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Chapter 21
ELECTRICAL PROPERTIES OF MATTER
GOALS
When you have mastered the contents of this chapter, you will be able to achieve the
following goals:
Definitions
Define each of the following terms, and use it in an operational definition:
dielectric constant
equipotential surfaces
electrical field
dipole
potential gradient
capacitance of a capacitor
potential difference
Coulomb's Law
Apply the basic model of an electrostatic field, and use Coulomb's law to calculate the
force on one, two, or three given point charges.
Potential Gradient
Apply the gradient to electrical field phenomena.
Moving Charged Particles
Explain the motion of charged particles in an electric field.
Capacitance
Solve capacitance and capacitor problems, including the use of a capacitor as a means of
storing electrical energy.
Applications of Electrostatics
List a number of applications of electrostatic principles to daily living and to medical
equipment.
PREREQUISITES
Before beginning this chapter you should have achieved the goals of Chapter 2,
Unifying Approaches, Chapter 4, Forces and Newton's Laws, Chapter 5, Energy, and
Chapter 9, Transport Phenomena.
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Chapter 21
ELECTRICAL PROPERTIES OF MATTER
21.1 Introduction
You have observed and experienced many phenomena that are examples of the
electrical nature of matter. During the winter you have experienced a shock by walking
over a rug and then touching a metal object or by sliding across your automobile seat
and touching the door handle. You have seen bolts of lightning. Clothes you take out of
a clothes dryer often cling together due to static electricity. The focusing and imaging
system of a television picture tube makes use of the force on electrons moving in an
electric field. Your heart is an electric device that keeps its steady beat with electric
synchronization.
These examples all involve the electrical nature of matter and electric interactions. In
this chapter, the electric field model will be introduced and used to explain electric
phenomena.
21.2 Electrical Forces
A dry glass rod after being rubbed with silk will pick up small bits of paper. A dry
rubber rod, after being rubbed with cat's fur, will pick up small bits of paper and will
attract the rubbed glass rod. An analysis of many experiments such as these has lead us
to postulate that there are two kinds of electricity which we call positive and negative
(Figure 21.1). Conventionally, positive electricity is defined as the electricity that
appears on a glass rod when it is rubbed with silk, and negative electricity is defined as
the electricity that appears on a hard rubber rod rubbed with cat's fur. If an object has
equal amounts of positive electricity and negative electricity, it is said to be neutral. If
an object has an excess of negative electricity, it is said to be negatively charged. If an
object has a deficiency of negative electricity, it is said to be positively charged.
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In 1909 Robert A. Millikan performed a series of experiments on charged oil drops.
He reported that these drops always contained an amount of electrical charge that was
an integer times a fundamental constant:
total charge = Ne
(21.1)
where N is an integer, 0,ñ1,ñ2,ñ3,...;and e is the magnitude of fundamental charge,
which we call the electronic charge. The best measurements of the size of the electronic
charge now indicates that the unit of electric charge in the SI units, which is called a
coulomb (C), is equal to a very large number of electron charges.
1 coulomb = 1 C = 6,241,450,300,000,000,000 e
1 C = 6.24 x 1018e
(21.2)
or
e = 1.60 x 10-19 C
(21.3)
In the eighteenth century, experiments like those illustrated in Figure 21.2 had
shown that like charges repel each other and unlike charges attract each other. The
inverse square law for electrical forces was verified by the experiments of Charles
Coulomb. According to Coulomb's law of electrical interaction, the magnitude of the
electrical force between two point charges is proportional to the product of the charges
and inversely proportional to the square of the distance between the charges,
(21.4)
where q1 and q2 are the amounts of electric charge of the two point charges and where r
is the distance between the point charges. This expression may be written as a vector
equation as follows:
(21.5)
where F12 is the force acting on charge 2 because of the presence of charge q1 and
is a
vector of unit magnitude which points in the direction from charge 1 to charge 2
(Figure 21.3). If q1 andq2 are of like sign, Equation 21.5 has the force between the charge
acting so as to push the two charges apart (repulsion). See Figure 21.3a and b. If the
charges are of opposite sign, the force between the charges acts to push the charges
together (attraction) as in Figure 21.3c and d. The value of the proportionality constant k
which appears in Equation 21.5 depends upon the system of units that is being used
and upon the medium in which the two charges are located. If the point charges are
located in a vacuum and the force is measured in Newtons, the charges in Coulombs
and the distance between the charges in meters, then k has the following value and
units:
(21.6)
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where ε is the permittivity of free space and has the numerical value of 8.85 x 10-12 C2/Nm2. For problems in this book you may use the value of k at 9.00 x 109 N- m2/C2 in SI
units. If charges are located in a medium that has a dielectric constant ε, then Equation
21.5 must be replaced by the following equation,
0
(21.7)
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Figure 21.3 Directions and magnitudes of forces between charges for repulsion and attraction. The dielectric constant accounts for the contribution of the molecular charges in the
medium in the determination of the electric field inside such materials. For many gases
the charge distribution of the molecules is symmetrical around the center of the
molecule. For gases with this spherical charge symmetry, e is about 1. Materials which
show a charge distribution that is not symmetric but is instead characterized by the
separation of its negative and positive charge centers, are called polar materials. Polar
molecules are electrically neutral but their nonsymmetric charge distribution results in
dielectric constants that are much greater than 1. A simple example of such a
distribution would be equal volumes of positive and negative spherical charges with
their centers slightly displaced from each other.
