it opposes the
decrease in the current and causes the current to decrease exponentially. A graph
of the current in the circuit versus time (Active Fig. 23.25) shows that the current is
continuously decreasing with time.
QUI C K QU IZ 23.7 The circuit in Figure 23.26 includes a power source that provides
a sinusoidal voltage. Therefore, the magnetic field in the inductor is constantly
changing. The inductor is a simple air-core solenoid. The switch in the circuit is
closed and the lightbulb glows steadily. An iron rod is inserted into the interior of
the solenoid, which increases the magnitude of the magnetic field in the solenoid.
As that happens, the brightness of the lightbulb (a) increases, (b) decreases, or
(c) is unaffected.
Figure 23.26 (Quick Quiz 23.7) A lightbulb
is powered by an AC source with an inductor in
the circuit. When the iron bar is inserted into
the coil, what happens to the brightness of the
lightbulb?
QUI C K QU IZ 23.8 Two circuits like the one shown in Active Figure 23.22 are
identical except for the value of L. In circuit A, the inductance of the inductor
is LA, and in circuit B, it is LB. Switch S2 has been in position b for both circuits
for a long time. At t 5 0, the switch is thrown to a in both circuits. At t 5 10 s,
the switch is thrown to b in both circuits. The resulting graphical representation
of the current as a function of time is shown in Figure 23.27. Assuming that the
time constant of each circuit is much less than 10 s, which of the following is true?
(a) LA . LB. (b) LA , LB. (c) There is not enough information to determine the
relative values.
Figure 23.27 (Quick Quiz 23.8)
Current–time graphs for two circuits with
different inductances.
23.7 | Energy Stored in a Magnetic Field
801
Exampl e 23.7 | Time Constant of an RL Circuit
Consider the circuit in Active Figure 23.22 again. Suppose the circuit elements have the following values: e 5 12.0 V,
R 5 6.00 V, and L 5 30.0 mH.
(A) Find the time constant of the circuit.
SOLUTION
Conceptualize You should understand the behavior of this circuit from the discussion in this section.
Categorize We evaluate the results using equations developed in this section, so this example is a substitution problem.
L
30.0 3 1023 H
5 5.00 ms
5
R
6.00 V
␶5
Evaluate the time constant from Equation 23.15:
(B) Switch S2 is at position a, and switch S1 is thrown closed at t 5 0. Calculate the current in the circuit at t 5 2.00 ms.
SOLUTION
Evaluate the current at t 5 2.00 ms from
Equation 23.14:
I5
␧ (1 2 e2t/␶
R
)5
12.0 V
(1 2 e22.00 ms/5.00 ms) 5 2.00 A (1 2 e20.400)
6.00 V
5 0.659 A
(C) Compare the potential difference across the resistor with that across the inductor.
SOLUTION
At the instant the switch is closed, there is no current and therefore no potential difference across the resistor. At this instant, the battery voltage appears entirely across
the inductor in the form of a back emf of 12.0 V as the inductor tries to maintain the
zero-current condition. (The top end of the inductor in Active Fig. 23.22 is at a higher
electric potential than the bottom end.) As time passes, the emf across the inductor
decreases and the current in the resistor (and hence the voltage across it) increases as
shown in Figure 23.28. The sum of the two voltages at all times is 12.0 V.
Figure 23.28 (Example 23.7)
The time behavior of the voltages
across the resistor and inductor in
Active Figure 23.22 given the values
provided in this example.
23.7 | Energy Stored in a Magnetic Field
In the preceding section, we found that the induced emf set up by an inductor
prevents a battery from establishing an instantaneous current. Part of the energy
supplied by the battery goes into internal energy in the resistor, and the remaining
energy is stored in the inductor. If we multiply each term in Equation 23.13 by the
current I and rearrange the expression, we have
I e 5 I 2R 1 LI
dI
dt
23.19b
This expression tells us that the rate I e at which energy is supplied by the battery equals the sum of the rate I 2R at which energy is delivered to the resistor
and the rate LI (dI/dt) at which energy is delivered to the inductor. Therefore,
Equation 23.19 is simply an expression of energy conservation for the isolated
system of the circuit. (Actually, energy can leave the circuit by thermal conduction into the air and by electromagnetic radiation, so the system need not be
completely isolated.) If we let U denote the energy stored in the inductor at
any time, the rate dU/dt at which energy is delivered to the inductor can be
written as
dU
dI
5 LI
dt
dt
Pitfall Prevention | 23.3
Capacitors, Resistors, and
Inductors Store Energy Differently
Different energy-storage mechanisms
are at work in capacitors, inductors,
and resistors. A charged capacitor
stores energy as electrical potential
energy. An inductor stores energy as
what we could call magnetic potential
energy when it carries current.
Energy delivered to a resistor is
transformed to internal energy.
802 CHAPTER 23 | Faraday’s Law and Inductance
To find the total energy stored in the inductor at any instant, we can rewrite this
expression as dU 5 LI dI and integrate:
U5
E
U
dU 5
0
E
I
LI dI
0
U 5 12LI 2
c Energy stored in an inductor
23.20b
where L is constant and so has been removed from the integral. Equation 23.20 represents the energy stored in the magnetic field of the inductor when the current is I.
Equation 23.20 is similar to the equation for the energy stored in the electric field
of a capacitor, U 5 12C (DV )2 (Eq. 20.29). In either case, we see that energy from a
battery is required to establish a field and that energy is stored in the field. In the
case of the capacitor, we can conceptually relate the energy stored in the capacitor to
the electric potential energy associated with the separated charge on the plates. We
have not discussed a magnetic analogy to electric potential energy, so the storage of
energy in an inductor is not as easy to conceptualize.
To argue that energy is stored in an inductor, consider the circuit in Figure 23.29a, which is the same circuit as in Active Figure 23.22, with the addition
of a switch S3 across the resistor R. With switch S2 set to position a and S3 closed
as shown, a current is established in the inductor. Now, as in Figure 23.29b, switch
S2 is thrown to position b. The current persists in this (ideally) resistance-free and
battery-free circuit (the right-hand loop in Fig. 23.29b), consisting of only the inductor and a conducting path between its ends. There is no current in the resistor
(because the path around it through S3 is resistance free), so no energy is being
delivered to it. The next step is to open switch S3 as shown in Figure 23.29c, which
puts the resistor into the circuit. There is now current in the resistor, and energy
is delivered to the resistor. Where is the energy coming from? The only other element in the circuit previous to opening switch S3 was the inductor. Energy must
therefore have been stored in the inductor and is now being delivered to the
resistor.
Now let us determine the energy per unit volume, or energy density, stored in a
magnetic field. For simplicity, consider a solenoid whose inductance is L 5 ␮0n2A,
(see Example 23.6). The magnetic field of the solenoid is B 5 ␮0nI. Substituting the
expression for L and I 5 B/␮0n into Equation 23.20 gives
U 5 12LI 2 5 12␮0n2A,
e
e
a
B2
B 2
5
(A,)
␮0n
2␮0
1 2
e
b
c
Figure 23.29 An RL circuit used for conceptualizing energy storage in an inductor.
23.21b
23.7 | Energy Stored in a Magnetic Field
803
Because A, is the volume of the solenoid, the energy stored per unit volume in a
magnetic field—in other words, the magnetic energy density—is
uB 5
U
B2
5
A,
2␮0
23.22b
c Magnetic energy density
Although Equation 23.22 was derived for the special case of a solenoid, it is valid
for any region of space in which a magnetic field exists. Note that it is similar to the
equation for the energy per unit volume stored in an electric field, given by 12 ⑀0E 2
(Eq. 20.31). In both cases, the energy density is proportional to the square of the
magnitude of the field.
Q U I CK QUI Z 23.9 You are performing an experiment that requires the highestpossible magnetic energy density in the interior of a very long current-carrying solenoid. Which of the following adjustments increases the energy density? (More than
one choice may be correct.) (a) increasing the number of turns per unit length on
the solenoid (b) increasing the cross-sectional area of the solenoid (c) increasing
only the length of the solenoid while keeping the number of turns per unit length
fixed (d) increasing the current in the solenoid
Exampl e 23.8 | What Happens to the Energy in the Inductor?
Consider once again the RL circuit shown in Active Figure 23.22, with switch S2 at position a and the current having
reached its steady-state value. When S2 is thrown to position b, the current in the right-hand loop decays exponentially
with time according to the expression I 5 Ii e2t/␶, where Ii 5 e/R is the initial current in the circuit and ␶ 5 L/R is the
time constant. Show that all the energy initially stored in the magnetic field of the inductor appears as internal energy in
the resistor as the current decays to zero.
SOLUTION
Conceptualize Before S2 is thrown to b, energy is being delivered at a constant rate to the resistor from the battery and
energy is stored in the magnetic field of the inductor. After t 5 0, when S2 is thrown to b, the battery can no longer provide
energy and energy is delivered to the resistor only from the inductor.
Categorize We model the right-hand loop of the circuit as an isolated system so that energy is transferred between compo-
nents of the system but does not leave the system.
Analyze The energy in the magnetic field of the inductor at any time is U. The rate dU/dt at which energy leaves the inductor and is delivered to the resistor is equal to I 2R, where I is the instantaneous current.
dU
Substitute the current given by Equation 23.18 into
5 I 2R 5 (Ii e2Rt/L)2R 5 Ii 2R e22Rt/L
dt
dU/dt 5 I 2R :
E
`
Solve for dU and integrate this expression over the limits
t 5 0 to t : `:
U5
The value of the definite integral can be shown to be
L/2R (see Problem 74). Use this result to evaluate U :
U 5 Ii 2R
0
Ii 2R e22Rt/L dt 5 Ii 2R
E
`
e22Rt/L dt
0
12RL 2 5 LI
1
2
i
2
Finalize This result is equal to the initial energy stored in the magnetic field of the inductor, given by Equation 23.20, as
we set out to prove.
Exampl e 23.9 | The Coaxial Cable
Coaxial cables are often used to connect electrical devices, such as your video system, and in receiving signals in television
cable systems. Model a long coaxial cable as a thin, cylindrical conducting shell of radius b concentric with a solid cylinder
of radius a as in Figure 23.30 (page 804). The conductors carry the same current I in opposite directions. Calculate the
inductance L of a length , of this cable.
continued
804
CHAPTER 23 | Faraday’s Law and Inductance
23.9 cont.
SOLUTION
Conceptualize Consider Figure 23.30. Although we do not have a visible coil in this
geometry, imagine a thin, radial slice of the coaxial cable such as the light gold rectangle in Figure 23.30. If the inner and outer conductors are connected at the ends
of the cable (above and below the figure), this slice represents one large conducting loop. The current in the loop sets up a magnetic field between the inner and
outer conductors that passes through this loop. If the current changes, the magnetic
field changes and the induced emf opposes the original change in the current in the
conductors.
ᐉ
Categorize We categorize this situation as one in which we must return to the fundamental definition of inductance, Equation 23.11.
Analyze We must find the magnetic flux through the light gold rectangle in FigFigure 23.30 (Example 23.9)
ure 23.30. Ampère’s law (see Section 22.9) tells us that the magnetic field in the
Section
of a long coaxial cable. The
region between the conductors is due to the inner conductor alone and that its maginner and outer conductors carry
nitude is B 5 ␮0I/2␲r, where r is measured from the common center of the cylinders.
equal currents in opposite directions.
A sample circular field line is shown in Figure 23.30, along with a field vector tangent
to the field line. The magnetic field is zero outside the outer shell because the net current passing through the area en:
closed by a circular path surrounding the cable is zero; hence, from Ampère’s law, r B ? d :
s 5 0.
The magnetic field is perpendicular to the light gold rectangle of length , and width b 2 a, the cross section of interest. Because the magnetic field varies with radial position across this rectangle, we must use calculus to find the total magnetic flux.
Divide the light gold rectangle into strips of width dr such
as the darker strip in Figure 23.30. Evaluate the magnetic
flux through such a strip:
d FB 5 B dA 5 B , dr
Substitute for the magnetic field and integrate over the
entire light gold rectangle:
FB 5
Use Equation 23.11 to find the inductance of the cable:
L5
E
b ␮
a
FB
I
0I
2␲r
5
, dr 5
␮0,
2␲
ln
␮0 I ,
2␲
E
b dr
a
r
5
␮0 I ,
2␲
ln
1ab 2
1ab 2
Finalize The inductance increases if , increases, if b increases, or if a decreases. This result is consistent with our conceptualization: any of these changes increases the size of the loop represented by our radial slice and through which the
magnetic field passes, increasing the inductance.
23.8 | Context Connection: The Use of Transcranial
Magnetic Stimulation in Depression
In Sections 20.7 and 21.9, we discussed the electrical characteristics of neurons and
the propagation of an action potential along a neuron. In this Context Connection,
we discuss a new treatment for depression that is early in its development and is
closely related to the propagation of impulses along and between nerves.
Depression is a mental disorder in which patients exhibit decreased self-esteem,
low moods, sadness, loss of interest in previously enjoyable activities, and increased
possibility of suicidal thoughts and behaviors. It is a complicated disorder whose
cause seems to include biological conditions, psychological effects, social interactions, drug and alcohol use, and even genetics. As a result of the large number of
possible influences causing depression, the particular treatment plan for a given
individual is not clear without extensive counseling of the patient and attempts to
use a variety of treatments.
In our discussion, we will focus on the possible biological origins of depression. One hypothesis suggests
that depression is related to low levels of neurotransmitters
(particularly serotonin, norepinephrine, and dopamine)
in the synapses between neurons. Antidepressant medications, such as sertraline, act to increase the levels of these
neurotransmitters.
One of the more controversial treatments for severe depression that has not responded to other treatments is electroconvulsive therapy (ECT), in which seizures are induced
in a patient under anesthesia. The seizures are induced
by placing electrodes on the patient’s head and passing a
pulsed current between the electrodes. While the effects of
this procedure on human brains cannot ethically be studied in detail, results from animal experiments suggest possible new synapse formation from the treatment. Because
of the role of levels of neurotransmitters in depression, this Figure 23.31 The magnetic coil of a Neurostar TMS apparatus is
held near the head of a patient.
growth of synapses may be the reason for the improvement
in some depressed patients after undergoing electroconvulsive therapy. ECT was used in the 1940s and 1950s on severely disturbed patients
in large mental institutions. Today, its main use is in psychiatric hospitals. There
is still continuing controversy about the usage of ECT for patients suffering from
mental disorders.
A newer method of introducing electric current into the brain is transcranial
Transcranial magnetic
stimulation
magnetic stimulation (TMS). This procedure induces currents in the brain by
means of magnetic induction rather than by the application of a large voltage from
contact electrodes. A large coil of wire is placed against the scalp of the patient. The
coil carries an alternating current, creating an oscillating magnetic field, which induces currents in the nerve cells of the brain. Unlike electroconvulsive therapy, the
patient is awake and does not experience seizures. Figure 23.31 shows the coil of a
Neurostar TMS machine being applied to a patient’s head. While the United States
Food and Drug Administration has not approved TMS as a generic procedure at
the time of this printing, the agency has cleared the specific Neurostar device that
performs TMS.
TMS has been used to perform motor cortex mapping, in which connections
between the primary motor cortex and various muscles are measured to determine
damage from spinal cord injuries, strokes, and motor neuron disease. With this technique, muscular responses in the index finger, forearm, biceps, jaw, and leg can be
observed as the magnetic coil is moved to different locations of the cortex.
When the coil is moved over the occipital cortex, some patients report magnetophosphenes, which are flashes of light seen even though the eyes are closed. The most
common method of inducing phosphenes is mechanical: by rubbing the closed eyes.
Phosphenes resulting from a blow to the head are the origin of the phrase “seeing
stars” associated with such trauma.
The use of TMS in depression is more recent. Figure 23.32 shows a patient being
treated with a TMS apparatus. The patient sits in a chair and the coil is placed against
the head. The coil is switched on and off at frequencies up to 10 Hz, inducing currents in the brain. The magnetic field is increased until the patient’s fingers begin
to twitch, indicating a therapeutic level. Once that level is attained, the treatment
portion of the experience continues for about 40 minutes. These treatments are
repeated on a daily basis for a period of several weeks.
Studies have reported some effectiveness of the procedure in treating depression,
but more studies need to be performed to validate the evidence. One of the factors
working against tests of effectiveness is the difficulty in establishing a “fake” TMS
experience to be used as a placebo for comparison to the actual treatment. The
treatment causes low-level neck pain, headache, and twitching in the scalp, which
Figure 23.32 A patient is treated
with the Neurostar TMS apparatus.
are difficult to reproduce in a placebo intervention.
© 2011 Neuronetics. All Rights Reserved
805
© 2011 Neuronetics. All Rights Reserved
23.8 | Context Connection: The Use of Transcranial Magnetic Stimulation in Depression
806
CHAPTER 23 | Faraday’s Law and Inductance
SUMMARY |
The magnetic flux through a surface associated with a mag:
netic field B is
FB 5
E
:
:
B ? dA
where L is the inductance of the coil. Inductance is a measure
of the opposition of a device to a change in current.
The inductance of a coil is
23.1b
L5
where the integral is over the surface.
Faraday’s law of induction states that the emf induced in
a circuit is directly proportional to the time rate of change of
magnetic flux through the circuit:
e 5 2N
d FB
dt
23.3b
where N is the number of turns and FB is the magnetic flux
through each turn.
When a conducting bar of length , moves through a mag:
netic field B with a velocity :
v so that :
v is perpendicular to
:
B, the emf induced in the bar (called the motional emf) is
e 5 2B ,v
23.5b
Lenz’s law states that the induced current and induced
emf in a conductor are in such a direction as to oppose the
change that produced them.
A general form of Faraday’s law of induction is
R
:
E ?d:
s 52
d FB
dt
23.9b
:
where E is a nonconservative electric field produced by the
changing magnetic flux.
When the current in a coil changes with time, an emf
is induced in the coil according to Faraday’s law. The selfinduced emf is described by the expression
dI
eL 5 2 L dt
23.10b
OBJECTIVE QUESTIONS |
1. Figure OQ23.1 is a graph of the magnetic flux through a
certain coil of wire as a function of time during an interval
while the radius of the coil is increased, the coil is rotated
through 1.5 revolutions, and the external source of the
magnetic field is turned off, in that order. Rank the emf
induced in the coil at the instants marked A through E from
the largest positive value to the largest-magnitude negative
value. In your ranking, note any cases of equality and also
any instants when the emf is zero.
Figure OQ23.1
2. A long, fine wire is wound into a coil with inductance 5 mH.
The coil is connected across the terminals of a battery, and
N FB
I
23.11b
where FB is the magnetic flux through the coil and N is the
total number of turns. Inductance has the SI unit the henry
(H), where 1 H 5 1 V ? s/A.
If a resistor and inductor are connected in series to a
battery of emf ␧ as shown in Active Figure 23.22, switch S2
is set at position a, and switch S1 is thrown closed at t 5 0,
the current in the circuit varies with time according to the
expression
I(t) 5
e
R
(1 2 e2t/␶)
23.14b
where ␶ 5 L/R is the time constant of the RL circuit.
If switch S2 in Active Figure 23.22 is thrown to position b,
the current decays exponentially with time according to the
expression
I(t) 5
e e2t/␶
R
23.18b
where e/R is the initial current in the circuit.
The energy stored in the magnetic field of an inductor
carrying a current I is
U 5 12 L I 2
23.20b
The energy per unit volume (or energy density) at a point
where the magnetic field is B is
uB 5
B2
2␮0
23.22b
denotes answer available in Student
Solutions Manual/Study Guide
the current is measured a few seconds after the connection
is made. The wire is unwound and wound again into a different coil with L 5 10 mH. This second coil is connected
across the same battery, and the current is measured in the
same way. Compared with the current in the first coil, is the
current in the second coil (a) four times as large, (b) twice
as large, (c) unchanged, (d) half as large, or (e) one-fourth
as large?
3. Two solenoids, A and B, are wound using equal lengths of
the same kind of wire. The length of the axis of each solenoid is large compared with its diameter. The axial length
of A is twice as large as that of B, and A has twice as many
turns as B. What is the ratio of the inductance of solenoid A
to that of solenoid B? (a) 4 (b) 2 (c) 1 (d) 12(e) 14
4. A circular loop of wire with a radius of 4.0 cm is in a uniform magnetic field of magnitude 0.060 T. The plane of the
loop is perpendicular to the direction of the magnetic field.
In a time interval of 0.50 s, the magnetic field changes to
| Objective Questions
the opposite direction with a magnitude of 0.040 T. What
is the magnitude of the average emf induced in the loop?
(a) 0.20 V (b) 0.025 V (c) 5.0 mV (d) 1.0 mV (e) 0.20 mV
5. A rectangular conducting
loop is placed near a long
wire carrying a current I as
shown in Figure OQ23.5.
If I decreases in time, what
can be said of the current induced in the loop? (a) The
Figure OQ23.5
direction of the current
depends on the size of the
loop. (b) The current is clockwise. (c) The current is counterclockwise. (d) The current is zero. (e) Nothing can be said
about the current in the loop without more information.
11. Initially, an inductor with no resistance carries a constant current. Then the current is brought to a new constant value
twice as large. After this change, when the current is constant
at its higher value, what has happened to the emf in the inductor? (a) It is larger than before the change by a factor of
4. (b) It is larger by a factor of 2. (c) It has the same nonzero
value. (d) It continues to be zero. (e) It has decreased.
12. In Figure OQ23.12, the switch is left in position a for a long
time interval and is then quickly thrown to position b. Rank
the magnitudes of the voltages across the four circuit elements a short time thereafter from the largest to the smallest.
6. A flat coil of wire is placed in a uniform magnetic field that
is in the y direction. (i) The magnetic flux through the coil
is a maximum if the plane of the coil is where? More than
one answer may be correct. (a) in the xy plane (b) in the yz
plane (c) in the xz plane (d) in any orientation, because it is
a constant (ii) For what orientation is the flux zero? Choose
from the same possibilities as in part (i).
7. A solenoidal inductor for a printed circuit board is being redesigned. To save weight, the number of turns is reduced by
one-half, with the geometric dimensions kept the same. By how
much must the current change if the energy stored in the inductor is to remain the same? (a) It must be four times larger.
(b) It must be two times larger. (c) It should be left the same.
(d) It should be one-half as large. (e) No change in the current
can compensate for the reduction in the number of turns.
8. If the current in an inductor is doubled, by what factor is the
stored energy multiplied? (a) 4 (b) 2 (c) 1 (d) 12 (e) 14
9. A square, flat loop of wire is pulled at
B
constant velocity through a region of
uniform magnetic field directed perpendicular to the plane of the loop
v
as shown in Figure OQ23.9. Which
of the following statements are correct? More than one statement may
be correct. (a) Current is induced in
Figure OQ23.9
the loop in the clockwise direction.
(b) Current is induced in the loop in the counterclockwise
direction. (c) No current is induced in the loop. (d) Charge
separation occurs in the loop, with the top edge positive.
(e) Charge separation occurs in the loop, with the top edge
negative.
10. The bar in Figure OQ23.10 moves
B
on rails to the right with a veloc:
ity v , and a uniform, constant
magnetic field is directed out of
the page. Which of the followv
ing statements are correct? More
than one statement may be correct. (a) The induced current in
the loop is zero. (b) The induced
Figure OQ23.10
current in the loop is clockwise.
(c) The induced current in the
loop is counterclockwise. (d) An external force is required
to keep the bar moving at constant speed. (e) No force is required to keep the bar moving at constant speed.
807
Figure OQ23.12
13. A bar magnet is held in a vertical oriv
entation above a loop of wire that lies
in the horizontal plane as shown in
Figure OQ23.13. The south end of the
magnet is toward the loop. After the
magnet is dropped, what is true of the
induced current in the loop as viewed
from above? (a) It is clockwise as the
Figure OQ23.13
magnet falls toward the loop. (b) It is
counterclockwise as the magnet falls toward the loop. (c) It
is clockwise after the magnet has moved through the loop
and moves away from it. (d) It is always clockwise. (e) It is
first counterclockwise as the magnet approaches the loop
and then clockwise after it has passed through the loop.
14. What happens to the amplitude of the induced emf when
the rate of rotation of a generator coil is doubled? (a) It
becomes four times larger. (b) It becomes two times larger.
(c) It is unchanged. (d) It becomes one-half as large. (e) It
becomes one-fourth as large.
15. Two coils are placed near each other as shown in Figure OQ23.15. The coil on the left is connected to a battery
and a switch, and the coil on the right is connected to a
resistor. What is the direction of the current in the resistor (i) at an instant immediately after the switch is thrown
closed, (ii) after the switch has been closed for several seconds, and (iii) at an instant after the switch has then been
thrown open? Choose each answer from the possibilities
(a) left, (b) right, or (c) the current is zero.
e
Figure OQ23.15
16. A circuit consists of a conducting movable bar and a
light bulb connected to two conducting rails as shown in
808
CHAPTER 23 | Faraday’s Law and Inductance
Figure OQ23.16. An external magnetic field is directed
perpendicular to the plane of the circuit. Which of the following actions will make the bulb light up? More than one
statement may be correct. (a) The bar is moved to the left.
(b) The bar is moved to the right. (c) The magnitude of
the magnetic field is increased. (d) The magnitude of the
magnetic field is decreased. (e) The bar is lifted off the rails.
B
17. Two rectangular loops of
wire lie in the same plane as
shown in Figure OQ23.17.
If the current I in the outer
loop is counterclockwise
and increases with time,
what is true of the current
Figure OQ23.17
induced in the inner loop?
More than one statement
may be correct. (a) It is zero. (b) It is clockwise. (c) It is counterclockwise. (d) Its magnitude depends on the dimensions of
the loops. (e) Its direction depends on the dimensions of the
loops.
Figure OQ23.16
CONCEPTUAL QUESTIONS |
1. A switch controls the current in a circuit that has a large inductance. The electric arc at the switch (Fig. CQ23.1) can melt
and oxidize the contact surfaces, resulting in high resistivity of
the contacts and eventual destruction of the switch. Is a spark
more likely to be produced at the switch when the switch is
being closed, when it is being opened, or does it not matter?
denotes answer available in Student
Solutions Manual/Study Guide
closed. (ii) Describe what the lightbulb does in each of circuits
(a) through (d) when, having been closed for a long time interval, the switch is opened.
3. What is the difference between magnetic flux and magnetic
field?
Alexandra Héder
4. Discuss the similarities between the energy stored in the
electric field of a charged capacitor and the energy stored
in the magnetic field of a current-carrying coil.
Figure CQ23.1
2. Consider the four circuits shown in Figure CQ23.2, each consisting of a battery, a switch, a lightbulb, a resistor, and either a
capacitor or an inductor. Assume the capacitor has a large capacitance and the inductor has a large inductance but no resistance. The lightbulb has high efficiency, glowing whenever
it carries electric current. (i) Describe what the lightbulb does
in each of circuits (a) through (d) after the switch is thrown
a
b
5. A spacecraft orbiting the Earth has a coil of wire in it. An astronaut measures a small current in the coil, although there
is no battery connected to it and there are no magnets in
the spacecraft. What is causing the current?
6. A circular loop of wire is located in a uniform and constant
magnetic field. Describe how an emf can be induced in the
loop in this situation.
7. A bar magnet is dropped toward a conducting ring lying on
the floor. As the magnet falls toward the ring, does it move
as a freely falling object? Explain.
8. In a hydroelectric dam, how is energy produced that is then
transferred out by electrical transmission? That is, how is
the energy of motion of the water converted to energy that
is transmitted by AC electricity?
9. A piece of aluminum is dropped vertically downward between the poles of an electromagnet. Does the magnetic
field affect the velocity of the aluminum?
10. The current in a circuit containing a coil, a resistor, and
a battery has reached a constant value. (a) Does the coil
have an inductance? (b) Does the coil affect the value of
the current?
11. (a) What parameters affect the inductance of a coil? (b) Does
the inductance of a coil depend on the current in the coil?
c
d
Figure CQ23.2
12. In Section 6.7, we defined conservative and nonconservative forces. In Chapter 19, we stated that an electric charge
creates an electric field that produces a conservative force.
Argue now that induction creates an electric field that produces a nonconservative force.
| Problems
13. When the switch in Figure CQ23.13a is closed, a current
is set up in the coil and the metal ring springs upward
(Fig. CQ23.13b). Explain this behavior.
© Cengage Learning/Charles D. Winters
14. Assume the battery in Figure CQ23.13a is replaced by an AC
source and the switch is held closed. If held down, the metal
ring on top of the solenoid becomes hot. Why?
a
b
Figure CQ23.13 Conceptual Questions 13 and 14.
15. A loop of wire is moving
near a long, straight wire
carrying a constant current I
as shown in Figure CQ23.15.
(a) Determine the direction
of the induced current in
the loop as it moves away
from the wire. (b) What
would be the direction of
the induced current in the
loop if it were moving toward
the wire?
16. After the switch is closed in
the LC circuit shown in Figure CQ23.16, the charge on
the capacitor is sometimes zero,
but at such instants the current
in the circuit is not zero. How is
this behavior possible?
809
v
Figure CQ23.15
Figure CQ23.16
PROBLEMS |
denotes Master It tutorial available in Enhanced
WebAssign
denotes asking for quantitative and conceptual reasoning
denotes symbolic reasoning problem
denotes “paired problems” that develop reasoning with
symbols and numerical values
denotes Watch It video solution available in Enhanced
WebAssign
The problems found in this chapter may be assigned
online in Enhanced WebAssign.
1. denotes straightforward problem; 2. denotes intermediate
problem; 3. denotes challenging problem
1. denotes full solution available in the Student Solutions Manual/
Study Guide
1. denotes problems most often assigned in Enhanced
WebAssign.
denotes biomedical problem
denotes guided problem
Section 23.1 Faraday’s Law of Induction
1.
A 30-turn circular coil of radius 4.00 cm and resistance
1.00 Ω is placed in a magnetic field directed perpendicular
to the plane of the coil. The magnitude of the magnetic field
varies in time according to the expression B 5 0.010 0t 1
0.040 0t 2, where B is in teslas and t is in seconds. Calculate
the induced emf in the coil at t 5 5.00 s.
2. An instrument based on induced emf has been used to
measure projectile speeds up to 6 km/s. A small magnet is
imbedded in the projectile as shown in Figure P23.2. The
projectile passes through two coils separated by a distance
d. As the projectile passes through each coil, a pulse of emf
is induced in the coil. The time interval between pulses can
be measured accurately with an oscilloscope, and thus the
speed can be determined. (a) Sketch a graph of ΔV versus
t for the arrangement shown. Consider a current that flows
counterclockwise as viewed from the starting point of the
projectile as positive. On your graph, indicate which pulse
is from coil 1 and which is from coil 2. (b) If the pulse separation is 2.40 ms and d 5 1.50 m, what is the projectile
speed?
Figure P23.2
3. A flat loop of wire consisting of a single turn of crosssectional area 8.00 cm2 is perpendicular to a magnetic field
that increases uniformly in magnitude from 0.500 T to 2.50 T
in 1.00 s. What is the resulting induced current if the loop
has a resistance of 2.00 V?
4.
Scientific work is currently underway to determine
whether weak oscillating magnetic fields can affect human
health. For example, one study found that drivers of trains
had a higher incidence of blood cancer than other railway workers, possibly due to long exposure to mechanical
810
CHAPTER 23 | Faraday’s Law and Inductance
around a circular return cord. Let n represent the number of turns in the toroid per unit distance along it. Let A
represent the cross-sectional area of the toroid. Let I(t) 5
Imax sin ␻t represent the current to be measured. (a) Show
that the amplitude of the emf induced in the Rogowski coil is
emax 5 ␮0nA␻Imax. (b) Explain why the wire carrying the
unknown current need not be at the center of the Rogowski
coil and why the coil will not respond to nearby currents
that it does not enclose.
devices in the train engine cab. Consider a magnetic field of
magnitude 1.00 3 1023 T, oscillating sinusoidally at 60.0 Hz.
If the diameter of a red blood cell is 8.00 ␮m, determine the
maximum emf that can be generated around the perimeter
of a cell in this field.
5.
6.
7.
8.
An aluminum ring of radius r1 5 5.00 cm and resistance
3.00 3 1024 Ω is placed around one end of a long air-core
solenoid with 1 000 turns per meter and radius r2 5 3.00 cm
as shown in Figure P23.5. Assume the axial component of
the field produced by the solenoid is one-half as strong over
the area of the end of the solenoid as at the center of the solenoid. Also assume the solenoid produces negligible field
outside its cross-sectional area. The current in the solenoid
is increasing at a rate
of 270 A/s. (a) What is
the induced current in
the ring? At the center
of the ring, what are
(b) the magnitude and
(c) the direction of
the magnetic field proFigure P23.5 Problems 5 and 6.
duced by the induced
current in the ring?
An aluminum ring of radius r1 and resistance R is placed
around one end of a long aircore solenoid with n turns per
meter and smaller radius r2 as shown in Figure P23.5. Assume the axial component of the field produced by the solenoid over the area of the end of the solenoid is one-half
as strong as at the center of the solenoid. Also assume the
solenoid produces negligible field outside its cross-sectional
area. The current in the solenoid is increasing at a rate of
ΔI/Δt. (a) What is the induced current in the ring? (b) At
the center of the ring, what is the magnetic field produced
by the induced current in the ring? (c) What is the direction
of this field?
A loop of wire in
the shape of a rectangle
of width w and length
L and a long, straight
wire carrying a current
I lie on a tabletop as
shown in Figure P23.7.
(a) Determine the magnetic flux through the
loop due to the curFigure P23.7
rent I. (b) Suppose the
current is changing with time according to I 5 a 1 bt,
where a and b are constants. Determine the emf that
is induced in the loop if b 5 10.0 A/s, h 5 1.00 cm,
w 5 10.0 cm, and L 5 1.00 m. (c) What is the direction of the
induced current in the rectangle?
When a wire carries an AC current with a known
frequency, you can use a Rogowski coil to determine the
amplitude Imax of the current without disconnecting
the wire to shunt the current through a meter. The e
Rogowski coil, shown in
Figure P23.8, simply clips
around the wire. It consists of
a toroidal conductor wrapped
Figure P23.8
9. A 25-turn circular coil of wire has diameter 1.00 m. It is
placed with its axis along the direction of the Earth’s magnetic field of 50.0 ␮T and then in 0.200 s is flipped 1808. An
average emf of what magnitude is generated in the coil?
10.
A long solenoid has n 5 400 turns per meter and carries
a current given I 5 30.0 (1 2 e 21.60t ), where I is in amperes
and t is in seconds. Inside the solenoid and coaxial with it
is a coil that has a radius of R 5 6.00 cm and consists of a
total of N 5 250 turns of fine wire (Fig. P23.10). What emf
is induced in the coil by the changing current?
Figure P23.10
11.
A strong electromagnet produces a uniform magnetic
field of 1.60 T over a cross-sectional area of 0.200 m2. A coil
having 200 turns and a total resistance of 20.0 Ω is placed
around the electromagnet. The current in the electromagnet is then smoothly reduced until it reaches zero in
20.0 ms. What is the current induced in the coil?
12. A piece of insulated wire is shaped into a
figure eight as shown in Figure P23.12.
For simplicity, model the two halves
of the figure eight as circles. The radius of the upper circle is 5.00 cm and
that of the lower circle is 9.00 cm. The
wire has a uniform resistance per unit
length of 3.00 Ω/m. A uniform magnetic field is applied perpendicular
Figure P23.12
to the plane of the two circles, in the
direction shown. The magnetic field
is increasing at a constant rate of 2.00 T/s. Find (a) the
magnitude and (b) the direction of the induced current
in the wire.
13.
A coil of 15 turns and radius 10.0 cm surrounds a long
solenoid of radius 2.00 cm
and 1.00 3 103 turns/meter
(Fig. P23.13). The current in
the solenoid changes as I 5
5.00 sin 120t, where I is in
amperes and t is in seconds.
Find the induced emf in the
15-turn coil as a function of
time.
Figure P23.13
| Problems
811
moves perpendicular to the field with a uniform amplitude of
1.50 cm. Find (a) the frequency and (b) the amplitude of the
emf induced between the ends of the string.
Section 23.2 Motional emf
Section 23.3 Lenz’s Law
Note: Problem 73 in Chapter 22 can be assigned with this section.
15.
ᐉ
Sascha Hahn/Shutterstock.com
14. A helicopter (Fig. P23.14)
has blades of length
3.00 m, extending out
from a central hub and
rotating at 2.00 rev/s. If
the vertical component of
the Earth’s magnetic field
is 50.0 ␮T, what is the
emf induced between the
blade tip and the center
hub?
Figure P23.14
Figure P23.15 shows
a top view of a bar that
B
can slide on two frictionless rails. The resisᐉ
F
tor is R 5 6.00 Ω, and
a 2.50-T magnetic field
is directed perpendicularly downward, into the
Figure P23.15 Problems 15
paper. Let , 5 1.20 m. through 18.
(a) Calculate the applied
force required to move the bar to the right at a constant
speed of 2.00 m/s. (b) At what rate is energy delivered to
the resistor?
16. Consider the arrangement shown in Figure P23.15. Assume
that R 5 6.00 V, , 5 1.20 m, and a uniform 2.50-T magnetic
field is directed into the page. At what speed should the bar
be moved to produce a current of 0.500 A in the resistor?
17. A conducting rod of length , moves on two horizontal, frictionless rails as shown in Figure P23.15. If a constant force
of 1.00 N moves the bar at 2.00 m/s through a magnetic
:
field B that is directed into the page, (a) what is the current
through the 8.00-V resistor R ? (b) What is the rate at which
energy is delivered to the resistor? (c) What is the mechani:
cal power delivered by the force Fapp ?
18.
A metal rod of mass m slides without friction along two
parallel horizontal rails, separated by a distance , and connected by a resistor R, as shown in Figure P23.15. A uniform
vertical magnetic field of magnitude B is applied perpendicular to the plane of the paper. The applied force shown
in the figure acts only for a moment, to give the rod a speed
v. In terms of m, ,, R, B, and v, find the distance the rod will
then slide as it coasts to a stop.
19. Review. After removing one string while restringing his acoustic guitar, a student is distracted by a video game. His experimentalist roommate notices his inattention and attaches one
end of the string, of linear density ␮ 5 3.00 3 1023 kg/m, to
a rigid support. The other end passes over a pulley, a distance
, 5 64.0 cm from the fixed end, and an object of mass m 5
27.2 kg is attached to the hanging end of the string. The
roommate places a magnet across the string as shown in
Figure P23.19. The magnet does not touch the string, but
produces a uniform field of 4.50 mT over a 2.00-cm length
of the string and negligible field elsewhere. Strumming the
string sets it vibrating vertically at its fundamental (lowest)
frequency. The section of the string in the magnetic field
Figure P23.19
20. Why is the following situation impossible? An automobile has a
vertical radio antenna of length , 5 1.20 m. The automobile travels on a curvy, horizontal road where the Earth’s
magnetic field has a magnitude of B 5 50.0 ␮T and is directed toward the north and downward at an angle of ␪ 5
65.08 below the horizontal. The motional emf developed
between the top and bottom of the antenna varies with the
speed and direction of the automobile’s travel and has a
maximum value of 4.50 mV.
21. The homopolar generator, also
called the Faraday disk, is a
low-voltage, high-current elecB
tric generator. It consists of a
rotating conducting disk with
one stationary brush (a sliding
electrical contact) at its axle
and another at a point on its
circumference as shown in Figure P23.21. A uniform magnetic
field is applied perpendicular
Figure P23.21
to the plane of the disk. Assume
the field is 0.900 T, the angular speed is 3.20 3 103 rev/min,
and the radius of the disk is 0.400 m. Find the emf generated
between the brushes. When superconducting coils are used
to produce a large magnetic field, a homopolar generator
can have a power output of several megawatts. Such a generator is useful, for example, in purifying metals by electrolysis.
If a voltage is applied to the output terminals of the generator, it runs in reverse as a homopolar motor capable of providing
great torque, useful in ship propulsion.
22.
A rectangular coil with resistance R has N turns, each of
length , and width w as shown in Figure P23.22. The coil
:
moves into a uniform magnetic field B with constant velocity
:
v . What are the magnitude and direction of the total magnetic force on the coil (a) as it enters the magnetic field,
(b) as it moves within the field, and (c) as it leaves the field?
B
v
ᐉ
Figure P23.22
812
CHAPTER 23 | Faraday’s Law and Inductance
23. A long solenoid, with its axis along the x axis, consists of 200
turns per meter of wire that carries a steady current of 15.0 A.
A coil is formed by wrapping 30 turns of thin wire around
a circular frame that has a radius of 8.00 cm. The coil is
placed inside the solenoid and mounted on an axis that is
a diameter of the coil and coincides with the y axis. The
coil is then rotated with an angular speed of 4.00␲ rad/s.
The plane of the coil is in the yz plane at t 5 0. Determine
the emf generated in the coil as a function of time.
24. Why is the following situation impossible?
A conducting rectangular loop of mass
M 5 0.100 kg, resistance R 5 1.00 Ω,
and dimensions w 5 50.0 cm by , 5
90.0 cm is held with its lower edge just
above a region with a uniform magnetic field of magnitude B 5 1.00 T as
shown in Figure P23.24.The loop is released from rest. Just as the top edge of
the loop reaches the region containing
the field, the loop moves with a speed
4.00 m/s.
Section 23.4 Induced emfs and Electric Fields
28.
ᐉ
Figure P23.24
Use Lenz’s law to answer the following questions concerning the direction of induced currents. Express your
answers in terms of the letter labels a and b in each part of
Figure P23.26. (a) What is the direction of the induced
current in the resistor R in Figure P23.26a when the bar
magnet is moved to the left? (b) What is the direction
of the current induced in the resistor R immediately
after the switch S in Figure P23.26b is closed? (c) What
is the direction of the induced current in the resistor R
when the current I in Figure P23.26c decreases rapidly
to zero?
B
Figure P23.28 Problems 28 and 29.
29.
30. A coiled telephone cord forms a spiral with 70 turns, a diameter of 1.30 cm, and an unstretched length of 60.0 cm.
Determine the self-inductance of one conductor in the unstretched cord.
31. A coil has an inductance of 3.00 mH, and the current in it
changes from 0.200 A to 1.50 A in a time interval of 0.200 s.
Find the magnitude of the average induced emf in the coil
during this time interval.
32.
A toroid has a major
radius R and a minor radius r and is tightly wound
with N turns of wire on a
hollow cardboard torus.
Figure P23.32 shows half
Figure P23.32
of this toroid, allowing
us to see its cross section. If R .. r, the magnetic field in the
region enclosed by the wire is essentially the same as the magnetic field of a solenoid that has been bent into a large circle of
radius R. Modeling the field as the uniform field of a long solenoid, show that the inductance of such a toroid is approximately
r2
L < 12␮ 0 N 2
R
33.
A 10.0-mH inductor carries a current I 5 Imax sin ␻t,
with Imax 5 5.00 A and f 5 ␻/2␲ 5 60.0 Hz. What is the selfinduced emf as a function of time?
b
c
Figure P23.26
Within the green dashed circle shown in Figure P23.28,
the magnetic field changes with time according to the expression B 5 2.00t 3 2 4.00t 2 1 0.800, where B is in teslas,
t is in seconds, and R 5 2.50 cm. When t 5 2.00 s, calculate (a) the magnitude and (b) the direction of the force
exerted on an electron located at point P1, which is at a
distance r1 5 5.00 cm from the center of the circular field
region. (c) At what instant is this force equal to zero?
Section 23.5 Inductance
e
a
A magnetic field directed into the page changes with
time according to B 5 0.030 0t 2 1 1.40, where B is in teslas
and t is in seconds. The field has a circular cross section of
radius R 5 2.50 cm (see Fig. P23.28). When t 5 3.00 s and
r2 5 0.020 0 m, what are (a) the magnitude and (b) the direction of the electric field at point P2?
B
25. Very large magnetic fields can be produced using a procedure called flux compression. A metallic cylindrical tube of radius R is placed coaxially in a long solenoid
of somewhat larger radius. The space between the tube and
the solenoid is filled with a highly explosive material. When
the explosive is set off, it collapses the tube to a cylinder of
radius r , R. If the collapse happens very rapidly, induced current in the tube maintains the magnetic flux nearly constant
inside the tube. If the initial magnetic field in the solenoid is
2.50 T and R/r 5 12.0, what maximum value of magnetic field
can be achieved?
26.
27. A coil of area 0.100 m2 is rotating at 60.0 rev/s with the
axis of rotation perpendicular to a 0.200-T magnetic field.
(a) If the coil has 1 000 turns, what is the maximum emf
generated in it? (b) What is the orientation of the coil with
respect to the magnetic field when the maximum induced
voltage occurs?
| Problems
34. An emf of 24.0 mV is induced in a 500-turn coil when the
current is changing at the rate of 10.0 A/s. What is the magnetic flux through each turn of the coil at an instant when
the current is 4.00 A?
35.
The current in a 90.0-mH inductor changes with
time as I 5 1.00t 2 2 6.00t, where I is in amperes and t
is in seconds. Find the magnitude of the induced emf at
(a) t 5 1.00 s and (b) t 5 4.00 s. (c) At what time is the
emf zero?
36.
An inductor in the form of a solenoid contains 420 turns
and is 16.0 cm in length. A uniform rate of decrease of current through the inductor of 0.421 A/s induces an emf of
175 ␮V. What is the radius of the solenoid?
Section 23.6 RL Circuits
37.
38.
A 12.0-V battery is connected into a series circuit containing a 10.0-V resistor and a 2.00-H inductor. In what time
interval will the current reach (a) 50.0% and (b) 90.0% of
its final value?