Since electrical charges interact with each other across some distance, they represent
an example of systems that interact-at-a-distance. An electric charge moving on one side
of the room seems to cause changes in the motion of an electrical charge on the other
side of the room. We then construct the model of an electric force field that fills all the
space in the vicinity of an electrical charge. Since there are electrical charges in all
molecules, the universe is more or less filled with electric fields. We construct electric
fields that obey linear relationships. We can use the principle of superposition to solve
problems that involve the interaction-at-a-distance of many different charges. In such
cases we solve the problem for each individual charge as if the others did not exist, and
then add up all of the individual results. Hence, in any region where an electric force is
acting upon an electric charge at rest, we say there exists an electric field. The direction
of the field is defined as the direction of the force that would act upon a positive charge
at that location in space. The magnitude of the field at that point is given by the force
divided by the charge. The electric field E at a point is given by
E = F/q
(21.8)
where F is the force acting on a charge q at that location. The electric field is a vector
quantity defined as the force upon a unit charge. The SI units for E are newtons per
coulomb. In the region where there is a single positive point charge q, the electric field is
given by,
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(21.9)
where is a vector of unit magnitude that points in a radial direction away from the
charge q. If there are several point charges in the region, the resultant electric field is
the vector sum of the separate fields resulting from each individual charge.
Figure 21.4 Determination of electric fields. EXAMPLE
In the diagram shown in Figure 21.4, determine the field at B resulting from a charge of
+1.6 x 10-12 C at A, 4 cm north of B, and a charge of -1.8 x 10-12 C 3 cm east of B.
The field at B resulting from the charge at A is a repulsive force whose magnitude is
given by,
and which is directed south. The field at B resulting from the charge at C is an attractive
field whose magnitude is given by,
directed to the east.
tan θ = 9/18 = 1/2
θ = 26.6° south of east.
A charge of +3 C located at point B would experience a force of 60.3 N acting 26.6°
south of east. A charge of -2 C at point B would experience a force of 40.2 N acting 26.6°
north of west.
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21.3 Electrical Forces Acting on Moving Charges
Now consider the behavior of a charged particle free to move in an electric field. We
first investigate the case of a charged body in a uniform electric field. Suppose we have
chargeq in a field of E. The force acting on the charge is q times E,
F = qE
(21.10)
in the direction of E. In an earlier chapter on Newton's laws (Chapter 4) we learned that
a constant force acting on a free body produces an acceleration given by Newton's
second law,
a = F/m =qE/m
(21.11)
where, in this case, the force is given by qE. We can apply the equations for uniformly
accelerated motion to this situation. So, if we start from rest at time t = 0, at any later
time t the velocity is given by the acceleration multiplied by the time,
v = at = (qE/m)t
(21.12)
The displacement y is given by one-half the acceleration times the square of the time,
y = 1/2 at2 = 1/2 (qE/m)t2
(21.13)
The square of the velocity is proportional to the product of the magnitudes of the field
and the displacement,
v2 = 2 | a | | y | = 2q | E | | y |/m
(21 .14)
The kinetic energy acquired after moving a distance y is given by the work done in
moving the charge, or the product of the force times the distance,
KE = 1/2 mv2 = q | E | | y |
(21.15)
Thus, we see that the motion of charged bodies can be influenced by the application of
electric fields. There are many applications of this principle.
EXAMPLES
1. An instrument that makes use of this phenomenon is the oscilloscope. Consider the
case of a beam of particles of charge -e moving with a velocity v in a horizontal
direction. They enter a vertical electric field E in the upward direction. What
happens to the particles? What is the shape of their path in the electric field?
2. Electrophoresis is a process that is used to separate different types of molecules. It is
very useful in separating proteins and amino acids for further analysis. The
electrophoresis process is based on the different motions of charged particles (ions)
in certain solvents (called buffers) under the influence of applied electric fields.
Different molecular constituents in the solution migrate at different speeds. For
example, when blood is subjected to this process, serum albumin and three globular
proteins are separated according to their speed of migration in the applied electric
field. What properties of the constituents would be involved in determining the
speed of migration in an electric field?