Show that I 5 Ii e2t/␶ is a solution of the differential equation
dI
50
dt
where Ii is the current at t 5 0 and ␶ 5 L/R.
IR 1 L
39. Consider the circuit in Figure P23.39, taking e 5 6.00 V,
L 5 8.00 mH, and R 5 4.00 V.
(a) What is the inductive time
constant of the circuit? (b) Cal- e
culate the current in the circuit
250 ␮s after the switch is closed.
(c) What is the value of the final
steady-state current? (d) After Figure P23.39 Problems
what time interval does the cur- 39, 40, 42 and 45.
rent reach 80.0% of its maximum value?
43. The switch in Figure
P23.43 is open for
t , 0 and is then
thrown closed at time e
t 5 0. Assume R 5
4.00 V, L 5 1.00 H,
and e 5 10.0 V.
Find (a) the current Figure P23.43 Problems 43 and 44.
in the inductor and
(b) the current in the switch as functions of time thereafter.
44.
The switch in Figure P23.43 is open for t , 0 and is then
thrown closed at time t 5 0. Find (a) the current in the inductor and (b) the current in the switch as functions of time
thereafter.
45.
In the circuit shown in Figure P23.39, let L 5 7.00 H,
R 5 9.00 V, and e 5 120 V. What is the self-induced emf
0.200 s after the switch is closed?
46. One application of an RL circuit is the generation of timevarying high voltage from a low-voltage source as shown in
Figure P23.46. (a) What is the current in the circuit a long
time after the switch has been in position a ? (b) Now the
switch is thrown quickly from a to b. Compute the initial
voltage across each resistor and across the inductor. (c) How
much time elapses before the voltage across the inductor
drops to 12.0 V?
Ω
Ω
Figure P23.46
47.
40. For the RL circuit shown in Figure P23.39, let the inductance be 3.00 H, the resistance 8.00 V, and the battery
emf 36.0 V. (a) Calculate ΔVR/e L, that is, the ratio of
the potential difference across the resistor to the emf
across the inductor when the current is 2.00 A. (b) Calculate the emf across the inductor when the current is
4.50 A.
41. A circuit consists of a coil, a switch, and a battery, all in
series. The internal resistance of the battery is negligible
compared with that of the coil. The switch is originally
open. It is thrown closed, and after a time interval Δt,
the current in the circuit reaches 80.0% of its final value.
The switch remains closed for a time interval much longer than Δt. Then the battery is disconnected and the
terminals of the coil are connected together to form a
short circuit. (a) After an equal additional time interval
Δt elapses, the current is what percentage of its maximum value? (b) At the moment 2Δt after the coil is shortcircuited, the current in the coil is what percentage of its
maximum value?
42. When the switch in Figure P23.39 is closed, the current
takes 3.00 ms to reach 98.0% of its final value. If R 5 10.0 V,
what is the inductance?
813
A 140-mH inductor and a 4.90-V
e
resistor are connected with a switch
to a 6.00-V battery as shown in Figure P23.47. (a) After the switch is
first thrown to a (connecting the
battery), what time interval elapses
before the current reaches 220 mA?
(b) What is the current in the inducFigure P23.47
tor 10.0 s after the switch is closed?
(c) Now the switch is quickly thrown
from a to b. What time interval elapses before the current in
the inductor falls to 160 mA?
Section 23.7 Energy Stored in a Magnetic Field
48. The magnetic field inside a superconducting solenoid is
4.50 T. The solenoid has an inner diameter of 6.20 cm and a
length of 26.0 cm. Determine (a) the magnetic energy density in the field and (b) the energy stored in the magnetic
field within the solenoid.
49.
An air-core solenoid with 68 turns is 8.00 cm long and
has a diameter of 1.20 cm. When the solenoid carries a current of 0.770 A, how much energy is stored in its magnetic
field?
814
50.
CHAPTER 23 | Faraday’s Law and Inductance
A flat coil of wire has an inductance of 40.0 mH
and a resistance of 5.00 V. It is connected to a 22.0-V battery
at the instant t 5 0. Consider the moment when the current
is 3.00 A. (a) At what rate is energy being delivered by the
battery? (b) What is the power being delivered to the resistance of the coil? (c) At what rate is energy being stored in
the magnetic field of the coil? (d) What is the relationship
among these three power values? (e) Is the relationship described in part (d) true at other instants as well? (f) Explain
the relationship at the moment immediately after t 5 0 and
at a moment several seconds later.
51. On a clear day at a certain location, a 100-V/m vertical electric field exists near the Earth’s surface. At the same place,
the Earth’s magnetic field has a magnitude of 0.500 3 1024 T.
Compute the energy densities of (a) the electric field and
(b) the magnetic field.
smaller than the exclusion level? (We will assume the magnetic field exists in empty space. Because of the effect of biological tissue, the exclusion level occurs in reality at about
30 cm from the coil.)
Additional Problems
55. A guitar’s steel string vibrates (see Fig. 23.6). The component of magnetic field perpendicular to the area of a pickup
coil nearby is given by
B 5 50.0 1 3.20 sin (1 046␲t)
where B is in milliteslas and t is in seconds. The circular
pickup coil has 30 turns and radius 2.70 mm. Find the emf
induced in the coil as a function of time.
56.
Section 23.8 Context Connection: The Use of
Transcranial Magnetic Stimulation in Depression
52.
Transcranial magnetic stimulation (TMS) is a noninvasive technique used to stimulate regions of the human
brain. In TMS, a small coil is placed on the scalp and a brief
burst of current in the coil produces a rapidly changing
magnetic field inside the brain. The induced emf can stimulate neuronal activity. (a) One such device generates an
upward magnetic field within the brain that rises from zero
to 1.50 T in 120 ms. Determine the induced emf around a
horizontal circle of tissue of radius 1.60 mm. (b) What If ?
The field next changes to 0.500 T downward in 80.0 ms.
How does the emf induced in this process compare with
that in part (a)?
53.
Consider a transcranial magnetic stimulation (TMS)
device containing a coil with several turns of wire, each
of radius 6.00 cm. In a circular area of the brain of radius
6.00 cm directly below and coaxial with the coil, the magnetic field changes at the rate of 1.00 3 104 T/s. Assume that
this rate of change is the same everywhere inside the circular
area. (a) What is the emf induced around the circumference
of this circular area in the brain? (b) What electric field is
induced on the circumference of this circular area?
54.
Suppose a transcranial magnetic stimulation (TMS)
device contains a single coil of wire of radius 6.00 cm as the
source of the magnetic field. A typical peak magnetic field
at the center of the coil in such a device is 1.50 T. (a) If the
field at the center of the coil has this peak value, what current exists in the coil? (b) The treatment area in the brain
is about 2.50 cm below the center of the coil. If the current
in the coil is changing at the rate of 1.00 3 107 A/s, what
is the rate of change of the magnetic field at the treatment
area and along the axis of the coil? (c) Some patients are
advised against TMS treatment if they have magnetic-sensitive metals in their heads, such as aneurysm clips, stents, or
bullet fragments. In addition, patients with electronic devices may also be advised against TMS: cochlear implants,
pacemakers, cardioverter defibrillators, insulin pumps, and
vagus nerve stimulators. The recommended magnetic field
exclusion level for magnetic-sensitive metals and electronic
devices is 5.00 3 1024 T. For the current in part (a), at what
distance from the coil, along the center axis, must these
metals or devices be in order to experience a magnetic field
Consider the apparatus shown in Figure P31.56 in
which a conducting bar can be moved along two rails connected to a lightbulb. The whole system is immersed in a
magnetic field of magnitude B 5 0.400 T perpendicular and
into the page. The distance between the horizontal rails is
, 5 0.800 m. The resistance of the lightbulb is R 5 48.0 Ω,
assumed to be constant. The bar and rails have negligible
resistance. The bar is moved toward the right by a constant
force of magnitude F 5 0.600 N. We wish to find the maximum power delivered to the lightbulb. (a) Find an expression for the current in the lightbulb as a function of B, ,, R,
and v, the speed of the bar. (b) When the maximum power
is delivered to the lightbulb, what analysis model properly
describes the moving bar? (c) Use the analysis model in
part (b) to find a numerical value for the speed v of the bar
when the maximum power is being delivered to the lightbulb. (d) Find the current in the lightbulb when maximum
power is being delivered to it.(e) Using P 5 I 2R, what is
the maximum power delivered to the lightbulb? (f) What is
the maximum mechanical input power delivered to the bar
by the force F ? (g) We have assumed the resistance of the
lightbulb is constant. In reality, as the power delivered to
the lightbulb increases, the filament temperature increases
and the resistance increases. Does the speed found in part
(c) change if the resistance increases and all other quantities are held constant? (h) If so, does the speed found in
part (c) increase or decrease? If not, explain. (i) With the
assumption that the resistance of the lightbulb increases
as the current increases, does the power found in part
(f) change? (j) If so, is the power found in part (f) larger or
smaller? If not, explain.
F
,
B
Figure P23.56
57.
Strong magnetic fields are used in such medical procedures as magnetic resonance imaging, or MRI. A technician wearing a brass bracelet enclosing area 0.005 00 m2
places her hand in a solenoid whose magnetic field is 5.00 T
directed perpendicular to the plane of the bracelet. The
electrical resistance around the bracelet’s circumference is
| Problems
0.020 0 Ω. An unexpected power failure causes the field to
drop to 1.50 T in a time interval of 20.0 ms. Find (a) the current induced in the bracelet and (b) the power delivered to
the bracelet. Note: As this problem implies, you should not
wear any metal objects when working in regions of strong
magnetic fields.
58. Figure P23.58 is a graph
e
of the induced emf versus time for a coil of
N turns rotating with
angular speed ␻ in a
uniform magnetic field
directed perpendicular
to the coil’s axis of rotation. What If ? Copy
this sketch (on a larger
Figure P23.58
scale) and on the same
set of axes show the graph of emf versus t (a) if the number
of turns in the coil is doubled, (b) if instead the angular
speed is doubled, and (c) if the angular speed is doubled
while the number of turns in the coil is halved.
59. Suppose you wrap wire onto the core from a roll of cellophane tape to make a coil. Describe how you can use a bar
magnet to produce an induced voltage in the coil. What is
the order of magnitude of the emf you generate? State the
quantities you take as data and their values.
60.
At t 5 0, the open switch in Figure P23.60 is thrown
closed. We wish to find a symbolic expression for the current in the inductor for time t . 0. Let this current be
called I and choose it to be downward in the inductor in
Figure P23.60. Identify I1 as the current to the right through
R 1 and I2 as the current downward through R 2. (a) Use
Kirchhoff’s junction rule to find a relation among the three
currents. (b) Use Kirchhoff’s loop rule around the left loop
to find another relationship. (c) Use Kirchhoff’s loop rule
around the outer loop to find a third relationship. (d) Eliminate I1 and I2 among the three equations to find an equation involving only the current I. (e) Compare the equation
in part (d) with Equation 23.13 in the text. Use this comparison to rewrite Equation 23.14 in the text for the situation in
this problem and show that
I(t) 5
e [1 2 e 2(R 9/L)t ]
R1
where R 9 5 R 1R 2/(R 1 1 R 2).
e
Figure P23.60
61.
The magnetic flux through a metal ring varies with time
t according to FB 5 at 3 2 bt 2, where FB is in webers, a 5
6.00 s23, b 5 18.0 s22, and t is in seconds. The resistance of
the ring is 3.00 V. For the interval from t 5 0 to t 5 2.00 s,
determine the maximum current induced in the ring.
62. Review. A particle with a mass of 2.00 3 10216 kg and a
charge of 30.0 nC starts from rest, is accelerated through a
815
potential difference ΔV, and is fired from a small source in
a region containing a uniform, constant magnetic field of
magnitude 0.600 T. The particle’s velocity is perpendicular
to the magnetic field lines. The circular orbit of the particle
as it returns to the location of the source encloses a magnetic flux of 15.0 ␮Wb. (a) Calculate the particle’s speed.
(b) Calculate the potential difference through which the
particle was accelerated inside the source.
63.
To monitor the breathing of a hospital patient, a thin
belt is girded around the patient’s chest. The belt is a 200turn coil. When the patient inhales, the area encircled by
the coil increases by 39.0 cm2. The magnitude of the Earth’s
magnetic field is 50.0 ␮T and makes an angle of 28.08 with
the plane of the coil. Assuming a patient takes 1.80 s to inhale, find the average induced emf in the coil during this
time interval.
64. A betatron is a device that accelerates electrons to energies
in the MeV range by means of electromagnetic induction.
Electrons in a vacuum chamber are held in a circular orbit
by a magnetic field perpendicular to the orbital plane. The
magnetic field is gradually increased to induce an electric
field around the orbit. (a) Show that the electric field is
in the correct direction to make the electrons speed up.
(b) Assume the radius of the orbit remains constant. Show
that the average magnetic field over the area enclosed by
the orbit must be twice as large as the magnetic field at the
circle’s circumference.
65.
A long, straight wire carries a current given by I 5
Imax sin (␻t 1 ␾). The wire lies in the plane of a rectangular
coil of N turns of wire as shown in Figure P23.65. The quantities Imax, ␻, and ␾ are all constants. Assume Imax 5 50.0 A,
␻ 5 200␲ s21, N 5 100, h 5 w 5 5.00 cm, and L 5 20.0 cm.
Determine the emf induced in the coil by the magnetic field
created by the current in the straight wire.
Figure P23.65
66. Figure P23.66 shows a stationary conductor whose shape
is similar to the letter e. The
radius of its circular portion
is a 5 50.0 cm. It is placed in
a constant magnetic field of
0.500 T directed out of the
θ
page. A straight conducting
rod, 50.0 cm long, is pivoted
about point O and rotates with
a constant angular speed of
2.00 rad/s. (a) Determine the
induced emf in the loop POQ.
Note that the area of the loop
Figure P23.66
is ␪a2/2. (b) If all the conducting material has a resistance per length of 5.00 Ω/m, what is
the induced current in the loop POQ at the instant 0.250 s
after point P passes point Q ?
816
CHAPTER 23 | Faraday’s Law and Inductance
that maintains a constant emf e is connected between the
:
rails, and a constant magnetic field B is directed perpendicularly out of the page. Assuming the bar starts from rest at time
t 5 0, show that at time t it moves with a speed
67. The toroid in Figure P23.67 consists of N turns and has a
rectangular cross section. Its inner and outer radii are a and
b, respectively. The figure shows half of the toroid to allow us
to see its cross section. Compute the inductance of a 500-turn
toroid for which a 5 10.0 cm, b 5 12.0 cm, and h 5 1.00 cm.
␧
v5
Bd
2 2
(1 2 e 2B d t/mR)
e
B
Figure P23.67 Problems 67 and 68.
68.
The toroid in Figure P23.67 consists of N turns and has a
rectangular cross section. Its inner and outer radii are a and
b, respectively. Find the inductance of the toroid.
69. (a) A flat, circular coil does not actually produce a uniform
magnetic field in the area it encloses. Nevertheless, estimate
the inductance of a flat, compact, circular coil with radius
R and N turns by assuming the field at its center is uniform
over its area. (b) A circuit on a laboratory table consists
of a 1.50-volt battery, a 270-V resistor, a switch, and three
30.0-cm-long patch cords connecting them. Suppose the
circuit is arranged to be circular. Think of it as a flat coil
with one turn. Compute the order of magnitude of its inductance and (c) of the time constant describing how fast
the current increases when you close the switch.
Review problems. Problems 70, 71, and 73 apply ideas from
this and earlier chapters to some properties of superconductors, which were introduced in Section 21.3.
70. Review. In an experiment carried out by S. C. Collins between 1955 and 1958, a current was maintained in a superconducting lead ring for 2.50 yr with no observed loss,
even though there was no energy input. If the inductance
of the ring were 3.14 3 1028 H and the sensitivity of the
experiment were 1 part in 109, what was the maximum resistance of the ring? Suggestion: Treat the ring as an RL circuit
carrying decaying current and recall that the approximation e2x < 1 2 x is valid for small x.
71. Review. A novel method of storing energy has been proposed. A huge underground superconducting coil, 1.00 km
in diameter, would be fabricated. It would carry a maximum
current of 50.0 kA in each winding of a 150-turn Nb3Sn
solenoid. (a) If the inductance of this huge coil were 50.0
H, what would be the total energy stored? (b) What would
be the compressive force per unit length acting between two
adjacent windings 0.250 m apart?
72.
A bar of mass m and resistance R slides without friction in
a horizontal plane, moving on parallel rails as shown in Figure P23.72. The rails are separated by a distance d. A battery
Figure P23.72
73. Review. The use of superconductors has been proposed
for
power
transmission
lines. A single coaxial cable
(Fig. P23.73) could carry
a power of 1.00 3 103 MW
(the output of a large power
plant) at 200 kV, DC, over a
Figure P23.73
distance of 1.00 3 103 km
without loss. An inner wire of
radius a 5 2.00 cm, made from the superconductor Nb3Sn,
carries the current I in one direction. A surrounding superconducting cylinder of radius b 5 5.00 cm would carry
the return current I. In such a system, what is the magnetic
field (a) at the surface of the inner conductor and (b) at
the inner surface of the outer conductor? (c) How much
energy would be stored in the magnetic field in the space
between the conductors in a 1.00 3 103 km superconducting line? (d) What is the pressure exerted on the outer conductor due to the current in the inner conductor?
74.
Complete the calculation in Example 23.8 by proving
that
E
`
0
75.
e22Rt/L dt 5
L
2R
The plane of a square loop of
wire with edge length a 5 0.200 m
is oriented vertically and along an
east–west axis. The Earth’s magnetic field at this point is of magnitude B 5 35.0 ␮T and is directed
northward at 35.08 below the horizontal. The total resistance of the
loop and the wires connecting it
to a sensitive ammeter is 0.500 V.
Figure P23.75
If the loop is suddenly collapsed
by horizontal forces as shown in Figure P23.75, what total
charge enters one terminal of the ammeter?
Context
7
CONC LU S I O N
Nuclear Magnetic Resonance
and Magnetic Resonance
Imaging
I
n this Context Conclusion, we discuss an application that is now widely used as a
noninvasive diagnostic tool in medical practice. This application is known as MRI,
for magnetic resonance imaging.
In Section 22.11, we discussed the spin angular momentum of an electron and the
associated magnetic moment of the electron. Spin is a general property of all particles. For example, protons and neutrons in the nucleus of an atom have spin and
:
. As we saw in Section 22.6, the potential
an associated magnetic dipole moment ␮
energy of a system consisting of a magnetic dipole moment in an external magnetic
: :
field is U 5 2␮
? B.
:
When the magnetic moment ␮
is aligned with the field as closely as quantum
physics allows, the potential energy of the dipole–field system has its minimum value
:
Emin. When ␮
is as antiparallel to the field as possible, the potential energy has its
maximum value Emax. Because the directions of the spin and the magnetic moment
for a particle are quantized (see Chapter 29), the energies of the dipole–field system are also quantized. We introduced the concept of quantized states of energy in
Chapter 11. In general, there are other allowed energy states between Emin and Emax
corresponding to the quantized directions of the magnetic moment with respect to
the field. These states are often called spin states because they differ in energy as a
result of the direction of the spin.
The number of spin states depends on the spin of the nucleus. The simplest situation is shown in Figure 1, for a nucleus with only two possible spin states having
energies Emin and Emax.
It is possible to observe transitions between these two spin states in a sample using
a technique known as nuclear magnetic resonance (NMR). A constant magnetic field
changes the energy associated with the spin states, splitting them apart in energy as
shown in Figure 1. The sample is also exposed to electromagnetic waves in the radio
range of the electromagnetic spectrum. When the frequency of the radio waves is adjusted such that the photon energy matches the separation energy between spin states,
a resonance condition exists and the photon is absorbed by a nucleus in the ground
state, raising the nucleus–magnetic field system to the higher energy spin state. This
results in a net absorption of energy by the system, which is detected by the experimental control and measurement system. A diagram of the apparatus used to detect
an NMR signal is illustrated in Figure 2 (page 818). The absorbed energy is supplied
by the oscillator producing the radio waves. Nuclear magnetic resonance and a related
technique called electron spin resonance are extremely important methods for studying
nuclear and atomic systems and how these systems interact with their surroundings.
A widely used medical diagnostic technique called MRI, for magnetic resonance
imaging, is based on nuclear magnetic resonance. In MRI, the patient is placed
B
m
m
Figure 1 A nucleus with spin 12 is
placed in a magnetic field.
c Nuclear magnetic resonance
Magnetic resonance imaging
(MRI)
817
818 CONTEXT 7 CONCLUSION | Nuclear Magnetic Resonance and Magnetic Resonance Imaging
Figure 2 Experimental
arrangement for nuclear magnetic
resonance. The radio-frequency
magnetic field created by the
coil surrounding the sample and
provided by the variable-frequency
oscillator is perpendicular to the
constant magnetic field created by
the electromagnet. When the nuclei
in the sample meet the resonance
condition, the nuclei absorb energy
from the radio-frequency field of
the coil; this absorption changes
the characteristics of the circuit in
which the coil is included. Most
modern NMR spectrometers use
superconducting magnets at fixed
field strengths and operate at
frequencies of approximately
200 MHz.
inside a large solenoid that supplies a spatially varying magnetic field. Because of
the variation in the magnetic field across the patient’s body, protons in hydrogen
atoms in water molecules in different parts of the body have different splittings in
energy between spin states, so that the resonance signal can be used to provide information on the positions of the protons. A computer is used to analyze the position information to provide data for constructing a final image. MRI scans showing
incredible detail in internal body structure are shown in Figure 3. The main advantage of MRI over other imaging techniques in medical diagnostics is that it does not
cause damage to cellular structures as x-rays or gamma rays do. Photons associated
with the radio-frequency signals used in MRI have energies of only about 10–7 eV.
Because molecular bond strengths are much larger (on the order of 1 eV), the radiofrequency radiation cannot cause cellular damage. In comparison, x-rays or gamma
rays have energies ranging from 104 to 106 eV and can cause considerable cellular
damage. Therefore, despite some individuals’ fears of the word nuclear associated
with magnetic resonance imaging, the radio-frequency radiation involved is much
safer than x-rays or gamma rays!
In this Context, we have seen a number of applications of magnetism to medical
procedures. The catheter ablation procedures (Sections 21.2 and 22.12) and MRI
scans discussed here have saved many lives through accurate diagnosis and treatment. Transcranial magnetic stimulation (Section 23.8) is a relatively new procedure that may turn out to be extremely helpful in its application to depression. Who
knows what other uses of magnetism in medicine may lie in the near future? Keep a
watch out in the newspapers and on the Internet and you may see the next application very soon!
Problems
1. The radio frequency at which a nucleus having a magnetic moment of magnitude ␮ displays resonance absorption between spin states is called the Larmor
frequency and is given by
2␮B
DE
5
h
h
© Tobkatrina | Dreamstime.com
PASIEKA/Science Photo Library/Getty Images
f5
a
b
Figure 3 Examples of MRI scans of the human brain. (a) A sagittal section of the human brain, showing highly detailed views of several brain
structures. (b) A computer-enhanced view of an axial section through the brain shows a metastatic tumor of the cerebrum, appearing in white.
| Nuclear Magnetic Resonance and Magnetic Resonance Imaging
Calculate the Larmor frequency for (a) free neutrons in a magnetic field of
1.00 T, (b) free protons in a magnetic field of 1.00 T, and (c) free protons in
the Earth’s magnetic field at a location where the magnitude of the field is
50.0 ␮T.
2.
In magnetic resonance imaging (MRI), the patient is placed inside a large
solenoid. Suppose an MRI solenoid is 2.40 m long, 0.900 m in diameter, and is
wound with a single layer of superconducting niobium-titanium wire of radius
1.00 mm. Each turn of the solenoid is laid right next to the previous turn so
that there is no space between the turns. The magnetic field produced by the
solenoid is 1.55 T. (a) What current exists in the solenoid to produce this field?
(b) What is the magnetic flux through the solenoid? (c) When the machine is
turned off, the field reduces linearly to zero in 5.00 s. What emf is induced in
the interior of the solenoid while the machine is turning off? (d) What is the
total mass of the niobium-titanium wire that is wound to make the solenoid?
Assume that the density of the wire is 6.00 3 103 kg/m3.
819
Context
8
Lasers
he invention of the laser was popularly credited to
Arthur L. Schawlow and Charles H. Townes for many
years after their publication of a proposal for the laser in
a 1958 issue of Physical Review. Schawlow and Townes received a patent for the device in 1959. In 1960, the first
laser was built and operated by Theodore Maiman. This
device used a ruby crystal to create the laser light, which
was emitted in pulses from the end of a ruby cylinder. A
flash lamp was used to excite the laser action.
In 1977, the first victory in a 30-year-long legal
battle was completed in which Gordon Gould, who
was a graduate student at Columbia University in the
late 1950s, received a patent for inventing the laser
in 1957 as well as coining the term. Believing erroneously that he had to have a working prototype before
he could file for a patent, he did not file until later in
1959 than had Schawlow and Townes. Gould’s legal
battles ended in 1987. By this time, Gould’s technology was being widely used in industry and medicine.
His victory finally resulted in his control of patent
rights to perhaps 90% of the lasers used and sold in
the United States.
© Lawrence Manning/CORBIS
T
Figure 2 The original ruby laser emitted red light, as did many lasers
Rolf Vennenbernd/dpa/Landov
developed soon afterward. Today, lasers are available in a variety of colors
and various regions of the electromagnetic spectrum. In this photograph,
a green laser is used to perform scientific research.
Figure 1 A patient is made ready to receive laser light during eye surgery.
820
Since the development of the first device, laser technology has experienced tremendous growth. Lasers that
cover wavelengths in the infrared, visible, and ultraviolet
regions are now available. Various types of lasers use
solids, liquids, and gases as the active medium. Although
the original laser emitted light over a very narrow range
around a fixed wavelength, tunable lasers are now available, in which the wavelength can be varied.
© 36clicks | Dreamstime.com
CONTEXT 8 | Lasers 821
The laser is an omnipresent technological tool in
our daily life. Applications include LASIK eye surgery
(covered in more detail in Chapter 26), surgical “welding” of detached retinas, precision surveying and length
measurement, a potential source for inducing nuclear
fusion reactions, precision cutting of metals and other
materials, and telephone communication along optical
fibers.
We also use lasers to read information from compact discs for use in audio entertainment and computer applications. DVD and Blu-ray disc players use
lasers to read video information. Lasers are used in
retail stores to read price and inventory information
from product labels. In the laboratory, lasers can be
used to trap atoms and cool them to microkelvins
above absolute zero and to move microscopic biological organisms around harmlessly.
These and other applications are possible because
of the unique characteristics of laser light. In addition
to its being almost monochromatic, laser light is also
highly directional and can therefore be sharply focused
to produce regions of extreme intensity.
In this Context, we shall investigate the physics of
electromagnetic radiation and optics and apply the
principles to an understanding of the behavior of laser
©iStockphoto.com/ozgurcankaya
Figure 3 A laser cutting machine cuts through a thick sheet of steel.
Figure 4 A bar code scanner uses light from a laser to identify products
being purchased. The reflections from the bar code on the package are
read and entered into the computer to determine the price
of the item.
light and its applications. A major focus of our study will
be on the technology of optical fibers and how they are
used in industry and medicine. We shall study the nature of light as we respond to our central question:
What is special about laser light, and how is it used
in technological applications?
Chapter
24
Electromagnetic Waves
Chapter Outline
24.1 Displacement Current and the
Generalized Form of Ampère’s Law
24.2 Maxwell’s Equations and Hertz’s
Discoveries
24.3 Electromagnetic Waves
24.4 Energy Carried by Electromagnetic
Waves
24.5 Momentum and Radiation Pressure
24.6 The Spectrum of Electromagnetic
Waves
24.7 Polarization of Light Waves
24.8 Context Connection: The Special
Properties of Laser Light
A
lthough we are not always aware
of their presence, electromagnetic
waves permeate our environment. In the
form of visible light, they enable us to
view the world around us with our eyes.
Electromagnetic waves cover a broad
Infrared waves from the surface of the
spectrum of wavelengths, with waves in
Earth warm our environment, radio-frequency waves carry our favorite radio
various wavelength ranges having distinct
entertainment, microwaves cook our food and are used in radar communicaproperties. This photo of the Crab Nebula
tion systems, and the list goes on and on. The waves described in Chapter 13
is made with x-rays, which are like visible
are mechanical waves, which require a medium through which to propagate.
light, except for their very short wavelength.
Electromagnetic waves, in contrast, can propagate through a vacuum. Despite
In this chapter, we shall study the common
features of x-rays, visible light, and other
this difference between mechanical and electromagnetic waves, much of the beforms of electromagnetic radiation.
havior in the wave models of Chapters 13 and 14 is similar for electromagnetic
waves.
The purpose of this chapter is to explore the properties of electromagnetic
waves. The fundamental laws of electricity and magnetism—Maxwell’s equations—
form the basis of all electromagnetic phenomena. One of these equations predicts
that a time-varying electric field produces a magnetic field just as a time-varying
magnetic field produces an electric field. From this generalization, Maxwell provided
the final important link between electric and magnetic fields. The most dramatic
prediction of his equations is the existence of electromagnetic waves that propagate
through empty space with the speed of light. This discovery led to many practical
822
NASA/CXC/SAO/F. Seward
SUMMARY
24.1 | Displacement Current and the Generalized Form of Ampère’s Law 823
applications, such as communication by radio, television, and cellular telephone, and to
the realization that light is one form of electromagnetic radiation.
24.1 | Displacement Current and the Generalized
Form of Ampère’s Law
We have seen that charges in motion, or currents, produce magnetic fields. When a
current-carrying conductor has high symmetry, we can calculate the magnetic field
using Ampère’s law, given by Equation 22.29:
R
Path P
q
:
B d:
s 5 0I
q
where the line integral is over any closed path through which the conduction current passes and the conduction current is defined by I 5 dq/dt.
In this section, we shall use the term conduction current to refer to the type of current that we have already discussed, that is, current carried by charged particles in a
wire. We use this term to differentiate this current from a different type of current
we will introduce shortly. Ampère’s law in this form is valid only if the conduction
current is continuous in space. Maxwell recognized this limitation and modified
Ampère’s law to include all possible situations.
This limitation can be understood by considering a capacitor being charged as in
Figure 24.1. When conduction current exists in the wires, the charge on the plates
changes, but no conduction current exists between the plates. Consider the two surfaces S1 (a circle, shown in blue) and S2 (a paraboloid, in tan, passing between the
plates) in Figure 24.1 bounded by the same path P. Ampère’s law says that the
:
line integral of B d :
s around this path must equal 0I, where I is the conduction current through any surface bounded by the path P.
When the path P is considered as bounding S1, the right-hand side of Equation 22.29 is 0I because the conduction current passes through S1 while the capacitor is charging. When the path bounds S2, however, the right-hand side of
Equation 22.29 is zero because no conduction current passes through S2. Therefore,
a contradictory situation arises because of the discontinuity of the current! Maxwell
solved this problem by postulating an additional term on the right side of Equation
22.29, called the displacement current Id , defined as
Id ; 0
dFE
dt
24.1b
:
S2
I
S1
The conduction current I in the
wire passes only through S1, which
leads to a contradiction in
Ampère’s law that is resolved only
if one postulates a displacement
current through S2.
Figure 24.1 Two surfaces S1 and
S2 near the plate of a capacitor are
bounded by the same path P.
c Displacement current
:
Recall that FE is the flux of the electric field, defined as FE ; r E d A (Eq. 19.14).
(The word displacement here does not have the same meaning as in Chapter 2; it is
historically entrenched in the language of physics, however, so we continue to use it.)
Equation 24.1 is interpreted as follows. As the capacitor is being charged (or
discharged), the changing electric field between the plates may be considered as
equivalent to a current between the plates that acts as a continuation of the conduction current in the wire. When the expression for the displacement current given
by Equation 24.1 is added to the conduction current on the right side of Ampère’s
law, the difficulty represented in Figure 24.1 is resolved. No matter what surface
bounded by the path P is chosen, either conduction current or displacement current
passes through it. With this new notion of displacement current, we can express the
general form of Ampère’s law (sometimes called the Ampère–Maxwell law) as1
R
:
B d:
s 5 0(I 1 Id) 5 0I 1 00
I
dFE
dt
24.2b
1Strictly speaking, this expression is valid only in a vacuum. If a magnetic material is present, a magnetizing current must
also be included on the right side of Equation 24.2 to make Ampère’s law fully general.
c Ampère–Maxwell law
824 CHAPTER 24 | Electromagnetic Waves
The electric field lines between
the plates create an electric flux
through surface S.
We can understand the meaning of this expression by referring to Figure 24.2.
:
:
The electric flux through surface S is FE 5 # E d A 5 EA, where A is the area
of the capacitor plates and E is the magnitude of the uniform electric field between
the plates. If q is the charge on the plates at any instant, then E 5 /0 5 q/(0A) (see
Example 19.12). Therefore, the electric flux through S is
I
FE 5 EA 5
S
q
q
0
Hence, the displacement current through S is
S
E
Id 5 0
q
I
North Wind Picture Archives
Figure 24.2 When a conduction
current exists in the wires, a changing
:
electric field E exists between the
plates of the capacitor.
James Clerk Maxwell
Scottish Theoretical Physicist
(1831–1879)
Maxwell developed the electromagnetic
theory of light and the kinetic theory of
gases and explained the nature of Saturn’s
rings and color vision. Maxwell’s successful
interpretation of the electromagnetic
field resulted in the field equations that
bear his name. Formidable mathematical
ability combined with great insight
enabled him to lead the way in the study of
electromagnetism and kinetic theory. He
died of cancer before he was 50.
24.3b
That is, the displacement current Id through S is precisely equal to the conduction
current I in the wires connected to the capacitor!
By considering surface S, we can identify the displacement current as the source
of the magnetic field on the surface boundary. The displacement current has its
physical origin in the time-varying electric field. The central point of this formalism
is that magnetic fields are produced both by conduction currents and by time-varying
electric fields. This result was a remarkable example of theoretical work by Maxwell,
and it contributed to major advances in the understanding of electromagnetism.
QUI C K QU IZ 24.1 In an RC circuit, the capacitor begins to discharge. (i) During
the discharge, in the region of space between the plates of the capacitor, is there
(a) conduction current but no displacement current, (b) displacement current but
no conduction current, (c) both conduction and displacement current, or (d) no
current of any type? (ii) In the same region of space, is there (a) an electric field but
no magnetic field, (b) a magnetic field but no electric field, (c) both electric and
magnetic fields, or (d) no fields of any type?
24.2 | Maxwell’s Equations and Hertz’s Discoveries
We now present four equations that are regarded as the basis of all electrical and
magnetic phenomena. These equations, developed by Maxwell, are as fundamental
to electromagnetic phenomena as Newton’s laws are to mechanical phenomena. In
fact, the theory that Maxwell developed was more far-reaching than even he imagined because it turned out to be in agreement with the special theory of relativity, as
Einstein showed in 1905.
Maxwell’s equations represent the laws of electricity and magnetism that we have
already discussed, but they have additional important consequences. For simplicity,
we present Maxwell’s equations as applied to free space, that is, in the absence of any
dielectric or magnetic material. The four equations are
R
R
c Gauss’s law
c Gauss’s law in magnetism
R
c Faraday’s law
c Ampère–Maxwell law
dq
dFE
5
dt
dt
R
:
:
q
0
24.4b
B dA 5 0
24.5b
:
:
:
:
E dA 5
E d:
s 52
dFB
dt
B d:
s 5 0I 1 0 0
24.6b
dFE
dt
24.7b
Equation 24.4 is Gauss’s law: the total electric flux through any closed surface
equals the net charge inside that surface divided by 0. This law relates an electric
field to the charge distribution that creates it.
24.2 | Maxwell’s Equations and Hertz’s Discoveries 825
Equation 24.5 is Gauss’s law in magnetism, and it states that the net magnetic flux
through a closed surface is zero. That is, the number of magnetic field lines that enter
a closed volume must equal the number that leave that volume, which implies that magnetic field lines cannot begin or end at any point. If they did, it would mean that isolated magnetic monopoles existed at those points. That isolated magnetic monopoles
have not been observed in nature can be taken as a confirmation of Equation 24.5.
Equation 24.6 is Faraday’s law of induction, which describes the creation of an
electric field by a changing magnetic flux. This law states that the emf, which is the
line integral of the electric field around any closed path, equals the rate of change
of magnetic flux through any surface bounded by that path. One consequence of
Faraday’s law is the current induced in a conducting loop placed in a time-varying
magnetic field.
Equation 24.7 is the Ampère–Maxwell law, and it describes the creation of a magnetic field by a changing electric field and by electric current: the line integral of the
magnetic field around any closed path is the sum of 0 multiplied by the net current
through that path and 00 multiplied by the rate of change of electric flux through
any surface bounded by that path.
Once the electric and magnetic fields are known at some point in space, the force
acting on a particle of charge q can be calculated from the expression
:
:
F 5 q E 1 q:
v
:
B
24.8b
This relationship is called the Lorentz force law. (We saw this relationship earlier as
Eq. 22.6.) Maxwell’s equations, together with this force law, completely describe all
classical electromagnetic interactions in a vacuum.
Notice the symmetry of Maxwell’s equations. Equations 24.4 and 24.5 are symmetric, apart from the absence of the term for magnetic monopoles in Equation 24.5.
:
Furthermore, Equations 24.6 and 24.7 are symmetric in that the line integrals of E
:
and B around a closed path are related to the rate of change of magnetic flux and
electric flux, respectively. Maxwell’s equations are of fundamental importance not
only to electromagnetism, but to all science. Hertz once wrote, “One cannot escape
the feeling that these mathematical formulas have an independent existence and
an intelligence of their own, that they are wiser than we are, wiser even than their
discoverers, that we get more out of them than we put into them.”
In the next section, we show that Equations 24.6 and 24.7 can be combined to
obtain a wave equation for both the electric field and the magnetic field. In empty
space, where q 5 0 and I 5 0, the solution to these two equations shows that the
speed at which electromagnetic waves travel equals the measured speed of light. This
result led Maxwell to predict that light waves are a form of electromagnetic radiation.
Hertz performed experiments that verified Maxwell’s prediction. The experimental apparatus Hertz used to generate and detect electromagnetic waves is shown
schematically in Figure 24.3. An induction coil is connected to a transmitter made
up of two spherical electrodes separated by a narrow gap. The coil provides short
voltage surges to the electrodes, making one positive and the other negative. A spark
is generated between the spheres when the electric field near either electrode surpasses the dielectric strength for air (3 3 106 V/m; see Table 20.1). Free electrons
in a strong electric field are accelerated and gain enough energy to ionize any molecules they strike. This ionization provides more electrons, which can accelerate and
cause further ionizations. As the air in the gap is ionized, it becomes a much better
conductor and the discharge between the electrodes exhibits an oscillatory behavior
at a very high frequency. From an electric-circuit viewpoint, this experimental apparatus is equivalent to an LC circuit in which the inductance is that of the coil and
the capacitance is due to the spherical electrodes. By applying Kirchhoff’s loop rule
to an LC circuit, similar to the way we applied it to an RC circuit in Section 21.9, we
can show that the current in an LC circuit oscillates in simple harmonic motion at
the frequency
5
1
LC
24.9b
c Lorentz force law
The transmitter consists of two
spherical electrodes connected to
an induction coil, which provides
short voltage surges to the
spheres, setting up oscillations in
the discharge between the
electrodes.
Induction
coil
q
q
Transmitter
Receiver
The receiver is a nearby loop
of wire containing a second
spark gap.
Figure 24.3 Schematic diagram of
Hertz’s apparatus for generating and
detecting electromagnetic waves.
© Bettmann/Corbis
826 CHAPTER 24 | Electromagnetic Waves
Heinrich Rudolf Hertz
German Physicist (1857–1894)
Hertz made his most important
discovery of electromagnetic waves in
1887. After finding that the speed of an
electromagnetic wave was the same as that
of light, Hertz showed that electromagnetic
waves, like light waves, could be reflected,
refracted, and diffracted. The hertz, equal to
one complete vibration or cycle per second,
is named after him.
Because L and C are small in Hertz’s apparatus, the frequency of oscillation is
high, on the order of 100 MHz. Electromagnetic waves are radiated at this frequency
as a result of the oscillation (and hence acceleration) of free charges in the transmitter circuit. Hertz was able to detect these waves using a single loop of wire with
its own spark gap (the receiver). Such a receiver loop, placed several meters from
the transmitter, has its own effective inductance, capacitance, and natural frequency
of oscillation. In Hertz’s experiment, sparks were induced across the gap of the receiving electrodes when the receiver’s frequency was adjusted to match that of the
transmitter. In this way, Hertz demonstrated that the oscillating current induced in
the receiver was produced by electromagnetic waves radiated by the transmitter. His
experiment is analogous to the mechanical phenomenon in which a tuning fork
responds to acoustic vibrations from an identical tuning fork that is oscillating.
In addition, Hertz showed in a series of experiments that the radiation generated
by his spark-gap device exhibited the wave properties of interference, diffraction,
reflection, refraction, and polarization, which are all properties exhibited by light
as we shall see in this chapter and Chapters 25–27. Therefore, it became evident
that the radio-frequency waves Hertz was generating had properties similar to those
of light waves and that they differed only in frequency and wavelength. Perhaps his
most convincing experiment was the measurement of the speed of this radiation.
Waves of known frequency were reflected from a metal sheet and created a standingwave interference pattern whose nodal points could be detected. The measured distance between the nodal points enabled determination of the wavelength . Using
the relationship v 5 f (Eq. 13.12), Hertz found that v was close to 3 3 108 m/s, the
known speed c of visible light.
THINKING PHYSICS 24.1
In radio transmission, a radio wave serves as a carrier
wave and the sound wave is superimposed on the carrier wave. In amplitude modulation (AM radio), the
amplitude of the carrier wave varies according to the
sound wave. (The word modulate means “to change.”) In
frequency modulation (FM radio), the frequency of the
carrier wave varies according to the sound wave. The
navy sometimes uses flashing lights to send Morse code
to neighboring ships, a process that is similar to radio
broadcasting. Is this process AM or FM? What is the carrier frequency? What is the signal frequency? What is the
broadcasting antenna? What is the receiving antenna?
Reasoning The flashing of the light according to
Morse code is a drastic amplitude modulation because
the amplitude is changing between a maximum value
and zero. In this sense, it is similar to the on-and-off
binary code used in computers and compact discs.
The carrier frequency is that of the visible light, on the
order of 1014 Hz. The signal frequency depends on the
skill of the signal operator, but is on the order of a few
hertz, as the light is flashed on and off. The broadcasting antenna for this modulated signal is the filament
of the lightbulb in the signal source. The receiving antenna is the eye. b
24.3 | Electromagnetic Waves
y
S
E
S
z
B
S
c
x
Active Figure 24.4 The fields in
an electromagnetic wave traveling at
velocity :
c in the positive x direction
at one point on the x axis. These
fields depend only on x and t.
In his unified theory of electromagnetism, Maxwell showed that time-dependent
electric and magnetic fields satisfy a linear wave equation. (The linear wave equation for mechanical waves is Equation 13.20.) The most significant outcome of this
theory is the prediction of the existence of electromagnetic waves.
Maxwell’s equations predict that an electromagnetic wave consists of oscillating
electric and magnetic fields. The changing fields induce each other, which maintains the propagation of the wave; a changing electric field induces a magnetic field,
:
:
and a changing magnetic field induces an electric field. The E and B vectors are
perpendicular to each other, and to the direction of propagation, as shown in Active Figure 24.4 at one instant of time and one point in space. The direction of
:
:
B, which we shall explore
propagation is the direction of the vector product E
more fully in Section 24.4.
In Active Figure 24.4, we have chosen the direction of propagation of the wave to
be the positive x axis. We have also chosen the y axis to be parallel to the electric field
24.3 | Electromagnetic Waves 827
:
vector. Given these choices, it is necessarily true that the magnetic field B is in the
z direction as in Active Figure 24.4. Waves in which the electric and magnetic fields
are restricted to being parallel to certain directions are said to be linearly polarized
waves. Furthermore, let us assume that at any point in space in Active Figure 24.4, the
magnitudes E and B of the fields depend on x and t only, not on the y or z coordinates.
Let us also imagine that the source of the electromagnetic waves is such that a
wave radiated from any position in the yz plane (not just from the origin as might be
suggested by Active Fig. 24.4) propagates in the x direction and that all such waves
are emitted in phase. If we define a ray as the line along which a wave travels, all rays
for these waves are parallel. This whole collection of waves is often called a plane
wave. A surface connecting points of equal phase on all waves, which we call a wave
front, is a geometric plane. In comparison, a point source of radiation sends waves
out in all directions. A surface connecting points of equal phase for this situation is
a sphere, so we call the radiation from a point source a spherical wave.
To generate the prediction of electromagnetic waves, we start with Faraday’s law,
Equation 24.6:
dFB
:
E d:
s 52
dt
Pitfall Prevention | 24.1
What Is “a” Wave?
What do we mean by a single wave?
The word wave represents both the
emission from a single point (“wave
radiated from any position in the yz
plane” in the text) and the collection
of waves from all points on the source
(“plane wave” in the text). You should
be able to use this term in both ways
and understand its meaning from the
context.
R
According to Equation 24.12,
S
this spatial variation in E gives
rise to a time-varying magnetic
field along the z direction.
Let’s again assume the electromagnetic wave is traveling in the x direction, with the
:
:
electric field E in the positive y direction and the magnetic field B in the positive
z direction.