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21.4 Electrical Potential
Consider two points, A and B, in an electric field. A test charge q is moved from A to B,
and we measure the work done in moving q from A to B without any acceleration. The
electric potential difference between A and B, VAB, is defined as the work done per unit
charge. So,
VAB = VB - VA = W / Q or W =qVAB
(21.16)
W may be positive, zero, or negative. If W is positive, B is said to be at a higher potential
than A, and an external agent is required to move a positive charge q from A to B
(Figure 21.5a) . If W is 0, A and B are at the same potential (moving a charge at right
angles to the electric field lines, for example, as in Figure 21.5b. If W is negative, B is
said to be at a lower potential than A and the work is done by the electric field (moving
a positive charge from A to a point B closer to a negative charge as in Figure 21.5c. In
the SI units, W is in joules, q in coulombs, and the potential is expressed in joules per
coulomb, which is called volts (V). One volt equals an energy of one joule expended on
one coulomb. The difference in potential between the two points is given by Equation
21.16. If one selects point A as a reference point and assigns it a value of zero potential,
then any other potential, such as that at any point P, is given by
Vp = Wp /q
(21.16)
whereWp represents the work required to take charge q from A to the point P under
consideration. Usually the potential is taken to be zero at some point an infinite distance
away.
An electric charge of positive q will gain kinetic energy if it is released at an electric
potential V and moves to zero potential. The work done on the charge is W = Vq . Since
work is done on the charge, it will gain kinetic energy,
KE = 1/2 mv2 = qV
(21.17)
The velocity of a charged particle, which starts from rest and undergoes a change of
electric potential V is given by
(21.18)
Since the electric field was developed as a linear model for interaction-at-a-distance,
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the principle of superposition holds for potential, and you can find the resultant
potential at a given point due to the separate individual charges by adding the
individual potentials,
Vtotal = V1 + V2 + V3 ... + Vn
It can be shown that the potential due to a point charge at a distance r from the charge q
is given by
V = k(q / r)
(21.19a)
(See Equation 21.40.) Hence the potential at a given point produced by a number of
point charges is given by
V = k (q1 / r1 + q2 / r2 +q3 / r3 ... + qn / rn )
(21.19b)
EXAMPLE
Find the potential at B resulting from a charge of + 1.6 x 10-12 C at A, 4 cm north of B,
and a charge of - 1.8 x 10-12 C at C, 3 cm east of B. See Figure 21.4. The potential at B due
to the charge A is
VA = kqA / rA = 9 x 109 x 1.6 x 10-12 / 4 x 10-2 = 0.36 V
The potential at B due to charge C is
VC = kqC /rC = 9 x 109( -1.8 x 10-12 / 3 x 10-2) = -0.54 V
V at B = VA + VC = 0.36 + (-0.54) = -0.18 V where the potential at infinity is taken as zero.
If no work is required to move a charge from one point to another, the two points
are at the same potential and are said to be on an equi-potential surface. Let us consider
the case of two large, parallel, flat conducting plates separated by a distance d with
plate A charged positively, and plate B charged negatively. The field is directed from
plate A to plate B, perpendicular to the plates, and is uniform (Figure 21.6). The work
required to take a positive charge q from plate B to plate A is
w = FD = + (-qEd)
(21.20)
because the force F must be equal but opposite to the electric field qE in order to move
the charge from plate B to plate A. We can also express the work in terms of the
potential difference between plates B and A,
w = (VA -VB)q
(21.21)
We can set Equations 21.20 and 21.21 equal to each other and solve for E
E = - (VA -VB )/ d
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The magnitude of the electric field E is the ratio of the change in potential (VA -VB) to
the change in distance d. In other words, the electric field is a potential gradient, and the
negative sign indicates that the electric field is oppositely directed to the potential
gradient (that is, E is directed from high to low potential). In the case of parallel plates
one notes that the direction of the electric field is perpendicular to the equipotential
surfaces. This result was derived for a special case; however, it can be shown that the
electric field is always equal to the negative of the potential gradient:
E = - ΔV / Δs (directed ⊥ to surfaces of constant potential)
(21.23)
where ΔV is the potential difference over the distance Δs. Also, the direction of the
electric field is always perpendicular to the equipotential surfaces, directed from high to
low potential.
Questions
Show that the electric field must always be perpendicular to the equipotential surfaces
by assuming that E is not perpendicular to an equipotential surface and showing that
this leads to a contradiction for an equipotential surface.