Consider a rectangle of width dx and height , lying in the xy plane as shown in Fig:
ure 24.5. To apply Equation 24.6, let’s first evaluate the line integral of E d :
s around
this rectangle in the counterclockwise direction at an instant of time when the wave
is passing through the rectangle. The contributions from the top and bottom of the
:
rectangle are zero because E is perpendicular to d :
s for these paths. We can express
the electric field on the right side of the rectangle as
E(x 1 dx) < E(x) 1
dE
dx
4
dx 5 E(x) 1
t constant
10E0x 2 dx
E d:
s 5 [E(x 1 dx)], 2 [E(x)], < ,
4
5 , dx
x constant
0B
0t
E(x dx)
ᐉ
S
B
E(x )
24.10b
Because the magnetic field is in the z direction, the magnetic flux through the rectangle of area ,dx is approximately FB 5 B ,dx (assuming dx is very small compared
with the wavelength of the wave). Taking the time derivative of the magnetic flux
gives
dFB
dB
5 , dx
dt
dt
S
z
line integral over this rectangle is approximately
:
dx
S
0E
dx
0x
where E(x) is the field on the left side of the rectangle at this instant.2 Therefore, the
R
y
24.11b
Substituting Equations 24.10 and 24.11 into Equation 24.6 gives
10E0x 2 dx 5 2,dx 0B0t
,
0E
0B
52
24.12b
0x
0t
In a similar manner, we can derive a second equation by starting with Maxwell’s
:
fourth equation in empty space (Eq. 24.7). In this case, the line integral of B d :
s is
evaluated around a rectangle lying in the xz plane and having width dx and length
, as in Figure 24.6 (page 828). Noting that the magnitude of the magnetic field
2Because dE/dx in this equation is expressed as the change in E with x at a given instant t, dE/dx is equivalent to the
partial derivative −E/−x. Likewise, dB/dt means the change in B with time at a particular position x ; therefore, in Equation 24.11, we can replace dB/dt with −B/−t.
x
Figure 24.5 At an instant when
a plane wave moving in the positive x direction passes through a
rectangular path of width dx lying
in the xy plane, the electric field in
:
the y direction varies from E (x) to
:
E (x 1 dx).
828 CHAPTER 24 | Electromagnetic Waves
According to Equation
24.15, this
S
spatial variation in B gives rise to
a time-varying electric field along
the y direction.
changes from B(x) to B(x 1 dx) over the width dx and that the direction for taking
the line integral is counterclockwise when viewed from above in Figure 24.6, the line
integral over this rectangle is found to be approximately
R
y
S
E
S
z
B(x)
x
S
dx
B(x dx)
ᐉ
B d:
s 5 [B(x)], 2 [B(x 1 dx)], < 2,
24.13b
The electric flux through the rectangle is FE 5 E ,dx, which, when differentiated
with respect to time, gives
0 FE
0E
5 , dx
24.14b
0t
0t
Substituting Equations 24.13 and 24.14 into Equation 24.7 gives
10B0x 2 dx 5 , dx 10E0t 2
2,
Figure 24.6 At an instant when
a plane wave passes through a
rectangular path of width dx lying in
the xz plane, the magnetic field in
:
the z direction varies from B(x) to
:
B(x 1 dx).
10B0x 2 dx
:
0 0
0B
0E
5 200
0x
0t
24.15b
Taking the derivative of Equation 24.12 with respect to x and combining the result
with Equation 24.15 gives
0 2E
0 0B
0 0B
0
0E
52
52
52
200
0x 2
0x 0t
0t 0x
0t
0t
1 2
1 2
1
2
0 2E
0 2E
5 00 2
2
0x
0t
24.16b
In the same manner, taking the derivative of Equation 24.15 with respect to x and
combining it with Equation 24.12 gives
0 2B
0 2B
5
0
0
0x 2
0t 2
24.17b
Equations 24.16 and 24.17 both have the form of the linear wave equation3 with the
wave speed v replaced by c, where
c5
c Speed of electromagnetic waves
1
00
24.18b
Let’s evaluate this speed numerically:
c5
1
(4 3 1027 T ? m/A)(8.854 19 3 10212 C2/N ? m2)
5 2.997 92 3 108 m/s
Because this speed is precisely the same as the speed of light in empty space, we are
led to believe (correctly) that light is an electromagnetic wave.
The simplest solution to Equations 24.16 and 24.17 is a sinusoidal wave for which
the field magnitudes E and B vary with x and t according to the expressions
c Sinusoidal electric and magnetic
fields
E 5 Emax cos (kx 2 t)
24.19b
B 5 Bmax cos (kx 2 t)
24.20b
where Emax and Bmax are the maximum values of the fields. The angular wave number
is k 5 2/, where is the wavelength. The angular frequency is 5 2f , where f is
the wave frequency. The ratio /k equals the speed of an electromagnetic wave, c :
2f
5
5 f 5 c
k
2/
3The linear wave equation is of the form (−2y/−x 2) 5 (1/v 2)(−2y/−t 2), where v is the speed of the wave and y is the wave
function. The linear wave equation was introduced as Equation 13.20, and we suggest you review Section 13.2.
24.3 | Electromagnetic Waves 829
where we have used Equation 13.12, v 5 c 5 f, which relates the speed, frequency,
and wavelength of any continuous wave. Therefore, for electromagnetic waves, the
wavelength and frequency of these waves are related by
5
c
3.00 3 108 m/s
5
f
f
24.21b
Active Figure 24.7 is a pictorial representation, at one instant, of a sinusoidal, linearly
polarized electromagnetic wave moving in the positive x direction.
Taking partial derivatives of Equations 24.19 (with respect to x) and 24.20 (with
respect to t) gives
y
S
E
z
S
S
B
c
x
Active Figure 24.7 A sinusoidal
electromagnetic wave moves in the
positive x direction with a speed c.
0E
5 2kEmax sin (kx 2 t)
0x
0B
5 Bmax sin (kx 2 t)
0t
Substituting these results into Equation 24.12 shows that, at any instant,
kEmax 5 Bmax
Emax
Bmax
5
5c
k
Using these results together with Equations 24.19 and 24.20 gives
Pitfall Prevention | 24.2
:
:
E Stronger Than B ?
Emax
Bmax
5
E
5c
B
24.22b
That is, at every instant, the ratio of the magnitude of the electric field to the
magnitude of the magnetic field in an electromagnetic wave equals the speed of
light.
Finally, note that electromagnetic waves obey the superposition principle (which
we discussed in Section 14.1 with respect to mechanical waves) because the differential equations involving E and B are linear equations. For example, we can add two
waves with the same frequency and polarization simply by adding the magnitudes of
the two electric fields algebraically.
Because the value of c is so large,
some students incorrectly interpret
Equation 24.22 as meaning that
the electric field is much stronger
than the magnetic field. Electric
and magnetic fields are measured
in different units, however, so they
cannot be directly compared. In
Section 24.4, we find that the electric
and magnetic fields contribute
equally to the wave’s energy.
Doppler Effect for Light
Another feature of electromagnetic waves is that there is a shift in the observed frequency of the waves when there is relative motion between the source of the waves
and the observer. This phenomenon, known as the Doppler effect, was introduced
in Chapter 13 as it pertains to sound waves. In the case of sound, the motion of the
source with respect to the medium of propagation can be distinguished from the
motion of the observer with respect to the medium. Light waves must be analyzed differently, however, because they require no medium of propagation, and no method
exists for distinguishing the motion of a light source from the motion of the observer.
If a light source and an observer approach each other with a relative speed v, the
frequency f 9 measured by the observer is
f9 5
c1v
f
c2v
24.23b
where f is the frequency of the source measured in its rest frame. This Doppler
shift equation, unlike the Doppler shift equation for sound, depends only on the
relative speed v of the source and observer and holds for relative speeds as great
as c. As you might expect, the equation predicts that f 9 . f when the source and
observer approach each other. We obtain the expression for the case in which the
source and observer recede from each other by substituting negative values for v in
Equation 24.23.
c Doppler effect for
electromagnetic waves
830 CHAPTER 24 | Electromagnetic Waves
The most spectacular and dramatic use of the Doppler effect for electromagnetic
waves is the measurement of shifts in the frequency of light emitted by a moving
astronomical object such as a galaxy. Light emitted by atoms and normally found in
the extreme violet region of the spectrum is shifted toward the red end of the spectrum for atoms in other galaxies, indicating that these galaxies are receding from us.
American astronomer Edwin Hubble (1889–1953) performed extensive measurements of this red shift to confirm that most galaxies are moving away from us, indicating that the Universe is expanding.
Example 24.1 | An Electromagnetic Wave
y
A sinusoidal electromagnetic wave of frequency 40.0 MHz
travels in free space in the x direction as in Figure 24.8.
(A) Determine the wavelength and period of the wave.
SOLUTION
Conceptualize Imagine the wave in Figure 24.8 moving to the
right along the x axis, with the electric and magnetic fields
oscillating in phase.
S
E 750ˆj N/C
Figure 24.8 (Example
24.1) At some instant, a
plane electromagnetic wave
moving in the x direction
has a maximum electric
field of 750 N/C in the
z
positive y direction.
S
B
S
c
x
Categorize We determine the results using equations developed in this section, so we categorize this example as a substitution problem.
c
3.00 3 108 m/s
5 5
5 7.50 m
Use Equation 24.21 to find the wavelength of the wave:
f
40.0 3 106 Hz
Find the period T of the wave as the inverse of the
frequency:
T5
1
1
5
5 2.50 3 1028 s
f
40.0 3 106 Hz
(B) At some point and at some instant, the electric field has its maximum value of 750 N/C and is directed along the y
axis. Calculate the magnitude and direction of the magnetic field at this position and time.
SOLUTION
Use Equation 24.22 to find the magnitude of the magnetic field:
:
Bmax 5
Emax
750 N/C
5 2.50 3 1026 T
5
c
3.00 3 108 m/s
:
Because E and B must be perpendicular to each other and perpendicular to the direction of wave propagation (x in this
:
case), we conclude that B is in the z direction.
(C) An observer on the x axis, far to the right in Figure 24.8, moves to the left along the x axis at 0.500c. What frequency
does this observer measure for the electromagnetic wave?
SOLUTION
Use Equation 24.23 for the Doppler effect to find the
observed frequency:
f9 5
c1v
f 5 40.0 MHz
c2v
c 1 (10.500c)
c 2 (10.500c)
5 69.3 MHz
We substituted v as a positive number because the observer is moving toward the source.
24.4 | Energy Carried by Electromagnetic Waves
In Section 13.5, we found that mechanical waves carry energy. Electromagnetic waves
also carry energy, and as they propagate through space they can transfer energy
to objects placed in their path. This notion was introduced in Chapter 7 when we
24.4 | Energy Carried by Electromagnetic Waves 831
discussed the transfer mechanisms in the conservation of energy equation, and it was
noted again in Chapter 17 in the discussion of thermal radiation. The rate of flow
:
of energy in an electromagnetic wave is described by a vector S , called the Poynting
vector, defined by the expression
:
S;
1 :
E
0
:
24.24b
B
The magnitude of the Poynting vector represents the rate at which energy flows
through a unit surface area perpendicular to the flow and its direction is along the
direction of wave propagation (Fig. 24.9). Therefore, the Poynting vector represents
power per unit area. The SI units of the Poynting vector are J/s ? m2 5 W/m2.
:
As an example, let us evaluate the magnitude of S for a plane electromagnetic
:
:
:
:
wave. We have u E
B u 5 EB because E and B are perpendicular to each other. In
this case,
EB
S5
24.25b
0
c Poynting vector
Pitfall Prevention | 24.3
An Instantaneous Value
The Poynting vector given by
Equation 24.24 is time dependent. Its
magnitude varies in time, reaching a
maximum value at the same instant as
:
:
the magnitudes of E and B do.
y
Because B 5 E/c, we can also express the magnitude of the Poynting vector as
S5
E2
0c
5
E
0
S
These equations for S apply at any instant of time.
What is of more interest for a sinusoidal electromagnetic wave (Eqs. 24.19 and
24.20) is the time average of S over one or more cycles, which is the intensity I, that
is, the average power per unit area. When this average is taken, we obtain an expression involving the time average of cos2(kx 2 t), which equals 12. Therefore, the average value of S (or the intensity of the wave) is
I 5 Savg 5
EmaxBmax
20
S
cB 2
5
2
E max
20c
5
2
cBmax
20
24.26b
B
z
S
S
S
c
x
:
Figure 24.9 The Poynting vector S
for an electromagnetic wave is along
the direction of wave propagation.
c Wave intensity
Recall that the energy per unit volume uE , which is the instantaneous energy
density associated with an electric field (Section 20.9), is given by Equation 20.31:
uE 5 12 0E 2
24.27b
and that the instantaneous energy density uB associated with a magnetic field (Section 23.7) is given by Equation 23.22:
uB 5
B2
20
24.28b
Because E and B vary with time for an electromagnetic wave, the energy densities
also vary with time. Using the relationships B 5 E/c and c 5 1/ 00, Equation 24.28
becomes
uB 5
00 2 1
(E/c)2
5
E 5 20E 2
20
20
Comparing this result with the expression for uE , we see that
uB 5 uE
That is, for an electromagnetic wave, the instantaneous energy density associated
with the magnetic field equals the instantaneous energy density associated with the
electric field. Therefore, in a given volume the energy is equally shared by the two
fields.
The total instantaneous energy density u is equal to the sum of the energy densities associated with the electric and magnetic fields:
u 5 uE 1 uB 5 0E 2 5
B2
0
c Total instantaneous energy
density of an electromagnetic wave
832 CHAPTER 24 | Electromagnetic Waves
When this expression is averaged over one or more cycles of an electromagnetic
wave, we again obtain a factor of 12. Therefore, the total average energy per unit volume of an electromagnetic wave is
uavg 5 0(E 2)avg 5 12 0 E 2max 5
c Average energy density of an
electromagnetic wave
2
B max
20
24.29b
Comparing this result with Equation 24.26 for the average value of S, we see that
I 5 Savg 5 cuavg
24.30b
In other words, the intensity of an electromagnetic wave equals the average energy
density multiplied by the speed of light.
QUI C K QU IZ 24.2 An electromagnetic wave propagates in the negative y direction.
The electric field at a point in space is momentarily oriented in the positive x direction. In which direction is the magnetic field at that point momentarily oriented?
(a) the negative x direction (b) the positive y direction (c) the positive z direction
(d) the negative z direction
QUI C K QU IZ 24.3 Which of the following quantities does not vary in time for plane
electromagnetic waves? (a) magnitude of the Poynting vector (b) energy density uE
(c) energy density uB (d) intensity I
Example 24.2 | Fields on the Page
Estimate the maximum magnitudes of the electric and magnetic fields of the light that is incident on this page because
of the visible light coming from your desk lamp. Treat the lightbulb as a point source of electromagnetic radiation that is
5% efficient at transforming energy coming in by electrical transmission to energy leaving by visible light.
SOLUTION
Conceptualize The filament in your lightbulb emits electromagnetic radiation. The brighter the light, the larger the magnitudes of the electric and magnetic fields.
Categorize Because the lightbulb is to be treated as a point source, it emits equally in all directions, so the outgoing electromagnetic radiation can be modeled as a spherical wave.
Analyze We mentioned that intensity is equivalent to the average power of radiation per unit area. For a point source
emitting uniformly in all directions, the power is distributed evenly over the surface area 4r 2 of an expanding sphere of
increasing radius r, centered on the source. Therefore, I 5 Pavg/4r 2, where P represents power.
Pavg
E 2max
4r
20c
Set this expression for I equal to the intensity of an electromagnetic wave given by Equation 24.26:
I5
Solve for the electric field magnitude:
Emax 5
5
2
0c Pavg
2r 2
Let’s make some assumptions about numbers to enter in this equation. The visible light output of a 60-W lightbulb operating at 5% efficiency is approximately 3.0 W by visible light. (The remaining energy transfers out of the lightbulb by conduction and invisible radiation.) A reasonable distance from the lightbulb to the page might be 0.30 m.
Substitute these values:
Emax 5
(4 3 1027 T ? m/A)(3.00 3 108 m/s)(3.0 W)
B
2(0.30 m)2
5 45 V/m
Use Equation 24.22 to find the magnetic field magnitude:
Bmax 5
Emax
45 V/m
5 1.5 3 1027 T
5
c
3.00 3 108 m/s
Finalize This value of the magnetic field magnitude is two orders of magnitude smaller than the Earth’s magnetic field.
24.5 | Momentum and Radiation Pressure 833
24.5 | Momentum and Radiation Pressure
Electromagnetic waves transport linear momentum as well as energy. Hence, it follows that pressure is exerted on a surface when an electromagnetic wave impinges
on it. In what follows, let us assume that the electromagnetic wave strikes a surface at
normal incidence and transports a total energy TER to a surface in a time interval Dt.
If the surface absorbs all the incident energy TER in this time, Maxwell showed that
p delivered to this surface has a magnitude
the total momentum :
p5
TER
c
(complete absorption)
24.31b
c Momentum delivered to a
perfectly absorbing surface
The radiation pressure P exerted on the surface is defined as force per unit area
F/A. Let us combine this definition with Newton’s second law:
Pitfall Prevention | 24.4
F
1 dp
P 5
5
A
A dt
So Many p’s
If we now replace p, the momentum transported to the surface by radiation, from
Equation 24.31, we have
P 5
1 dp
1 d TER
5
c
A dt
A dt
1 2 5 1c
We have p for momentum and P for
pressure, and they are both related to
P for power! Be sure to keep all these
symbols straight.
(dTER/dt)
A
We recognize (dTER/dt)/A as the rate at which energy is arriving at the surface per
unit area, which is the magnitude of the Poynting vector. Therefore, the radiation
pressure P exerted on the perfectly absorbing surface is
P5
S
c
(complete absorption)
24.32b
c Radiation pressure exerted on a
perfectly absorbing surface
An absorbing surface for which all the incident energy is absorbed (none is reflected) is called a black body. A more detailed discussion of a black body will be
presented in Chapter 28.
As we found in the last section, the intensity of an electromagnetic wave I is equal
to the average value of S (Eq. 24.26), so we can express the average radiation pressure as
Savg
I
Pavg 5
5
(complete absorption)
24.33b
c
c
Furthermore, because Savg represents power per unit area, we find that the average
power delivered to a surface of area A is (using “Power ” to represent power because
we also have P for pressure here)
(Power)avg 5 IA
(complete absorption)
24.34b
If the surface is a perfect reflector, the momentum delivered in a time interval
Dt for normal incidence is twice that given by Equation 24.31, or p 5 2TER /c. That
is, a momentum TER /c is delivered first by the incident wave and then again by the
reflected wave, a situation analogous to a ball colliding elastically with a wall.4 Finally,
the radiation pressure exerted on a perfect reflecting surface for normal incidence
of the wave is twice that given by Equation 24.32, or
2S
(complete reflection)
24.35b
c
Although radiation pressures are very small (about 5 3 1026 N/m2 for direct
sunlight), they have been measured using torsion balances such as the one shown in
Figure 24.10. Light is allowed to strike either a mirror or a black disk, both of which
are suspended from a fine fiber. Light striking the black disk is completely absorbed,
so all its momentum is transferred to the disk. Light striking the mirror (normal
Light
Mirror
P5
4For oblique incidence, the momentum transferred is 2T
2
ER cos /c and the pressure is given by P 5 2S cos /c, where is
the angle between the normal to the surface and the direction of propagation.
Black
disk
Figure 24.10 An apparatus for
measuring the pressure exerted
by light. In practice, the system is
contained in a high vacuum.
834 CHAPTER 24 | Electromagnetic Waves
incidence) is totally reflected and hence the momentum transfer is twice as great as
that transferred to the disk. The radiation pressure is determined by measuring the
angle through which the horizontal connecting rod rotates. The apparatus must be
placed in a high vacuum to eliminate the effects of air currents.
QUI C K QU IZ 24.4 In an apparatus such as that in Figure 24.10, suppose the
black disk is replaced by one with half the radius. Which of the following are
different after the disk is replaced? (a) radiation pressure on the disk (b) radiation force on the disk (c) radiation momentum delivered to the disk in a given
time interval
THINKING PHYSICS 24.2
A large amount of dust occurs in the interplanetary
space in the Solar System. Although this dust can theoretically have a variety of sizes, from molecular size upward, very little of it is smaller than about 0.2 m in our
Solar System. Why? (Hint: The Solar System originally
contained dust particles of all sizes.)
Reasoning Dust particles in the Solar System are sub-
ject to two forces: the gravitational force toward the Sun
and the force from radiation pressure due to sunlight,
which is away from the Sun. The gravitational force is
proportional to the cube of the radius of a spherical
dust particle because it is proportional to the particle’s
mass. The radiation force is proportional to the square
of the radius because it depends on the circular crosssection of the particle. For large particles, the gravitational force is larger than the force from radiation
pressure. For small particles, less than about 0.2 m,
the larger force from radiation pressure sweeps these
particles out of the Solar System. b
Example 24.3 | Solar Energy
The Sun delivers about 1 000 W/m2 of energy to the Earth’s surface.
(A) Calculate the total power incident on a roof of dimensions 8.00 m 3 20.0 m.
SOLUTION
Conceptualize It should be easy to imagine energy streaming in from the Sun and striking the roof. This transfer of energy
is represented by TER in Equation 7.2.
Categorize We will use equations developed in this section for this problem, so we categorize this as a substitution problem.
The Poynting vector has an average magnitude
I 5 Savg 5 1 000 W/m2, which represents the power per
unit area. Assuming that the radiation is incident normal
to the roof, find the power delivered to the whole roof
using Equation 24.34:
(Power)avg 5 IA 5 (1 000 W/m2)(8.00 m)(20.0 m)
5 1.60 3 105 W
(B) Determine the radiation pressure and radiation force on the roof, assuming that the roof covering is a perfect
absorber.
SOLUTION
I
1 000 W/m2
5 3.33 3 1026 N/m2
5
c
3.00 3 108 m/s
Using Equation 24.33 with I 5 1 000 W/m2, find the average radiation pressure on the roof:
Pavg 5
Noting that pressure is defined as force per unit area, find
the radiation force on the roof:
F 5 PavgA 5 (3.33 3 1026 N/m2)(8.00 m)(20.0 m)
5 5.33 3 1024 N
24.5 | Momentum and Radiation Pressure 835
Exampl e 24.4 | Pressure from a Laser Pointer
When giving presentations, many people use a laser pointer to direct the attention of the audience to information on a
screen. If a 3.0-mW pointer creates a spot on a screen that is 2.0 mm in diameter, determine the radiation pressure on a
screen that reflects 70% of the light that strikes it. The power 3.0 mW is a time-averaged value.
SOLUTION
Conceptualize Imagine the waves striking the screen and exerting a radiation pressure on it. The pressure should not be
very large.
Categorize This problem involves a calculation of radiation pressure using an approach like that leading to Equation 24.32
or Equation 24.35, but it is complicated by the 70% reflection.
Analyze We begin by determining the magnitude of the beam’s Poynting vector.
Divide the time-averaged power delivered via the
electromagnetic wave by the cross-sectional area
of the beam:
Savg 5
(Power)avg
A
5
(Power)avg
r 2
5
3.0 3 1023 W
2.0 3 1023 m 2
2
1
5 955 W/m2
2
Now let’s determine the radiation pressure from the laser beam. Equation 24.35 indicates that a completely reflected beam
would apply an average pressure of Pavg 5 2Savg /c. We can model the actual reflection as follows. Imagine that the surface
absorbs the beam, resulting in pressure Pavg 5 Savg /c. Then the surface emits the beam, resulting in additional pressure
Pavg 5 Savg/c. If the surface emits only a fraction f of the beam (so that f is the amount of the incident beam reflected),
the pressure due to the emitted beam is Pavg 5 f Savg /c.
Savg
Savg
Use this model to find the total pressure on the surface
due to absorption and re-emission (reflection):
Pavg 5
Evaluate this pressure for a beam that is 70% reflected:
Pavg 5 (1 1 0.70)
c
1f
c
5 (1 1 f )
Savg
c
955 W/m2
5 5.4 3 1026 N/m2
3.0 3 108 m/s
Finalize The pressure has an extremely small value, as expected. (Recall from Section 15.1 that atmospheric pressure is
approximately 105 N/m2). Consider the magnitude of the Poynting vector, Savg 5 955 W/m2. It is about the same as the
intensity of sunlight at the Earth’s surface. For this reason, it is not safe to shine the beam of a laser pointer into a person’s
eyes, which may be more dangerous than looking directly at the Sun.
Space Sailing
When imagining a trip to another planet, we normally think of traditional rocket
engines that convert chemical energy in fuel carried on the spacecraft to kinetic
energy of the spacecraft. An interesting alternative to this approach is that of space
sailing. A space-sailing craft includes a very large sail that reflects light. The motion
of the spacecraft depends on pressure from light, that is, the force exerted on the sail
by the reflection of light from the Sun. Calculations performed (before U.S. government budget cutbacks shelved early space-sailing projects) showed that sailing craft
could travel to and from the planets in times similar to those for traditional rockets,
but for less cost.
Calculations show that the radiation force from the Sun on a practical sailcraft
with large sails could be equal to or slightly larger than the gravitational force on the
sailcraft. If these two forces are equal, the sailcraft can be modeled as a particle in
equilibrium because the inward gravitational force of the Sun balances the outward
force exerted by the light from the Sun. If the sailcraft has an initial velocity in some
direction away from the Sun, it would move in a straight line under the action of
these two forces, with no necessity for fuel. A traditional spacecraft with its rocket engines turned off, on the other hand, would slow down as a result of the gravitational
force on it due to the Sun. Both the force on the sail and the gravitational force from
the Sun fall off as the inverse square of the Sun–sailcraft separation. Therefore, in
836 CHAPTER 24 | Electromagnetic Waves
Raymond A. Serway
theory, the straight-line motion of the sailcraft would continue forever with no fuel
requirement.
By using just the motion imparted to a sailcraft by the Sun, the craft could reach
Alpha Centauri in about 10 000 years. This time interval can be reduced to 30 to 100
years using a beamed power system. In this concept, light from the Sun is gathered by
a transformation device in orbit around the Earth and is converted to a laser beam
or microwave beam aimed at the sailcraft. The force from this intense beam of radiation increases the acceleration of the craft, and the transit time is significantly
reduced. Calculations indicate that the sailcraft could achieve design speeds of up to
20% of the speed of light using this technique.
24.6 | The Spectrum of Electromagnetic Waves
Electromagnetic waves travel through vacuum with speed c, frequency f, and wavelength . The various types of electromagnetic waves, all produced by accelerating
charges, are shown in Figure 24.11. Note the wide range of frequencies and wavelengths. Let us briefly describe the wave types shown in Figure 24.11.
Radio waves are the result of charges accelerating, for example, through conducting wires in a radio antenna. They are generated by such electronic devices as LC
oscillators and are used in radio and television communication systems.
Microwaves (short-wavelength radio waves) have wavelengths ranging between
about 1 mm and 30 cm and are also generated by electronic devices. Because of their
short wavelengths, they are well suited for radar systems used in aircraft navigation
and for studying the atomic and molecular properties of matter. Microwave ovens
are a domestic application of these waves.
Infrared waves have wavelengths ranging from about 1 mm to the longest wavelength of visible light, 7 3 1027 m. These waves, produced by objects at room temperature and by molecules, are readily absorbed by most materials. Infrared radiation
has many practical and scientific applications, including physical therapy, infrared photography,
Wavelength
Frequency, Hz
and vibrational spectroscopy. Your remote control
22
10
for your TV or DVD player likely uses an infrared
The visible light
1021
Gamma rays
beam to communicate with the video device.
spectrum is
1 pm
1020
Visible light, the most familiar form of electroexpanded to show
details of the colors.
magnetic waves, is that part of the spectrum the
1019
human eye can detect. Light is produced by hot
X-rays
1018
⬃400 nm
objects like lightbulb filaments and by the rear1 nm
1017
rangement of electrons in atoms and molecules.
Violet
Blue
The wavelengths of visible light are classified by
1016
Ultraviolet
Green
color, ranging from violet ( < 4 3 1027 m) to red
15
10
Yellow
1 mm
( < 7 3 1027 m). The eye’s sensitivity is a function
1014
Visible light
of wavelength and is a maximum at a wavelength
Orange
1013
Infrared
of about 5.5 3 1027 m (yellow–green). Table 24.1
Red
provides approximate correspondences between
1012
⬃700 nm
1 mm
the wavelength of visible light and the color as11
10
signed to it by humans. Light is the basis of the
1 cm Adjacent wave types
1010
Microwaves
science of optics, to be discussed in Chapters 25
exhibit some overlap
of frequencies.
109
through 27.
1m
8
Ultraviolet light covers wavelengths ranging
TV,
FM
10
from
about 4 3 1027 m (400 nm) down to about
Radio waves
107
6 3 10210 m (0.6 nm). The Sun is an important
106
AM
source of ultraviolet waves, which are the main
1 km
cause of suntans and sunburns. Atoms in the strato105
sphere absorb most of the ultraviolet waves from the
4
10
Long wave
Sun (which is fortunate because ultraviolet waves in
103
large quantities have harmful effects on humans).
One important constituent of the stratosphere is
Figure 24.11 The electromagnetic spectrum.
Wearing sunglasses that do not block
ultraviolet (UV) light is worse for
your eyes than wearing no sunglasses.
The lenses of any sunglasses absorb
some visible light, thus causing the
wearer’s pupils to dilate. If the glasses
do not also block UV light, more
damage may be done to the eye’s lens
because of the dilated pupils. If you
wear no sunglasses at all, your pupils
are contracted, you squint, and much
less UV light enters your eyes. Highquality sunglasses block nearly all the
eye-damaging UV light.
24.7 | Polarization of Light Waves 837
ozone (O3), which results from reactions of oxygen with ultraviolet radiation. This
ozone shield converts lethal high-energy ultraviolet radiation to harmless infrared
radiation. A great deal of concern has arisen concerning the depletion of the protective ozone layer by the use of a class of chemicals called chlorofluorocarbons (e.g.,
Freon) in aerosol spray cans and as refrigerants.
X-rays are electromagnetic waves with wavelengths in the range of about 1028 m
(10 nm) down to 10213 m (1024 nm). The most common source of x-rays is the acceleration of high-energy electrons bombarding a metal target. X-rays are used as a
diagnostic tool in medicine and as a treatment for certain forms of cancer. Because
x-rays damage or destroy living tissues and organisms, care must be taken to avoid
unnecessary exposure and overexposure. X-rays are also used in the study of crystal structure; x-ray wavelengths are comparable to the atomic separation distances
(< 0.1 nm) in solids.
Gamma rays are electromagnetic waves emitted by radioactive nuclei and during certain nuclear reactions. They have wavelengths ranging from about 10210 m
to less than 10214 m. Gamma rays are highly penetrating and produce serious damage when absorbed by living tissues. Consequently, those working near such dangerous radiation must be protected with heavily absorbing materials, such as layers
of lead.
Q U I CK QUI Z 24.5 In many kitchens, a microwave oven is used to cook food. The
frequency of the microwaves is on the order of 1010 Hz. Are the wavelengths of
these microwaves on the order of (a) kilometers, (b) meters, (c) centimeters, or
(d) micrometers?
Q UI C K QUI Z 24.6 A radio wave of frequency on the order of 105 Hz is used to
carry a sound wave with a frequency on the order of 103 Hz. Is the wavelength
of this radio wave on the order of (a) kilometers, (b) meters, (c) centimeters, or
(d) micrometers?
THINKING PHYSICS 24.3
The center of sensitivity of our eyes is close to the same
frequency as the center of the wavelength distribution
of light from the Sun. Is that an amazing coincidence?
Reasoning It is not a coincidence; rather, it is the re-
sult of biological evolution. Humans have evolved so as
to be most visually sensitive to the wavelengths that are
TABLE 24.1 |
Approximate Correspondence
Between Wavelengths of
Visible Light and Color
Wavelength Range
(nm)
Color
Description
400–430
Violet
430–485
Blue
485–560
Green
560–590
Yellow
590–625
Orange
625–700
Red
Note: The wavelength ranges here are
approximate. Different people will describe
colors differently.
Pitfall Prevention | 24.5
Heat Rays
Infrared rays are often called “heat
rays.” This terminology is a misnomer.
Although infrared radiation is used to
raise or maintain temperature, as in
the case of keeping food warm with
“heat lamps” at a fast-food restaurant,
all wavelengths of electromagnetic
radiation carry energy that can
cause the temperature of a system
to increase. As an example, consider
using your microwave oven to bake a
potato, whose temperature increases
because of microwaves.
Center of eyesight sensitivity
strongest from the Sun. It is an interesting conjecture
to imagine aliens from another planet, with a Sun with
a different temperature, arriving at Earth. Their eyes
would have the center of sensitivity at different wavelengths than ours. How would their vision of the Earth
compare with ours? b
24.7 | Polarization of Light Waves
As we learned in Section 24.3, the electric and magnetic vectors associated with an
electromagnetic wave are perpendicular to each other and also to the direction of
wave propagation as shown in Active Figure 24.4. The phenomenon of polarization
described in this section is a property that specifies the directions of the electric and
magnetic fields associated with an electromagnetic wave.
An ordinary beam of light consists of a large number of waves emitted by the
atoms of the light source. Each atom produces a wave with its own orientation of
:
the electric field E , corresponding to the direction of vibration in the atom. The
direction of polarization of the electromagnetic wave is defined to be the direction
:
in which E is vibrating. Because all directions of vibration are possible in a group of
atoms emitting a beam of light, however, the resultant beam is a superposition of
waves produced by the individual atomic sources. The result is an unpolarized light
838 CHAPTER 24 | Electromagnetic Waves
beam, represented schematically in Figure 24.12a. The direction of wave propagation in this figure is perpendicular to the page. The figure suggests that all directions
of the electric field vector lying in a plane perpendicular to the direction of propagation are equally probable.
S
S
:
E
E
A beam of light is said to be linearly polarized if the orientation of E is the
same for all individual waves at all times at a particular point as suggested in Figure 24.12b. (Sometimes such a wave is described as plane polarized.) The wave
described in Active Figure 24.4 is an example of a wave linearly polarized along the
:
y axis. As the field propagates in the x direction, E is always along the y axis. The
:
plane formed by E and the direction of propagation is called the plane of polarization of the wave. In Active Figure 24.4, the plane of polarization is the xy plane. It is
possible to obtain a linearly polarized wave from an unpolarized wave by removing
from the unpolarized wave all components of electric field vectors except those
a
b
that lie in a single plane.
The most common technique for polarizing light is to send it through a material
Figure 24.12 (a) An unpolarized
light beam viewed along the direction
that passes only components of electric field vectors that are parallel to a characterof propagation (perpendicular to
istic direction of the material called the polarizing direction. In 1938, E. H. Land
the page). The time-varying electric
discovered such a material, which he called Polaroid, that polarizes light through
field vector can be in any direction
selective
absorption by oriented molecules. This material is fabricated in thin sheets
in the plane of the page with equal
of long-chain hydrocarbons, which are stretched during manufacture so that the
probability. (b) A linearly polarized
light beam with the time-varying
molecules align. After a sheet is dipped into a solution containing iodine, the molelectric field vector in the vertical
ecules become good electric conductors. The conduction, however, takes place pridirection.
marily along the hydrocarbon chains because the valence electrons of the molecules
can move easily only along the chains (valence electrons are “free” electrons that
can readily move through the conductor). As a result, the molecules readily absorb
light whose electric field vector is parallel to their length and transmit light whose
electric field vector is perpendicular to their length. It is common
to refer to the direction perpendicular to the molecular chains as
The analyzer allows
The polarizer polarizes
the component of the
the incident light along
the transmission axis. An ideal polarizer passes the components of
light parallel to its axis
its transmission axis.
electric vectors that are parallel to the transmission axis. Compoto pass through.
nents perpendicular to the transmission axis are absorbed. If light
Unpolarized
passes through several polarizers, whatever is transmitted has the
light
plane of polarization parallel to the polarizing direction of the last
S
E0
polarizer through which it passed.
Let us now obtain an expression for the intensity of light that
u
passes through a polarizing material. In Active Figure 24.13, an
unpolarized light beam is incident on the first polarizing sheet,
Transmission Polarized
called the polarizer, where the transmission axis is as indicated.
axis
light
The light that passes through this sheet is polarized vertically,
:
and the transmitted electric field vector is E 0. A second polarizing
Active Figure 24.13 Two polarizing sheets whose
sheet, called the analyzer, intercepts this beam with its transmistransmission axes make an angle with each other. Only a
sion axis at an angle of to the axis of the polarizer. The comfraction of the polarized light incident on the analyzer is
:
transmitted through it.
ponent of E 0 that is perpendicular to the axis of the analyzer is
completely absorbed, and the component parallel to that axis is
E0 cos . We know from Equation 24.26 that the transmitted intensity varies as the
square of the transmitted amplitude, so we conclude that the intensity of the transmitted (polarized) light varies as
The red dot signifies the
velocity vector for the wave
coming out of the page.
c Malus’s law
I 5 Imax cos2 24.36b
where Imax is the intensity of the polarized wave incident on the analyzer. This expression, known as Malus’s law, applies to any two polarizing materials whose transmission axes are at an angle of to each other. From this expression, note that the
transmitted intensity is a maximum when the transmission axes are parallel ( 5 0
or 1808) and zero (complete absorption by the analyzer) when the transmission axes
are perpendicular to each other. This variation in transmitted intensity through a
pair of polarizing sheets is illustrated in Figure 24.14. Because the average value of
cos2 is 12, the intensity of initially unpolarized light is reduced by a factor of one-half
as the light passes through a single ideal polarizer.
24.8 | Context Connection: The Special Properties of Laser Light 839
The transmitted light has
lesser intensity when the
transmission axes are at an
angle of 45 with each other.
The transmitted light
intensity is a minimum when
the transmission axes are
perpendicular to each other.
a
b
c
Figure 24.14 The intensity of light
transmitted through two polarizers
depends on the relative orientation
of their transmission axes. The red
arrows indicate the transmission axes
of the polarizers.
Henry Leap and Jim Lehman
The transmitted light has
maximum intensity when
the transmission axes are
aligned with each other.
Q U I CK QUI Z 24.7 A polarizer for microwaves can be made as a grid of parallel metal
wires approximately 1 cm apart. Is the electric field vector for microwaves transmitted
through this polarizer (a) parallel or (b) perpendicular to the metal wires?
24.8 | Context Connection: The Special Properties
of Laser Light
In this chapter and the next three, we shall explore the nature of laser light and a
variety of applications of lasers in our technological society. The primary properties
of laser light that make it useful in these applications are the following:
• The light is coherent. The individual rays of light in a laser beam maintain a fixed phase relationship with one another, resulting in no destructive
interference.
• The light is monochromatic. Laser light has a very small range of wavelengths.
• The light has a small angle of divergence. The beam spreads out very little, even
over large distances.
ENERGY
To understand the origin of these properties, let us combine our knowledge of
atomic energy levels from Chapter 11 with some special requirements for the atoms
that emit laser light.
As we found in Chapter 11, the energies of an atom are quantized. We used a
semigraphical representation called an energy level diagram in that chapter to help us
understand the quantized energies in an atom. The production
of laser light depends heavily on the properties of these energy
The incoming photon stimulates
levels in the atoms, the source of the laser light.
the atom to emit a second photon
The word laser is an acronym for light amplification by stimof energy hf E 2 E 1.
ulated emission of radiation. The full name indicates one of
the requirements for laser light, that the process of stimulated
E2
E2
emission must occur to achieve laser action.
hf
Suppose an atom is in the excited state E 2 as in Active Figure 24.15 and a photon with energy hf 5 E 2 2 E1 is incident
hf E 2 E 1
on it. The incoming photon can stimulate the excited atom to
return to the ground state and thereby emit a second photon
having the same energy hf and traveling in the same direction.
hf
Note that the incident photon is not absorbed, so after the
E1
E1
stimulated emission, two identical photons exist: the incident
photon and the emitted photon. The emitted photon is in
After
Before
phase with the incident photon. These photons can stimulate
other atoms to emit photons in a chain of similar processes.
Active Figure 24.15 Stimulated emission of a photon by
The many photons produced in this fashion are the source of an incoming photon of energy hf. Initially, the atom is in the
the intense, coherent light in a laser.
excited state.
840 CHAPTER 24 | Electromagnetic Waves
Figure 24.16 Schematic diagram
of a laser design.
The tube contains
the atoms that are
the active medium.
Due to spontaneous
emission, some photons
leave the side of the tube.
The parallel end mirrors confine
the photons to the tube, but mirror
2 is only partially reflective.
Laser
output
Mirror 1
The stimulating wave is
that moving parallel to
the axis of the tube.
Mirror 2
Energy input
An external source of
energy “pumps” the
atoms to excited states.
For the stimulated emission to result in laser light, we must have a buildup of
photons in the system. The following three conditions must be satisfied to achieve
this buildup:
The atom emits 632.8-nm photons
through stimulated emission in the
transition E 3* E 2. That is the
source of coherent light in the laser.
Metastable state
E 3*
hf
l 632.8 nm
E2
ENERGY
Output
energy
Input
energy
E1
Figure 24.17 Energy-level diagram
for a neon atom in a helium–neon
laser.
• The system must be in a state of population inversion. More atoms must be in
an excited state than in the ground state. Atoms in the ground state can absorb
photons, raising them to the excited state. The population inversion assures that
we have more emission of photons from excited atoms than absorption by atoms
in the ground state.
• The excited state of the system must be a metastable state, which means that its
lifetime must be long compared with the usually short lifetime of excited states,
which is typically 1028 s. In this case, stimulated emission is likely to occur before
spontaneous emission. The energy of a metastable state is indicated with an asterisk, E *.
• The emitted photons must be confined in the system long enough to enable
them to stimulate further emission from other excited atoms, which is achieved
by using reflecting mirrors at the ends of the system. One end is made totally
reflecting, and the other is slightly transparent to allow the laser beam to escape
(Fig. 24.16).
One device that exhibits stimulated emission of radiation is the helium–neon gas
laser. Figure 24.17 is an energy-level diagram for the neon atom in this system. The
mixture of helium and neon is confined to a glass tube that is sealed at the ends by
mirrors. A voltage applied across the tube causes electrons to sweep through the
tube, colliding with the atoms of the gases and raising them into excited states. Neon
atoms are excited to state E 3* through this process and also as a result of collisions
with excited helium atoms. Stimulated emission occurs, causing neon atoms to make
transitions to state E2. Neighboring excited atoms are also stimulated. The result is
the production of coherent light at a wavelength of 632.8 nm.
Applications
Since the development of the first laser in 1960, tremendous growth has occurred
in laser technology. Lasers that cover wavelengths in the infrared, visible, and ultraviolet regions are now available. Applications include surgical “welding” of detached retinas, precision surveying and length measurement, precision cutting of
metals and other materials, and telephone communication along optical fibers.
These and other applications are possible because of the unique characteristics
of laser light. In addition to being highly monochromatic, laser light is also highly
directional and can be sharply focused to produce regions of extremely intense
light energy (with energy densities 1012 times the density in the flame of a typical
cutting torch).
Lasers are used in precision long-range distance measurement (range finding).
In recent years, it has become important in astronomy and geophysics to measure
24.8 | Context Connection: The Special Properties of Laser Light 841
as precisely as possible the distances from various points on the surface of the Earth
to a point on the Moon’s surface. To facilitate these measurements, the Apollo astronauts set up a 0.5-m square of reflector prisms on the Moon, which enables
laser pulses directed from an Earth-based station to be retroreflected to the same
station. Using the known speed of light and the measured round-trip travel time of
a laser pulse, the Earth–Moon distance can be determined to a precision of better
than 10 cm.
Because various laser wavelengths can be absorbed in specific biological tissues,
lasers have a number of medical applications. For example, certain laser procedures
have greatly reduced blindness in patients with glaucoma and diabetes. Glaucoma is
a widespread eye condition characterized by a high fluid pressure in the eye, a condition that can lead to destruction of the optic nerve. A simple laser operation (iridectomy) can “burn” open a tiny hole in a clogged membrane, relieving the destructive
pressure. A serious side effect of diabetes is neovascularization, the proliferation of
weak blood vessels, which often leak blood. When neovascularization occurs in the
retina, vision deteriorates (diabetic retinopathy) and finally is destroyed. Today, it is
possible to direct the green light from an argon ion laser through the clear eye lens
and eye fluid, focus on the retina edges, and photocoagulate the leaky vessels. Even
people who have only minor vision defects such as nearsightedness are benefiting
from the use of lasers to reshape the cornea, changing its focal length and reducing
the need for eyeglasses.
Laser surgery is now an everyday occurrence at hospitals and medical clinics around the world. Infrared light at 10 m from a carbon dioxide laser can cut
through muscle tissue, primarily by vaporizing the water contained in cellular material. Laser power of approximately 100 W is required in this technique. The advantage of the “laser knife” over conventional methods is that laser radiation cuts tissue
and coagulates blood at the same time, leading to a substantial reduction in blood
loss. In addition, the technique virtually eliminates cell migration, an important consideration when tumors are being removed.
In biological and medical research, it is often important to isolate and collect unusual cells for study and growth. A laser cell separator exploits the tagging of specific
cells with fluorescent dyes. All cells are then dropped from a tiny charged nozzle and
laser-scanned for the dye tag. If triggered by the correct light-emitting tag, a small
voltage applied to parallel plates deflects the falling electrically charged cell into a
collection beaker.