21.5 Gauss' Law
Another formulation of the inverse square law for electric fields due to point charges is
known as Gauss' law. First we must define the electric flux φE. To compute the electric
flux through some small area ΔA, multiply the component of the electric field that is
perpendicular to that area times the area ΔA; that is, ΔφE =E⊥ΔA(Figure 21.7). For a
closed surface, Gauss' law states that the total flux passing through the area of that
closed surface will be equal to the total electric charge enclosed divided by the
permittivity of free space, ε0,
φ =qtotal/ε
ε
0
(21.24)
where φ is the total flux through the surface and qtotal is the total charge enclosed by the
surface. The total flux, the sum of all the small fluxes ΔφE that pass through all the small
areas ΔA needed to make up the total enclosed surface,
ε
Therefore
(21.25)
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EXAMPLES
1. Consider an isolated point charge. We know by symmetry that the electric field must
be radial and depend only upon the distance R from the point charge. So we choose a
spherical surface which has an area 4πR2 (Figure 21.8). Thus the electric flux is
φE =E(4πR2)
and
φE =E(4πR2) = q/ε
0
from Gauss' law. Solving for the value of the electric field we find:
(directed radially outward)
(21.26)
This is Coulomb's law for a point charge q. Gauss' law is primarily useful when
dealing with problems where we know from symmetry that the electric field is constant
over a properly chosen surface around the given charge distribution. Gauss' law shows
that the field outside any spherical distribution of charge is identical to that of a single
point charge q total located at the center of the spherical distribution.
2. Let us apply Gauss' law to a long cylinder of positive charge neglecting end effects.
(This is usually stated by saying the length is much greater than the radius of cylindrical
charge.)
Again the symmetry of the problem is such that we know there can only be an
electric field pointing radially outward from the center of the cylinder. Therefore, if we
pick a cylindrical surface surrounding the line charge (see Figure 21.9), the field is
perpendicular to the surface and constant over the whole cylindrical area. Then from
Gauss' law we get:
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φE = E (area of cylindrical surface) = qtotal /ε.
E = (2πRL) =qtotal / ε
0
(21.27)
where qtotal/L is the linear charge density inside the cylinder. We see that a long line of
charge produces a field that is proportional to 1/R as compared with 1/R2 for point or
uniformly spherically distributed charges.
3. In the case of electrostatic equilibrium the electric field inside a conductor must be
zero. If this were not the case, the free charge of the conductor would move under the
influence of the electric field. Consequently, conductors are equipotential surfaces for
electrostatic phenomena. We can use Gauss' law to show that the field inside an empty
cavity in a conductor must be zero. Imagine a closed surface inside the cavity. From
Gauss' law it follows that the field everywhere on the surface must be zero since there is
no charge inside the surface. Thus, everywhere inside the cavity the electric field is zero.
This result is completely independent of any charge on or outside the conductor. Such a
space becomes an electrostatic shield, as motion of charge outside the conductor will
have no affect on apparatus inside the cavity (Figure 21.10).
Figure 21.10 The field inside a hollow conductor is zero everywhere, and thus this space is shielded from effects due to charges outside of the conductor. This is the basis of electrostatic shielding that is used for many biological experimental set ups. 21.6 Electric Dipoles
Recall that polar materials result from neutral charge distributions that have their
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positive and negative charge centers separated by a distance we will call s. Such charge
distributions are called electric dipoles (Figure 21.11). The product of the charge and the
distance between charges qs gives the magnitude of the electric dipole and its vectors
directed from the negative charge to the positive charge. It can be shown that the
electric field due to a dipole is proportional to 1/r 3 where r is the distance from the
dipole to the field point and r >> s. Substances with large molecular electric dipoles
have high dielectric constants, for example, for water, ε = 80.36 at 20°C.
21.7 Electrostatic Applications
The basic principles of electrostatic phenomena given above are sufficient to provide
qualitative understanding of the electrical activity of the heart. One can ignore the
electrical conductivity of the body and use Coulomb's law in order to show that the
electrocardiogram (ECG) measures the potential distribution of a set of dipoles within
the heart. See Footnote
The dielectric constant of a material is a measure of the internal electric field for the
material. An external electric field applied to a material with a high dielectric constant
aligns the molecular dipoles inside the material producing a high internal electric field.
The external field does work on the molecular dipoles, and this work is stored as
potential energy of orientation within the material. The potential energy is a function of
temperature. Can you propose a model to predict its qualitative temperature
dependence? When the external field is removed, the molecular dipoles will gradually
return to random orientations (what effect will temperature have on this process?) and
produce very small depolarization currents in the materials.
The dielectric constant of water is ~ 80. This high value is typical of biological cells
and tissues. The interactions between such dielectric materials and externally applied
fields are important areas of study in biophysics.
The effectiveness of microwave ovens is a result of the resonant absorption of
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electromagnetic energy by molecular dipoles in the food being cooked in the oven. The
frequency of these resonances is in the microwave region of the electromagnetic
spectrum. Recent research shows that the mapping of absorption of microwaves by
human tissue may be very useful in early detection of cancer.
21.8 Capacitance
Electrical energy can be stored in a device called a capacitor. A capacitor can be made of
two conducting plates separated by a nonconductor (dielectric). In an uncharged
condition both plates are neutral. In a charged condition one plate is positively charged,
and the other is negatively charged; that is, the two plates are at different potentials.
The ratio of the charge on plates to the potential difference between the plates is defined
as the capacitance.
C=q/V
(21.28)
where q is the charge in coulombs, V is the potential in volts, and C is the capacitance in
farads. The farad is a very large unit of capacitance, and the capacitance of typical
devices is given in micro (10- 6) farads or pico (10-12) farads.
The capacitance of a capacitor depends upon the area of the plates, the thickness, the
properties of the dielectric, and the geometric configuration. For a parallel plate
capacitor
C = εε A /d
(21.29)
0
where C is in farads, ε is the dielectric constant, A is the area in square meters, d is the
thickness in meters, and ε is the permittivity of free space (8.85 x 10-12 F/m). Note that
the charge stored is the dielectric constant for a constant applied voltage.
0
EXAMPLE
What is the capacitance of an air capacitor whose parallel plates have an area of 1.00
cm2 and are spaced 1.00 mm apart?
εair = 1.00
C = 1 x 1.00 x 10-4 x 8.85 x 10-12/1.00 x 10-3
= 8.85 x 10-13 F = 0.885 pF
21.9 Combinations of Capacitors
In a variety of applications it is desirable to use a combination of capacitors. We can
connect two charged capacitors together in two ways. We can connect the two
oppositely charged plates or the two similarly charged plates.
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When the unlike plates of charged capacitors are connected together, the
arrangement is called a series connection (Figure 21.12). Charges will move from one
capacitor to the other until the positive charge on one plate is equal to the negative
charge on the plate connected to it,
q1 = q2 = q3 = Q = total charge in the combination
(21.30)
In this configuration the potential across the total combination is the sum of the
individual potentials,
V = V1 + V2 + V3
(21.31)
but the potential across a capacitor is defined as the ratio of the charge to the
capacitance. We can make use of that definition to derive an expression for the effective
capacitance of the three used in series,
(21.32)
By making use of Equation 21.30, we find that the effective capacitance C of a series
configuration of capacitors is calculated by taking the reciprocal of the sum of the
reciprocals of the individual capacitances. For three capacitors,C1, C2, andC3,
1/C = 1/C1 + 1/C2 + 1/C3
(21.33)
The configuration in which the like plates of charged capacitors are connected
together is called a parallel connection (Figure 21.13). In this configuration the charges
rearrange themselves until the potential V is the same across all of the capacitors
V = V1 = V2 = V3
(21.34)
and the total charge in the combination is the sum of the individual charges,
Q = q1 + q2 + q3
(21.35)
We can use the definition of the charge on a capacitor, CV, to derive an equation for the
effective capacitance for a number of capacitors connected in parallel,
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CV = C1V1 +C2V2 +C3V3 = (C1 + C2 +C3)V
C = C1 + C2 + C3
(21.36)
The effective capacitance of a number of capacitors connected in parallel is the sum of
the individual capacitances.
The charged capacitor has the ability to do work, and thus possesses energy. The
energy, which is stored in the dielectric, can be dissipated by connecting the two plates
by conductor. If we do so a redistribution of charges occurs. The energy stored in a
charged capacitor can be calculated by finding the increment of work required to charge
a capacitor from zero to a small charge Δq,
Δw = VΔq
(21.37)
So the total energy stored in the capacitor will be equal to the total work required to
build up the charge to Q,
w = Vave Q
where
(21.38)
.
(21.39)
where C is in farads, Q is in coulombs, V is in volts, and w is in joules.
The capacitor is an important component of many electrical instruments and
appliances. One example is the direct current defibrillator. Defibrillation is the
application of an electric shock to the heart to stop the rapid, uncoordinated
contractions of the heart muscle called fibrillation. In a defibrillator, the capacitor is
usually of the order of 10 to 20mF, the charging potential may be of the order of a few
thousand volts, and the energy delivered to the patient may be as much as 400 J. A
capacitor is also used in the electrical circuit of an electrocardiograph. The capacitor is
used to block any direct current component of the ECG signal. A third application is in
a heart pacemaker. A pacemaker is an electronic device which provides a regular
periodic electrical stimulus to the heart to make regular the rhythmic performance of
the heart. The energy requirement for a pacemaker is many orders of magnitude
smaller than that for a defibrillator.