An exciting area of research and technological applications arose in the 1990s
with the development of laser trapping of atoms. One scheme, called optical molasses
and developed by Steven Chu of Stanford University and his colleagues, involves
focusing six laser beams onto a small region in which atoms are to be trapped. Each
pair of lasers is along one of the x, y, and z axes and emits light in opposite directions (Fig. 24.18). The frequency of the laser light is tuned to be slightly below the
absorption frequency of the subject atom. Imagine that an atom has been placed
into the trap region and moves along the positive x axis toward the laser that is emitting light toward it (the rightmost laser on the x axis in Fig. 24.18). Because the atom
is moving, the light from the laser appears Doppler-shifted upward in frequency in
the reference frame of the atom. Therefore, a match between the Doppler-shifted
laser frequency and the absorption frequency of the atom exists and the atom absorbs photons.5 The momentum carried by these photons results in the atom being
pushed back to the center of the trap. By incorporating six lasers, the atoms are
pushed back into the trap regardless of which way they move along any axis.
In 1986, Chu developed optical tweezers, a device that uses a single tightly focused
laser beam to trap and manipulate small particles. In combination with microscopes,
optical tweezers have opened up many new possibilities for biologists. Optical tweezers have been used to manipulate live bacteria without damage, move chromosomes
within a cell nucleus, and measure the elastic properties of a single DNA molecule.
5The laser light traveling in the same direction as the atom is Doppler-shifted further downward in frequency, so there is
no absorption. Therefore, the atom is not pushed out of the trap by the diametrically opposed laser.
Uses of lasers in
ophthalmology
Laser surgery
Laser cell separator
y
S
c
S
c
S
c
S
c
S
c
z
S
c
x
Figure 24.18 An optical trap for
atoms is formed at the intersection
point of six counterpropagating laser
beams along mutually perpendicular
axes.
Optical tweezers
842 CHAPTER 24 | Electromagnetic Waves
Chu shared the 1997 Nobel Prize in Physics with two of his colleagues for the development of the techniques of optical trapping.
An extension of laser trapping, laser cooling, is possible because the normal high
speeds of the atoms are reduced when they are restricted to the region of the trap.
As a result, the temperature of the collection of atoms can be reduced to a few microkelvins. The technique of laser cooling allows scientists to study the behavior of
atoms at extremely low temperatures (Fig. 24.19).
In the 1920s, Satyendra Nath Bose (1894–1974) was studying photons and investigating collections of identical photons, which can all be in the same quantum
state. Einstein followed up on the work of Bose and predicted that a collection of
atoms could all be in the same quantum state if the temperature were low enough.
The proposed collection of atoms is called a Bose–Einstein condensate. In 1995, using
laser cooling supplemented with evaporative cooling, the first Bose–Einstein condensate was created in the laboratory by Eric Cornell and Carl Wieman, who won
the 2001 Nobel Prize in Physics for their work. Many laboratories are now creating Bose–Einstein condensates and studying their properties and possible applications. One interesting result was reported by a Harvard University group led by Lene
Vestergaard Hau in 2001. She and her colleagues announced that they were able to
bring a light pulse to a complete stop by using a Bose–Einstein condensate.6
More recently, scientists have discovered a new type of Bose–Einstein condensate
based on a quasiparticle called the polariton.7 The polariton, which is a superposition of a photon and an electronic excitation in a solid, exists typically for only a
few picoseconds in an optical cavity. These condensates are unique because they are
extremely light compared to atomic condensates and therefore exhibit quantum
effects at higher temperatures.
We have explored general properties of laser light in this chapter. In the Context Connection of Chapter 25, we shall explore the technology of optical fibers, in
which lasers are used in a variety of applications.
Courtesy of National Institute of Standards
and Technology, U.S. Dept. of Commerce
The orange dot is the sample
of trapped sodium atoms.
Figure 24.19 A staff member of
the National Institute of Standards
and Technology views a sample
of trapped sodium atoms cooled
to a temperature measured in
microkelvins.
SUMMARY |
Displacement current Id is defined as
Id ; 0
Electromagnetic waves, which are predicted by Maxwell’s
equations, have the following properties:
d FE
24.1b
dt
and represents an effective current through a region of space
in which an electric field is changing in time.
:
:
:
When used with the Lorentz force law (F 5 q E 1 q:
v
B),
Maxwell’s equations describe all electromagnetic phenomena:
R
R
R
R
• The electric and magnetic fields satisfy the following wave
equations, which can be obtained from Maxwell’s third
and fourth equations:
0 2E
0 2E
5 0 0 2
2
0x
0t
24.16b
24.17b
q
0
24.4b
0 2B
0 2B
5
0
0
0x 2
0t 2
B dA 5 0
24.5b
• Electromagnetic waves travel through a vacuum with the
speed of light c 5 3.00 3 108 m/s, where
:
:
:
:
E dA 5
:
E d:
s 52
:
d FB
24.6b
dt
B d:
s 5 0I 1 00
d FE
dt
24.7b
c5
1
00
24.18b
• The electric and magnetic fields of an electromagnetic
wave are perpendicular to each other and perpendicular
to the direction of wave propagation; hence, electromagnetic waves are transverse waves. The electric and magnetic
6C. Liu, Z. Dutton, C. H. Behroozi, and L. V. Hau, “Observation of Coherent Optical Information Storage in an Atomic
Medium Using Halted Light Pulses,” Nature, 409, 490–493, January 25, 2001.
7D. Snoke and P. Littlewood, “Polariton Condensates,” Physics Today, 42–47, August 2010.
| Objective Questions 843
fields of a sinusoidal plane electromagnetic wave propagating in the positive x direction can be written
E 5 Emax cos(kx 2 t)
24.19b
B 5 Bmax cos(kx 2 t)
24.20b
where is the angular frequency of the wave and k is the
angular wave number. These equations represent special
:
:
solutions to the wave equations for E and B.
:
:
• The instantaneous magnitudes of E and B in an electromagnetic wave are related by the expression
E
5c
B
24.22b
• Electromagnetic waves carry energy. The rate of flow of
energy crossing a unit area is described by the Poynting
:
vector S , where
:
S;
1 :
E
0
:
B
24.24b
The average value of the Poynting vector for a plane
electromagnetic wave has the magnitude
OBJECTIVE QUESTIONS |
1. Which of the following statements are true regarding electromagnetic waves traveling through a vacuum? More
than one statement may be correct. (a) All waves have the
same wavelength. (b) All waves have the same frequency.
(c) All waves travel at 3.00 3 108 m/s. (d) The electric
and magnetic fields associated with the waves are perpendicular to each other and to the direction of wave propagation. (e) The speed of the waves depends on their
frequency.
2. An electromagnetic wave with a peak magnetic field magnitude of 1.50 3 1027 T has an associated peak electric
field of what magnitude? (a) 0.500 3 10215 N/C (b) 2.00 3
1025 N/C (c) 2.20 3 104 N/C (d) 45.0 N/C (e) 22.0 N/C
3. Assume you charge a comb by running it through your
hair and then hold the comb next to a bar magnet. Do the
electric and magnetic fields produced constitute an electromagnetic wave? (a) Yes they do, necessarily. (b) Yes they do
because charged particles are moving inside the bar magnet. (c) They can, but only if the electric field of the comb
and the magnetic field of the magnet are perpendicular.
(d) They can, but only if both the comb and the magnet are
moving. (e) They can, if either the comb or the magnet or
both are accelerating.
4. If plane polarized light is sent through two polarizers, the
first at 458 to the original plane of polarization and the second at 908 to the original plane of polarization, what fraction of the original polarized intensity passes through the
1
last polarizer? (a) 0 (b) 14 (c) 12 (d) 18 (e) 10
I 5 Savg 5
E maxB max
20
5
E 2max
20c
5
2
cB max
20
24.26b
The average power per unit area (intensity) of a sinusoidal plane electromagnetic wave equals the average value of
the Poynting vector taken over one or more cycles.
• Electromagnetic waves carry momentum and hence can
exert pressure on surfaces. If an electromagnetic wave
whose intensity is I is completely absorbed by a surface on
which it is normally incident, the radiation pressure on
that surface is
S
(complete absorption)
24.32b
c
If the surface totally reflects a normally incident wave, the
pressure is doubled.
P5
The electromagnetic spectrum includes waves covering a
broad range of frequencies and wavelengths.
When polarized light of intensity Imax is incident on a polarizing film, the light transmitted through the film has an
intensity equal to Imax cos2 , where is the angle between the
transmission axis of the polarizing film and the electric field
vector of the incident light.
denotes answer available in Student
Solutions Manual/Study Guide
5. A typical microwave oven operates at a frequency of 2.45 GHz.
What is the wavelength associated with the electromagnetic
waves in the oven? (a) 8.20 m (b) 12.2 cm (c) 1.20 3 108 m
(d) 8.20 3 1029 m (e) none of those answers
6. A student working with a transmitting apparatus like
Heinrich Hertz’s wishes to adjust the electrodes to generate electromagnetic waves with a frequency half as large as
before. (i) How large should she make the effective capacitance of the pair of electrodes? (a) four times larger than
before (b) two times larger than before (c) one-half as large
as before (d) one-fourth as large as before (e) none of those
answers (ii) After she makes the required adjustment, what
will the wavelength of the transmitted wave be? Choose
from the same possibilities as in part (i).
7. A small source radiates an electromagnetic wave with a
single frequency into vacuum, equally in all directions.
(i) As the wave moves, does its frequency (a) increase,
(b) decrease, or (c) stay constant? Using the same choices,
answer the same question about (ii) its wavelength, (iii) its
speed, (iv) its intensity, and (v) the amplitude of its electric
field.
8. A plane electromagnetic wave with a single frequency
moves in vacuum in the positive x direction. Its amplitude
is uniform over the yz plane. (i) As the wave moves, does its
frequency (a) increase, (b) decrease, or (c) stay constant?
Using the same choices, answer the same question about
(ii) its wavelength, (iii) its speed, (iv) its intensity, and (v) the
amplitude of its magnetic field.
844 CHAPTER 24 | Electromagnetic Waves
9. (i) Rank the following kinds of waves according to their
wavelength ranges from those with the largest typical or average wavelength to the smallest, noting any cases of equality:
(a) gamma rays (b) microwaves (c) radio waves (d) visible
light (e) x-rays (ii) Rank the kinds of waves according to
their frequencies from highest to lowest. (iii) Rank the kinds
of waves according to their speeds from fastest to slowest.
Choose from the same possibilities as in part (i).
10. Assume the amplitude of the electric field in a plane electromagnetic wave is E1 and the amplitude of the magnetic
field is B1. The source of the wave is then adjusted so that
the amplitude of the electric field doubles to become 2E1.
(i) What happens to the amplitude of the magnetic field in
this process? (a) It becomes four times larger. (b) It becomes
two times larger. (c) It can stay constant. (d) It becomes onehalf as large. (e) It becomes one-fourth as large. (ii) What
happens to the intensity of the wave? Choose from the same
possibilities as in part (i).
11. A spherical interplanetary grain of dust of radius 0.2 mm
is at a distance r1 from the Sun. The gravitational force exerted by the Sun on the grain just balances the force due
to radiation pressure from the Sun’s light. (i) Assume the
grain is moved to a distance 2r1 from the Sun and released.
At this location, what is the net force exerted on the grain?
(a) toward the Sun (b) away from the Sun (c) zero (d) impossible to determine without knowing the mass of the
grain (ii) Now assume the grain is moved back to its original
CONCEPTUAL QUESTIONS |
iStockPhoto.com/Kdow
1. Despite the advent of digital television, some viewers still
use “rabbit ears” atop their sets (Fig. CQ24.1) instead
of purchasing cable television service or satellite dishes.
Certain orientations of the receiving antenna on a television set give better reception than others. Furthermore, the best orientation varies from station to station.
Explain.
location at r1, compressed so that it crystallizes into a sphere
with significantly higher density, and then released. In this
situation, what is the net force exerted on the grain? Choose
from the same possibilities as in part (i).
12. Consider an electromagnetic wave traveling in the positive
y direction. The magnetic field associated with the wave at
some location at some instant points in the negative x direction as shown in Figure OQ24.12. What is the direction of
the electric field at this position and at this instant? (a) the
positive x direction (b) the positive y direction (c) the positive z direction (d) the negative z direction (e) the negative
y direction
z
S
B
S
y
c
x
Figure OQ24.12
denotes answer available in Student
Solutions Manual/Study Guide
5. Radio stations often advertise “instant news.” If that means
you can hear the news the instant the radio announcer
speaks it, is the claim true? What approximate time interval
is required for a message to travel from Maine to California
by radio waves? (Assume the waves can be detected at this
range.)
6. For a given incident energy of an electromagnetic wave, why
is the radiation pressure on a perfectly reflecting surface
twice as great as that on a perfectly absorbing surface?
7. If a high-frequency current exists in a solenoid containing a metallic core, the core becomes warm due to induction. Explain why the material rises in temperature in this
situation.
8. An empty plastic or glass dish being removed from a microwave oven can be cool to the touch, even when food on an
adjoining dish is hot. How is this phenomenon possible?
9. Describe the physical significance of the Poynting vector.
Figure CQ24.1 Conceptual Question 1 and
Problem 62.
2. List at least three differences between sound waves and light
waves.
3. What new concept did Maxwell’s generalized form of
Ampère’s law include?
4. When light (or other electromagnetic radiation) travels
across a given region, (a) what is it that oscillates? (b) What
is it that is transported?
10. What does a radio wave do to the charges in the receiving
antenna to provide a signal for your car radio?
11. Why should an infrared photograph of a person look different from a photograph taken with visible light?
12. Suppose a creature from another planet has eyes that are
sensitive to infrared radiation. Describe what the alien
would see if it looked around your library. In particular,
what would appear bright and what would appear dim?
| Problems 845
PROBLEMS |
The problems found in this chapter may be assigned
online in Enhanced WebAssign.
denotes Master It tutorial available in Enhanced
WebAssign
denotes asking for quantitative and conceptual reasoning
denotes symbolic reasoning problem
denotes “paired problems” that develop reasoning with
symbols and numerical values
denotes Watch It video solution available in Enhanced
WebAssign
1. denotes straightforward problem; 2. denotes intermediate
problem; 3. denotes challenging problem
1. denotes full solution available in the Student Solutions Manual/
Study Guide
1. denotes problems most often assigned in Enhanced
WebAssign.
denotes biomedical problem
denotes guided problem
of the LC circuit, (b) the maximum charge that appears on
the capacitor, (c) the maximum current in the inductor, and
(d) the total energy the circuit possesses at t 5 3.00 s?
Section 24.1 Displacement Current and the Generalized
Form of Ampère’s Law
S
1. Consider the situation shown in
Eout
Figure P24.1. An electric field of
r
300 V/m is confined to a circular area
P
d 5 10.0 cm in diameter and directed
outward perpendicular to the plane of
d
the figure. If the field is increasing at
a rate of 20.0 V/m ? s, what are (a) the
Figure P24.1
direction and (b) the magnitude of
the magnetic field at the point P, r 5 15.0 cm from the center of the circle?
2.
3.
A 0.200-A current is charging a capacitor that has circular plates 10.0 cm in radius. If the plate separation is
4.00 mm, (a) what is the time rate of increase of electric
field between the plates? (b) What is the magnetic field between the plates 5.00 cm from the center?
A 0.100-A current is charging a capacitor that has
square plates 5.00 cm on each side. The plate separation is
4.00 mm. Find (a) the time rate of change of electric flux
between the plates and (b) the displacement current between the plates.
Section 24.2 Maxwell’s Equations and Hertz’s Discoveries
4. A 1.05-H inductor is connected in series with a variable capacitor in the tuning section of a short wave radio set. What
capacitance tunes the circuit to the signal from a transmitter broadcasting at 6.30 MHz?
5.
A proton moves through a region containing a uniform
:
electric field given by E 5 50.0 ĵ V/m and a uniform
:
magnetic field B 5 (0.200 î 1 0.300 ĵ 1 0.400k̂) T. Determine the acceleration of the proton when it has a velocity
:
v 5 200 î m/s.
6. A very long, thin rod carries electric charge with the linear
density 35.0 nC/m. It lies along the x axis and moves in the
x direction at a speed of 1.50 3 107 m/s. (a) Find the electric field the rod creates at the point (x 5 0, y 5 20.0 cm,
z 5 0). (b) Find the magnetic field it creates at the same
point. (c) Find the force exerted on an electron at this
point, moving with a velocity of (2.40 3 108) î m/s.
7.
The switch in Figure P24.7 is connected to position a for
a long time interval. At t 5 0, the switch is thrown to position b. After this time, what are (a) the frequency of oscillation
10.0 a
b
0.100 H
S
12.0 V
1.00 mF
Figure P24.7
8.
An electron moves through a uniform electric field
E 5 (2.50 î 1 5.00 ĵ ) V/m and a uniform magnetic field
:
B 5 0.400k̂ T. Determine the acceleration of the electron
when it has a velocity :
v 5 10.0 î m/s.
:
Section 24.3 Electromagnetic Waves
Note: Assume that the medium is vacuum unless specified
otherwise.
9. The speed of an electromagnetic wave traveling in a transparent nonmagnetic substance is v 5 1/ 00 , where
is the dielectric constant of the substance. Determine the
speed of light in water, which has a dielectric constant of
1.78 at optical frequencies.
10. (a) The distance to the North Star, Polaris, is approximately
6.44 3 1018 m. If Polaris were to burn out today, in what
year would we see it disappear? (b) How long does it take
for sunlight to reach the Earth? (c) How long does it take
for a microwave radar signal to travel from the Earth to
the Moon and back? (d) How long does it take for a radio
wave to travel once around the Earth in a great circle, close
to the planet’s surface? (e) How long does it take for light to
reach you from a lightning stroke 10.0 km away?
11. Review. A standing-wave pattern is set up by radio waves between two metal sheets 2.00 m apart, which is the shortest
distance between the plates that produces a standing-wave
pattern. What is the frequency of the radio waves?
12.
An electromagnetic wave in vacuum has an electric field
amplitude of 220 V/m. Calculate the amplitude of the corresponding magnetic field.
13.
Figure P24.13 shows a plane electromagnetic sinusoidal wave propagating in the x direction. Suppose the wavelength is 50.0 m and the electric field vibrates in the xy plane
846 CHAPTER 24 | Electromagnetic Waves
with an amplitude of 22.0 V/m. Calculate (a) the frequency
:
of the wave and (b) the magnetic field B when the electric
field has its maximum value in the negative y direction.
:
(c) Write an expression for B with the correct unit vector,
with numerical values for Bmax, k, and , and with its magnitude in the form
microwave signals. The beat frequency is measured. (a) For
an electromagnetic wave reflected back to its source from a
mirror approaching at speed v, show that the reflected wave
has frequency
c1v
f9 5
f
c2v
where f is the source frequency. (b) Noting that v is much
less than c, show that the beat frequency can be written
as fbeat 5 2v/. (c) What beat frequency is measured for
a car speed of 30.0 m/s if the microwaves have frequency
10.0 GHz? (d) If the beat frequency measurement in
part (c) is accurate to 65.0 Hz, how accurate is the speed
measurement?
B 5 Bmax cos (kx 2 t)
y
S
E
z
S
S
B
c
19. Review. An alien civilization occupies a planet circling a
brown dwarf, several light-years away. The plane of the planet’s orbit is perpendicular to a line from the brown dwarf to
the Sun, so the planet is at nearly a fixed position relative to
the Sun. The extraterrestrials have come to love broadcasts
of MacGyver, on television channel 2, at carrier frequency
57.0 MHz. Their line of sight to us is in the plane of the
Earth’s orbit. Find the difference between the highest and
lowest frequencies they receive due to the Earth’s orbital
motion around the Sun.
x
Figure P24.13 Problems 13 and 64.
14. Verify by substitution that the following equations are solutions to Equations 24.16 and 24.17, respectively:
E 5 Emax cos (kx 2 t)
B 5 Bmax cos (kx 2 t)
15. Review. A microwave oven is powered by a magnetron, an
electronic device that generates electromagnetic waves of
frequency 2.45 GHz. The microwaves enter the oven and
are reflected by the walls. The standing-wave pattern produced in the oven can cook food unevenly, with hot spots
in the food at antinodes and cool spots at nodes, so a turntable is often used to rotate the food and distribute the energy. If a microwave oven intended for use with a turntable
is instead used with a cooking dish in a fixed position, the
antinodes can appear as burn marks on foods such as carrot
strips or cheese. The separation distance between the burns
is measured to be 6 cm 6 5%. From these data, calculate the
speed of the microwaves.
16.
20. A light source recedes from an observer with a speed vS that
is small compared with c. (a) Show that the fractional shift
in the measured wavelength is given by the approximate
expression
vS
D
<
c
This phenomenon is known as the red shift because the visible light is shifted toward the red. (b) Spectroscopic measurements of light at 5 397 nm coming from a galaxy in
Ursa Major reveal a redshift of 20.0 nm. What is the recessional speed of the galaxy?
21. A Doppler weather radar station broadcasts a pulse of radio
waves at frequency 2.85 GHz. From a relatively small batch
of raindrops at bearing 38.68 east of north, the station receives a reflected pulse after 180 s with a frequency shifted
upward by 254 Hz. From a similar batch of raindrops at bearing 39.68 east of north, the station receives a reflected pulse
after the same time delay, with a frequency shifted downward by 254 Hz. These pulses have the highest and lowest
frequencies the station receives. (a) Calculate the radial
velocity components of both batches of raindrops. (b) Assume that these raindrops are swirling in a uniformly rotating vortex. Find the angular speed of their rotation.
In SI units, the electric field in an electromagnetic wave
is described by
Ey 5 100 sin (1.00 3 107x 2 t)
Find (a) the amplitude of the corresponding magnetic field
oscillations, (b) the wavelength , and (c) the frequency f.
17. A physicist drives through a stop light. When he is pulled over,
he tells the police officer that the Doppler shift made the red
light of wavelength 650 nm appear green to him, with a wavelength of 520 nm. The police
officer writes out a traffic citation for speeding. How fast was
the physicist traveling, according to his own testimony?
E 5 9.00 3 103 cos [(9.00 3 106)x 2 (3.00 3 1015)t]
B 5 3.00 3 1025 cos [(9.00 3 106)x 2 (3.00 3 1015)t]
© Cengage Learning/Ed Dodd
18. Police radar detects the speed
of a car (Fig. P24.18) as follows. Microwaves of a precisely
known frequency are broadcast
toward the car. The moving car
reflects the microwaves with a
Doppler shift. The reflected
waves are received and combined with an attenuated version of the transmitted wave.
Beats occur between the two
22. Why is the following situation impossible? An electromagnetic
wave travels through empty space with electric and magnetic fields described by
Figure P24.18
where all numerical values and variables are in SI units.
Section 24.4 Energy Carried by Electromagnetic Waves
23.
What is the average magnitude of the Poynting vector 5.00 mi from a radio transmitter broadcasting isotropically (equally in all directions) with an average power of
250 kW?
| Problems 847
24. In a region of free space, the electric field at an instant of
:
time is E 5 (80.0 î 1 32.0 ĵ 2 64.0k̂) N/C and the magnetic
:
field is B 5 (0.200 î 1 0.080 ĵ 1 0.290k̂) T. (a) Show that
the two fields are perpendicular to each other. (b) Determine the Poynting vector for these fields.