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ENRICHMENT
21.10 Calculus Derivations of Electrostatic Relationships
In Chapter 5 on work and energy you learned
that work is given by
Δw = F cosθΔ s or
where ds is the incremental element of
displacement. Let us apply this to an electric
field. The field is radial from a charge q (Figure
21.14). The line ab represents some arbitrary path
between these two points. Consider some
element of path where E makes an angle with the
path. The work done to take unit charge along
length ds is then E cosθ ds = dw. The integral of
the portion
is the integral along the line from a to b. For the field in the vicinity of a point charge,
ds cos θ = dr
where θ is the angle between ds and dr .
Hence the work is equal to the integral of electric intensity along the line ab,
w = kq (1/rb -1/ra) =Vb -Va for a unit charge
where
Va = kq /ra
Vb = kq /rb
(21.40)
If we choose ra = ∞ as reference point and Va = 0, then in general the potential due to a
point q of a point r meters away is
(21.41)
To show that the work done in moving a unit charge around any closed path in an
electrostatic field is zero, we proceed as follows: If a point charge q' is moved in an
electric field, the force is q'E. The work for moving q' a distance ds is then
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dw = (q'E cos θ) (ds)
Again,
but ds cosθ = dr. Then,
where ra is the starting and ending point of the loop.
In charging a capacitor, charge is transferred from a lower potential to a higher
potential. This requires an expenditure of energy. The work done to transfer charge dq
is dw = V Adq. But,V = q/C; so
Because Q = CV,
This result was introduced in Section 21.9 in an intuitive way.
SUMMARY
Use these questions to evaluate how well you have achieved the goals of this chapter.
The answers to these questions are given at the end of the summary with the number of
the section where you can find related content material.
Definitions
1. The electric field model for electrostatics is analogous to the gravitational field model.
Give the analogous electrostatic term for the given gravitational term:
source of field: mass, ___________ distance dependence: 1/r2, ___________.
2. The dielectric constant for a material is defined by which of the following ratios (Ee =
external electric field in free space, Ei = internal electric field)
a. Ee/Ei
b. Ei/Ee
c. qi/qe
d. qe/qi
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3. The relationship between the potential gradient and electric field can be written as
gradient of V(r) equals
a. energy
b. E
c. -E
d. qtotal
e. none of these
4. Equipotential surfaces are satisfied by the following
a. conductors
b. E = 0 surfaces
c. potential gradient = 0
d. charges at rest
e. all of these
5. The unit for potential differences is the volt; this is equivalent to:
a. joule- second
b. watt- second
c. joule/second
d. joule/coulomb
e. joule- coulomb
6. The electric dipole consists of a neutral charge distribution with a separation between
the centers of the positive and negative charge. The electric field of the dipole is
proportional to (s = charge separation)
a. s2
b. 1/s2
c. s
d. 1/s3
e. 1/s
7. The capacitor is a device that stores energy in its electric field. This energy depends
on the capacitor's
a. geometry
b. applied voltage
c. total charge
d. dielectric
e. all of these
8. Capacitance has units of farads. One farad is equal to
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a. volt/coulomb
b. coulomb/volt
c. volt x coulomb
d. volt/meter
e. none of these
9. By Coulomb's law the force on a unit positive charge half way between identical
charges q (in terms of the force F due to one charge) is
a. 2F
b. F
c. zero
d. F/2
e. 4F
10. The magnitude of the force on a unit positive charge half way between +q and -q
charge separated by a distance 2d is given as:
a. k(q/d2)
b. 2(kq/d2)
c. kq/2d2
d. zero
11. The direction of the force in question 10 is toward
a. +q
b. origin
c. -q
d. undetermined
Potential Gradient
12. The constant electric field in a region is known to be 10 N/C. The potential
difference between two points 0.5 m apart is
a. 10 V
b. 5 V
c. 20 V
d. 0.05 V
e. cannot tell from these data
13. The direction of the electric field is opposite the direction of the potential gradient
where the potential gradient is:
a. minimum
b. zero
c. maximum
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d. unknown
e. any value
Moving Charged Particles
14. The energy gained by a particle of charge +q and mass m accelerating through a
potential difference V is given by
a. qV
b. (qV)1/2
c. 2qV1/2/m
d. V/q
e. q/m
15. The energy gained by a charged particle passing through a potential difference V is
independent of the particle's
a. charge
b. mass
c. path length
d. initial speed
e. none of these
Capacitance
16. The energy stored by a capacitor can be written in terms of C (capacitance), V
(voltage), and q(charge) as
a. 1/2 qV
b. 1/2 CV2
c. 1/2 (q2/C)
d. none of these
e. 1/2 q2C
Applications of Electrostatics
17. Gauss' law suggests that shielding equipment from electric fields can be
accomplished by putting it inside
a. a cavity in dielectric
b. a vacuum
c. a cavity in conductors
d. none of these
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Answers
1. charge, 1/r2 (Section 21.2)
10. b (Section 21.2)
2. a (Section 21.2)
3. c (Section 21.4)
11. c (Section 21.2)
12. b (Section 21.4)
4. a,c (Section 21.4)
13. e (Section 21.4)
5. d (Section 21.4)
14. a (Section 21.4)
6. c (Section 21.6)
15. b,c,d (Section 21.4)
7. e (Sections 21.8 and 21.9)
16. a,c,b, (Section 21.9)
8. b (Section 21.7)
9. c (Section 21.2)
17. c (Section 21.5)
ALGORITHMIC PROBLEMS
Listed below are the important equations from this chapter. The problems following the
equations will help you learn to translate words into equations and to solve singleconcept problems.
Equations
(21.5)
(21.6)
(21.8, 21.9)
KE = 1/2 mv2 = q | E | | y |
(21.15)
W = qV
(21.16)
KE = qV
(21.17)
E = - ΔV / Δs (directed ⊥ to surfaces of constant potential)
(21.23)
(21.25)
C=q/V
(21.28)
C = εε A /d
(21.29)
1/C = 1/C1 + 1/C2 + 1/C3 (series)
(21.33)
C = C1 + C2 + C3 (parallel)
(21.36)
0
Chapter 21- Electrical Properties of Matter
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(21.39)
V = kq / r
Problems
(21.19a)
1. What is the electrostatic force between two ions in vacuum if one has a charge of 1.6 x
10-19 C and the other has a charge of 3.2 x 10-19 C and the charges are separated by a
distance of 4 x 10-10 m?
2. What is the kinetic energy of an ion that has a charge of 1.6 x 10-19 C and is accelerated
through a potential of 1.0 x 106 V?
3. Two parallel plates separated by 1.0 mm in a vacuum have a potential difference of
1000 V. What is the electric field of the capacitor?
4. A 2-mF capacitor is charged to a potential difference of 100 V. What is the charge on
the capacitor?
5. What is the energy stored in the charged capacitor of problem 4?
6. A 2.0-µF and a 4.0-µF capacitor are connected in series. What is the capacitance of an
equivalent single capacitor?
Answers
1. 2.88 x 10-9 N
2. 1.6 x 10
-13
J
4. 200 mC
5. 10-2 J
3. E = 1.0 x 106 N/C from positive to negative plate; E is 6. 1.3 µF
a vector quantity perpendicular to the plates
EXERCISES
These exercises are designed to help you apply the ideas of a section to physical
situations. When appropriate, the numerical answer is given in brackets at the end of
the exercise.
Section 21.2
1. Two point charges +4.00 C and -2.00 C are located along the x-axis at the origin and at
20 cm respectively. Sketch the electric field in the region of these charges. What is
the field far, far away from the origin? Locate all the points on the x-axis where the
electric field is zero. [2k/r2 N/C; x = 68.3 cm]
Section 21.3
2. A hydrogen ion of charge +e with a mass of 1.67 x 10-27 kg is initially at rest in an electric
field of 1.00 x 10-6 N/C. What is the velocity of the ion after 5 sec? How far has it
traveled? What is its kinetic energy? [u=479 m/sec;s = 1.2 km; ΔK = 19.2 x 10-25 J]
Section 21.4
3. For the charges and locations given in exercise 1, find the location of zero potential
points on the x-axis. What is the potential far, far away from the origin? [40 cm, 13.3
cm, ~ 2k/r J/C]
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4. In atomic and molecular experiments a unit of energy called an electron volt (eV) is
used. One electron volt is the energy gained by a charge e as it changes its electric
potential by one volt. Calculate the value of an electron volt in joules. [1 eV = 1.6 x
10-19 J]
5. In a given medical x-ray tube, electrons are accelerated through a potential of 10,000
V. How much energy does the accelerated electron have? What is the velocity of the
electron after acceleration? [1.6 x 10-15 J or 104 eV, 5.9 x 107 m/sec]
Section 21.5
6. Given a hollow spherical conducting shell with a +Q charge at the center of its inner
cavity. Show that the charge on the inner surface of the conductor is -Q and that a
charge of +Q is on the outer surface.
7. Show that if a copper ball is given a charge Q, the entire charge resides on its outer
surface.
Section 21.8
8. A parallel plate capacitor consists of two flat plates 20 cm square separated by a
dielectric 0.2 mm thick.
a. Find the capacitance if the dielectric is air.
b. Find capacitance if the dielectric is mica (dielectric constant =6). [a. C = 17.8 x 1010
F; b. 106.2x 10-10 F]
Section 21.9
9. If one has three capacitors with capacitances of 0.5, 1.0, and 2.0µF, what capacitance
can be produced by connecting these in various parallel and series combinations?