33.
A 15.0-mW helium–neon laser emits a beam of circular cross section with a diameter of 2.00 mm. (a) Find the
maximum electric field in the beam. (b) What total energy
is contained in a 1.00-m length of the beam? (c) Find the
momentum carried by a 1.00-m length of the beam.
25.
34.
A helium–neon laser emits a beam of circular cross section with a radius r and a power P. (a) Find the maximum
electric field in the beam. (b) What total energy is contained in a length , of the beam? (c) Find the momentum
carried by a length , of the beam.
If the intensity of sunlight at the Earth’s surface under
a fairly clear sky is 1 000 W/m2, how much electromagnetic
energy per cubic meter is contained in sunlight?
26. Consider a bright star in our night sky. Assume its distance
from the Earth is 20.0 light-years (ly) and its power output
is 4.00 3 1028 W, about 100 times that of the Sun. (a) Find
the intensity of the starlight at the Earth. (b) Find the power
of the starlight the Earth intercepts. One light-year is the
distance traveled by light through a vacuum in one year.
27.
Review. An AM radio station broadcasts isotropically
(equally in all directions) with an average power of 4.00 kW.
A receiving antenna 65.0 cm long is at a location 4.00 mi
from the transmitter. Compute the amplitude of the emf
that is induced by this signal between the ends of the receiving antenna.
28. When a high-power laser is used in the Earth’s atmosphere,
the electric field associated with the laser beam can ionize
the air, turning it into a conducting plasma that reflects the
laser light. In dry air at 08C and 1 atm, electric breakdown
occurs for fields with amplitudes above about 3.00 MV/m.
(a) What laser beam intensity will produce such a field?
(b) At this maximum intensity, what power can be delivered
in a cylindrical beam of diameter 5.00 mm?
29.
A community plans to build a facility to convert solar
radiation to electrical power. The community requires
1.00 MW of power, and the system to be installed has an efficiency of 30.0% (that is, 30.0% of the solar energy incident
on the surface is converted to useful energy that can power
the community). Assuming sunlight has a constant intensity
of 1 000 W/m2, what must be the effective area of a perfectly
absorbing surface used in such an installation?
Section 24.6 The Spectrum of Electromagnetic Waves
35.
The human eye is most sensitive to light having a wavelength of 5.50 3 1027 m, which is in the green–yellow region of the visible electromagnetic spectrum. What is the
frequency of this light?
36.
An important news announcement is transmitted by
radio waves to people sitting next to their radios 100 km
from the station and by sound waves to people sitting across
the newsroom 3.00 m from the newscaster. Taking the speed
of sound in air to be 343 m/s, who receives the news first?
Explain.
37. What are the wavelengths of electromagnetic waves in
free space that have frequencies of (a) 5.00 3 1019 Hz and
(b) 4.00 3 109 Hz?
38. Classify waves with frequencies of 2 Hz, 2 kHz, 2 MHz,
2 GHz, 2 THz, 2 PHz, 2 EHz, 2 ZHz, and 2 YHz on the electromagnetic spectrum. Classify waves with wavelengths of
2 km, 2 m, 2 mm, 2 m, 2 nm, 2 pm, 2 fm, and 2 am.
39. Review. Accelerating charges radiate electromagnetic
waves. Calculate the wavelength of radiation produced by a
proton in a cyclotron with a radius of 0.500 m and magnetic
field of 0.350 T.
40.
30. At what distance from a 100-W electromagnetic wave point
source does Emax 5 15.0 V/m?
31. The filament of an incandescent lamp has a 150-Ω resistance and carries a direct current of 1.00 A. The filament
is 8.00 cm long and 0.900 mm in radius. (a) Calculate the
Poynting vector at the surface of the filament, associated
with the static electric field producing the current and the
current’s static magnetic field. (b) Find the magnitude of
the static electric and magnetic fields at the surface of the
filament.
Section 24.5 Momentum and Radiation Pressure
32.
A possible means of space flight is to place a perfectly
reflecting aluminized sheet into orbit around the Earth and
then use the light from the Sun to push this “solar sail.” Suppose a sail of area A 5 6.00 3 105 m2 and mass m 5 6.00 3
103 kg is placed in orbit facing the Sun. Ignore all gravitational effects and assume a solar intensity of 1 370 W/m2.
(a) What force is exerted on the sail? (b) What is the sail’s
acceleration? (c) Assuming the acceleration calculated in
part (b) remains constant, find the time interval required
for the sail to reach the Moon, 3.84 3 108 m away, starting
from rest at the Earth.
Compute an order-of-magnitude estimate for the frequency of an electromagnetic wave with wavelength equal
to (a) your height and (b) the thickness of a sheet of
paper. How is each wave classified on the electromagnetic
spectrum?
41. In addition to cable and satellite broadcasts, television
stations still use VHF and UHF bands for digitally broadcasting their signals. Twelve VHF television channels
(channels 2 through 13) lie in the range of frequencies
between 54.0 MHz and 216 MHz. Each channel is assigned
a width of 6.00 MHz, with the two ranges 72.0–76.0 MHz
and 88.0–174 MHz reserved for non-TV purposes. (Channel 2, for example, lies between 54.0 and 60.0 MHz.) Calculate the broadcast wavelength range for (a) channel 4,
(b) channel 6, and (c) channel 8.
42.
A diathermy machine, used in physiotherapy, generates electromagnetic radiation that gives the effect of “deep
heat” when absorbed in tissue. One assigned frequency
for diathermy is 27.33 MHz. What is the wavelength of this
radiation?
43. Suppose you are located 180 m from a radio transmitter.
(a) How many wavelengths are you from the transmitter if
the station calls itself 1 150 AM? (The AM band frequencies
are in kilohertz.) (b) What if this station is 98.1 FM? (The
FM band frequencies are in megahertz.)
848 CHAPTER 24 | Electromagnetic Waves
Section 24.8 Context Connection: The Special Properties
of Laser Light
51.
Section 24.7 Polarization of Light Waves
45.
Plane-polarized light is incident on a single polar:
izing disk with the direction of E 0 parallel to the direction of the transmission axis. Through what angle should
the disk be rotated so that the intensity in the transmitted beam is reduced by a factor of (a) 3.00, (b) 5.00, and
(c) 10.0?
46.
In Figure P24.46, suppose the transmission axes of the
left and right polarizing disks are perpendicular to each
other. Also, let the center disk be rotated on the common
axis with an angular speed . Show that if unpolarized light
is incident on the left disk with an intensity Imax, the intensity of the beam emerging from the right disk is
1
I 5 16
Imax(1 2 cos 4t)
This result means that the intensity of the emerging beam
is modulated at a rate four times the rate of rotation of the
center disk. Suggestion: Use the trigonometric identities
cos2 5 12 (1 1 cos 2) and sin2 5 12 (1 2 cos 2)
High-power lasers in factories are used to cut through
cloth and metal (Fig. P24.51).
One such laser has a beam diameter of 1.00 mm and generates an electric field having
an amplitude of 0.700 MV/m
at the target. Find (a) the amplitude of the magnetic field
produced, (b) the intensity of
the laser, and (c) the power
delivered by the laser.
52. Figure P24.52 shows portions of the energy-level
diagrams of the helium and Figure P24.51 A laser cutting
neon atoms. An electrical dis- device mounted on a robot
charge excites the He atom arm is being used to cut
from its ground state (arbi- through a metallic plate.
trarily assigned the energy
E 1 5 0) to its excited state of 20.61 eV. The excited He atom
collides with a Ne atom in its ground state and excites this
atom to the state at 20.66 eV. Lasing action takes place for
electron transitions from E 3* to E 2 in the Ne atoms. From
the data in the figure, show that the wavelength of the red
He–Ne laser light is approximately 633 nm.
Transmission axis
u vt
Imax
20.61 eV
Lasing
18.70 eV
ENERGY
E2
I
Collision
Ground state
Figure P24.46
47.
20.66 eV
E 3*
Transmission
axis
Unpolarized light passes through two ideal Polaroid
sheets. The axis of the first is vertical, and the axis of the second is at 30.08 to the vertical. What fraction of the incident
light is transmitted?
48. Two handheld radio transceivers with dipole antennas
are separated by a large fixed distance. If the transmitting antenna is vertical, what fraction of the maximum
received power will appear in the receiving antenna when
it is inclined from the vertical by (a) 15.08, (b) 45.08, and
(c) 90.08?
49. You use a sequence of ideal polarizing filters, each with its
axis making the same angle with the axis of the previous
filter, to rotate the plane of polarization of a polarized light
beam by a total of 45.08. You wish to have an intensity reduction no larger than 10.0%. (a) How many polarizers do you
need to achieve your goal? (b) What is the angle between
adjacent polarizers?
50. Two polarizing sheets are placed together with their transmission axes crossed so that no light is transmitted. A third
sheet is inserted between them with its transmission axis at
an angle of 45.08 with respect to each of the other axes. Find
the fraction of incident unpolarized light intensity transmitted by the three-sheet combination. (Assume each polarizing sheet is ideal.)
E1
Ground state
Ne
He
Figure P24.52
53.
A neodymium–yttrium–aluminum garnet laser used
in eye surgery emits a 3.00-mJ pulse in 1.00 ns, focused to
a spot 30.0 m in diameter on the retina. (a) Find (in SI
units) the power per unit area at the retina. (In the optics
industry, this quantity is called the irradiance.) (b) What energy is delivered by the pulse to an area of molecular size,
taken as a circular area 0.600 nm in diameter?
54. A pulsed ruby laser emits light at 694.3 nm. For a 14.0-ps
pulse containing 3.00 J of energy, find (a) the physical
length of the pulse as it travels through space and (b) the
number of photons in it. (c) Assuming that the beam has a
circular cross section of 0.600 cm diameter, find the number of photons per cubic millimeter.
55. The carbon dioxide laser is one of the most powerful developed. The energy difference between the two laser levels
is 0.117 eV. Determine (a) the frequency and (b) the wavelength of the radiation emitted by this laser. (c) In what portion of the electromagnetic spectrum is this radiation?
56. Review. Figure 24.16 represents the light bouncing between
two mirrors in a laser cavity as two traveling waves. These
traveling waves moving in opposite directions constitute a
Philippe Plailly/SPL/Photo Researchers
44. A radar pulse returns to the transmitter-receiver after a total
travel time of 4.00 3 1024 s. How far away is the object that
reflected the wave?
| Problems 849
standing wave. If the reflecting surfaces are metallic films, the
electric field has nodes at both ends. The electromagnetic
standing wave is analogous to the standing string wave represented in Active Figure 14.9. (a) Assume that a helium–neon
laser has precisely flat and parallel mirrors 35.124 103 cm
apart. Assume that the active medium can efficiently amplify only light with wavelengths between 632.808 40 nm and
632.809 80 nm. Find the number of components that constitute the laser light, and the wavelength of each component,
precise to eight digits. (b) Find the root-mean-square speed
for a neon atom at 1208C. (c) Show that at this temperature
the Doppler effect for light emission by moving neon atoms
should realistically make the bandwidth of the light amplifier
larger than the 0.001 40 nm assumed in part (a).
57.
58.
A ruby laser delivers a 10.0-ns pulse of 1.00-MW average
power. If the photons have a wavelength of 694.3 nm, how
many are contained in the pulse?
atmosphere of the Earth allows 74.6% of the energy of sunlight to pass though it in clear weather. What is the power
received by a city when the space mirror is reflecting light to
it? (b) The plan is for the reflected sunlight to cover a circle
of diameter 8.00 km. What is the intensity of light (the average magnitude of the Poynting vector) received by the city?
(c) This intensity is what percentage of the vertical component of sunlight at St. Petersburg in January, when the sun
reaches an angle of 7.008 above the horizon at noon?
61. The intensity of solar radiation at the top of the Earth’s atmosphere is 1 370 W/m2. Assuming 60% of the incoming
solar energy reaches the Earth’s surface and you absorb
50% of the incident energy, make an order-of-magnitude
estimate of the amount of solar energy you absorb if you
sunbathe for 60 minutes.
62.
Review. In the absence of cable input or a satellite dish,
a television set can use a dipole-receiving antenna for VHF
channels and a loop antenna for UHF channels. In Figure CQ24.1, the “rabbit ears” form the VHF antenna and
the smaller loop of wire is the UHF antenna. The UHF
antenna produces an emf from the changing magnetic flux
through the loop. The television station broadcasts a signal with a frequency f, and the signal has an electric field
amplitude Emax and a magnetic field amplitude Bmax at the
location of the receiving antenna. (a) Using Faraday’s law,
derive an expression for the amplitude of the emf that appears in a single-turn, circular loop antenna with a radius
r that is small compared with the wavelength of the wave.
(b) If the electric field in the signal points vertically, what
orientation of the loop gives the best reception?
63.
A dish antenna having a diameter of 20.0 m receives (at
normal incidence) a radio signal from a distant source as
shown in Figure P24.63. The radio signal is a continuous sinusoidal wave with amplitude
Emax 5 0.200 V/m. Assume
the antenna absorbs all the radiation that falls on the dish.
(a) What is the amplitude
of the magnetic field in this
wave? (b) What is the intensity of the radiation received
by this antenna? (c) What
is the power received by the
antenna? (d) What force is
exerted by the radio waves on
Figure P24.63
the antenna?
64.
You may wish to review Section 13.5 on the transport
of energy by sinusoidal waves on strings. Figure P24.13 is a
graphical representation of an electromagnetic wave moving in the x direction. We wish to find an expression for the
intensity of this wave by means of a different process from
that by which Equation 24.26 was generated. (a) Sketch a
graph of the electric field in this wave at the instant t 5 0,
letting your flat paper represent the xy plane. (b) Compute
the energy density uE in the electric field as a function of x
at the instant t 5 0. (c) Compute the energy density in the
magnetic field uB as a function of x at that instant. (d) Find
the total energy density u as a function of x, expressed
in terms of only the electric field amplitude. (e) The energy in a “shoebox” of length and frontal area A is E 5
e0 uA dx. (The symbol E for energy in a wavelength imitates
The number N of atoms in a particular state is called
the population of that state. This number depends on the
energy of that state and the temperature. In thermal equilibrium, the population of atoms in a state of energy En is
given by a Boltzmann distribution expression
N 5 Ng e 2(En 2Eg)/k BT
ENERGY
where Ng is the population of the ground state of energy Eg ,
kB is Boltzmann’s constant, and T is the absolute temperature.
For simplicity, assume each energy level has only one quantum
state associated with it. (a) Before the power is switched on,
the neon atoms in a laser
E 4*
20.66 eV
are in thermal equilibrium
Red
at 27.08C. Find the equilibGreen
E3
18.70 eV
rium ratio of the populaE2
tions of the states E4* and
E3 shown for the red transition in Figure P24.58.
Lasers operate by a clever
Ground state
E1
0
artificial production of a
“population inversion” beFigure P24.58
tween the upper and lower
atomic energy states involved in the lasing transition. This
term means that more atoms are in the upper excited state
than in the lower one. Consider the E4* 2 E3 transition in Figure P24.58. Assume 2% more atoms occur in the upper state
than in the lower. (b) To demonstrate how unnatural such a
situation is, find the temperature for which the Boltzmann
distribution describes a 2.00% population inversion. (c) Why
does such a situation not occur naturally?
Additional Problems
59. Assume the intensity of solar radiation incident on the
cloud tops of the Earth is 1 370 W/m2. (a) Taking the average Earth–Sun separation to be 1.496 3 1011 m, calculate
the total power radiated by the Sun. Determine the maximum values of (b) the electric field and (c) the magnetic
field in the sunlight at the Earth’s location.
60. One goal of the Russian space program is to illuminate dark
northern cities with sunlight reflected to the Earth from a
200-m diameter mirrored surface in orbit. Several smaller
prototypes have already been constructed and put into
orbit. (a) Assume that sunlight with intensity 1 370 W/m2
falls on the mirror nearly perpendicularly and that the
850 CHAPTER 24 | Electromagnetic Waves
the notation of Section 13.5.) Perform the integration to
compute the amount of this energy in terms of A, , Emax, and
universal constants. (f) We may think of the energy transport
by the whole wave as a series of these shoeboxes going past as
if carried on a conveyor belt. Each shoebox passes by a point
in a time interval defined as the period T 5 1/f of the wave.
Find the power the wave carries through area A. (g) The intensity of the wave is the power per unit area through which
the wave passes. Compute this intensity in terms of Emax and
universal constants. (h) Explain how your result compares
with that given in Equation 24.26.
65. Consider a small, spherical particle of radius r located in
space a distance R 5 3.75 3 1011 m from the Sun. Assume
the particle has a perfectly absorbing surface and a mass
density of 5 1.50 g/cm3. Use S 5 214 W/m2 as the value
of the solar intensity at the location of the particle. Calculate the value of r for which the particle is in equilibrium
between the gravitational force and the force exerted by
solar radiation.
66.
Consider a small, spherical particle of radius r located
in space a distance R from the Sun, of mass MS. Assume the
particle has a perfectly absorbing surface and a mass density
. The value of the solar intensity at the particle’s location is
S. Calculate the value of r for which the particle is in equilibrium between the gravitational force and the force exerted
by solar radiation. Your answer should be in terms of S, R, ,
and other constants.
67. Review. A 1.00-m-diameter circular mirror focuses the Sun’s
rays onto a circular absorbing plate 2.00 cm in radius, which
holds a can containing 1.00 L of water at 20.08C. (a) If the
solar intensity is 1.00 kW/m2, what is the intensity on the
absorbing plate? At the plate, what are the maximum mag:
:
nitudes of the fields (b) E and (c) B? (d) If 40.0% of the
energy is absorbed, what time interval is required to bring
the water to its boiling point?
(For simplicity, ignore the cats’ absorption of radiation
from the environment.)
70.
Review. Gliese 581c is the first Earth-like extrasolar
terrestrial planet discovered. Its parent star, Gliese 581, is
a red dwarf that radiates electromagnetic waves with power
5.00 3 1024 W, which is only 1.30% of the power of the Sun.
Assume the emissivity of the planet is equal for infrared
and for visible light and the planet has a uniform surface
temperature. Identify (a) the projected area over which the
planet absorbs light from Gliese 581 and (b) the radiating
area of the planet. (c) If an average temperature of 287 K
is necessary for life to exist on Gliese 581c, what should the
radius of the planet’s orbit be?
71. Review. (a) A homeowner has a solar water heater installed
on the roof of his house (Fig. P24.71). The heater is a flat,
closed box with excellent thermal insulation. Its interior
is painted black, and its front face is made of insulating
glass. Its emissivity for visible light is 0.900, and its emissivity for infrared light is 0.700. Light from the noontime Sun
is incident perpendicular to the glass with an intensity of
1 000 W/m2, and no water enters or leaves the box. Find
the steady-state temperature of the box’s interior. (b) What
If? The homeowner builds an identical box with no water
tubes. It lies flat on the ground in front of the house. He
uses it as a cold frame, where he plants seeds in early spring.
Assuming the same noontime Sun is at an elevation angle of
50.08, find the steady-state temperature of the interior of the
box when its ventilation slots are tightly closed.
Review problems. Section 17.10 discussed electromagnetic
radiation as a mode of energy transfer. Problems 69 through
71 use ideas introduced both there and in this chapter.
69. Review. A 5.50-kg black cat and her four black kittens, each
with mass 0.800 kg, sleep snuggled together on a mat on
a cool night, with their bodies forming a hemisphere. Assume the hemisphere has a surface temperature of 31.08C,
an emissivity of 0.970, and a uniform density of 990 kg/m3.
Find (a) the radius of the hemisphere, (b) the area of its
curved surface, (c) the radiated power emitted by the cats
at their curved surface and, (d) the intensity of radiation at
this surface. You may think of the emitted electromagnetic
wave as having a single predominant frequency. Find (e) the
amplitude of the electric field in the electromagnetic wave
just outside the surface of the cozy pile and (f) the amplitude of the magnetic field. (g) What If? The next night, the
kittens all sleep alone, curling up into separate hemispheres
like their mother. Find the total radiated power of the family.
easyshoot/Shutterstock.com
68. In 1965, Arno Penzias and Robert Wilson discovered the
cosmic microwave radiation left over from the big bang expansion of the Universe. Suppose the energy density of this
background radiation is 4.00 3 10214 J/m3. Determine the
corresponding electric field amplitude.
Figure P24.71
72. A microwave source produces pulses of 20.0-GHz radiation,
with each pulse lasting 1.00 ns. A parabolic reflector with a
face area of radius 6.00 cm
is used to focus the microwaves into a parallel beam
12.0 cm
of radiation as shown in
Figure P24.72. The average power during each
Figure P24.72
pulse is 25.0 kW. (a) What
is the wavelength of these microwaves? (b) What is the total
energy contained in each pulse? (c) Compute the average
energy density inside each pulse. (d) Determine the amplitude of the electric and magnetic fields in these microwaves.
(e) Assuming that this pulsed beam strikes an absorbing
surface, compute the force exerted on the surface during
the 1.00-ns duration of each pulse.
| Problems 851
73. A linearly polarized microwave of wavelength 1.50 cm is directed along the positive x axis. The electric field vector has
a maximum value of 175 V/m and vibrates in the xy plane.
Assuming the magnetic field component of the wave can
be written in the form B 5 Bmax sin (kx 2 t), give values
for (a) Bmax, (b) k, and (c) . (d) Determine in which plane
the magnetic field vector vibrates. (e) Calculate the average
value of the Poynting vector for this wave. (f) If this wave
were directed at normal incidence onto a perfectly reflecting sheet, what radiation pressure would it exert? (g) What
acceleration would be imparted to a 500-g sheet (perfectly
reflecting and at normal incidence) with dimensions of
1.00 m 3 0.750 m?
74. The electromagnetic power radiated by a non-relativistic
particle with charge q moving with acceleration a is
P5
q 2a 2
60c 3
where 0 is the permittivity of free space (also called the permittivity of vacuum) and c is the speed of light in vacuum.
(a) Show that the right side of this equation has units of
watts. An electron is placed in a constant electric field of
magnitude 100 N/C. Determine (b) the acceleration of the
electron and (c) the electromagnetic power radiated by this
electron. (d) What If? If a proton is placed in a cyclotron
with a radius of 0.500 m and a magnetic field of magnitude
0.350 T, what electromagnetic power does this proton radiate just before leaving the cyclotron?
75. Review. An astronaut, stranded in space 10.0 m from her
spacecraft and at rest relative to it, has a mass (including
equipment) of 110 kg. Because she has a 100-W flashlight
that forms a directed beam, she considers using the beam
as a photon rocket to propel herself continuously toward
the spacecraft. (a) Calculate the time interval required for
her to reach the spacecraft by this method. (b) What If?
Suppose she throws the 3.00-kg flashlight in the direction
away from the spacecraft instead. After being thrown, the
flashlight moves at 12.0 m/s relative to the recoiling astronaut. After what time interval will the astronaut reach the
spacecraft?