[all series: 0.29µF; all parallel: 3.5 µF; six other arrangements: 2.33 µF; 1.40 µF; 1.17
µF; 0.86 µF; 0.71 µF; 0.43 µF]
PROBLEMS
The following problems may involve more than one physical concept. Where
appropriate, the numerical answer is given in brackets at the end of the problem.
10. Find the force of attraction between an electron and a proton (hydrogen nucleus) at
a distance of 5.3 x 10-11 m. How does this compare with their gravitational attraction?
[82 x 10-9 N, 3.6 x 10-47 N]
11. Given that the dielectric constant of sodium chloride is 6.12, find the force of
attraction between a Na+ ion and Cl- ion in a salt crystal if the separation is 2.8 x 1010
m. Calculate the energy of the ionic bond in a vacuum and in water (e = 80). [0.48 x
10-9 N; evac = 8.2 x 10-19 J, ewater = evac/80]
12. A square ABCD is 10.0 cm on a side. A charge of 2.00 x 10-10 C is placed at B, and a
charge of -3.00 x 10-10 C is placed at C. Find:
a. the field at D
b. the potential at D [a. 216 N/C down to right making angle of 17° with CD; b. -14.3
volt]
13. If a charge of 5.00 x 10-10 C is moved from D in problem 12 to the center of the
square, how much work is done? [7.55 x 10-10 J]
Chapter 21- Electrical Properties of Matter
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14. Given that the capacitance of 1.0 cm2 of a cell membrane is 1.0 µF, find the number of
ions necessary to charge the membrane to 70 mV (resting potential for an axon).
Assume that the ions are singly charged with q = 1.6 x 10-19 C. [44 x 1010 ions]
15. An air capacitor (0.1 µF) is charged with 20 µC of charge. The separation of the
plates is 1 mm.
a. Find the electric field between the plates of the capacitor.
b. Find the energy needed to charge the capacitor as given. [a. E = 2 x 105 V/m; b.
w = 2x 10- 3 J]
16. The electrical potential difference between the inside and the outside of a heart
muscle is about 90 mV. If the wall of each cell is an insulating layer of thickness of
5.0 x 10-9 m, what is the electric field in the cell membrane? Compare this with other
electrical fields, such as when an electrical breakdown occurs in air (104 V/cm). [18 x
106 V/m, 3 x 106 V/m]
17. A direct current defibrillator has a maximum energy output of 400 J. The capacitance
of the capacitor is 20µF. What is the maximum charging potential required? What is
the electrical charge impulse sent through the body for maximum charging
potential? [6300 V, 0.13 C]
18. If the energy for each impulse of a pacemaker, 2.4 x 10-4 J, is stored in a capacitor
charged to a potential of 6.0 V, what is the capacitance of the capacitor, and what is
the charge of each impulse? If 70 impulses are given per minute, how much energy
is needed per day? [13.3µF, 80 µC, 24 J]
19. Specially designed capacitors called ion chambers are used to detect ionizing
radiation. The incoming radiation produces ion pairs in the capacitor. These pairs
tend to neutralize the charged capacitor. The battery attached to the capacitor
recharges the capacitor. Outline a method that could be used to determine the
energy deposited in the capacitor.
20. A 3.0-µF and 6.0-µF capacitor are connected in series to a 120-V source of potential.
The capacitors are disconnected and reconnected with positive plates together and
negative plates together. What is:
a. The original charge on each capacitor?
b. The initial potential difference for each capacitor?
c. The final charge on each capacitor?
d. The final potential difference for each?
e. The change in energy of the charged capacitors? How do you account for the
difference
[a. Q3 = Q6 = 240 µC; b. V3> = 80 V, V6 = 40 V; c. Q3 = 160 µC, Q6 = 320 µC; d. V3 = V6 =
53V; e. 1600 µJ]
21. A 2.00-µF and 3.00-µF capacitor are connected in parallel across a 100-V line. They
are disconnected and reconnected with the positive plate of each capacitor
connected to the negative plate of the other. What is the final charge on each
capacitor and potential difference across each? What is the change in energy?
Explain the difference. [Q2 = 40 µC, Q3 = 60 µC, V2 = V3 = 20 V; ΔE = 24.0 x 10-3 J]
FOOTNOTES
1) For a complete discussion of electrostatic phenomena as applied to electrocardiograms, see R.
K. Hobbie, "The Electrocardiogram as an Example of Electrostatics," American Journal of Physics
41 (June 1973): 824-831.
Chapter 21- Electrical Properties of Matter