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CHAPTER NINETEEN
Electrical Field and Electrical Potential
In Chapter 18, we explained a range of observations by
developing a model in which charged carriers exert electrostatic forces on one another. In this chapter, we will formulate
the concepts of electrostatic force and electric charge quantitatively. As we did for gravitational forces moving bodies over
distances, we will develop related ideas of work and potential
energy. But suppose we want to know, say, how a battery
affects the charge passing through it. We don’t know what
the battery will be connected to or for how long. In cases
such as this, we must think about what happens to each unit
of charge. We will introduce quantities that tell how much force
is exerted on each unit of charge and how much potential energy
it gains or loses.
“. . . suppose we
want to know, say,
how a battery
affects the charge
passing through it.”
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19-1 Making Electrostatic Force and Charge Concepts Quantitative ◆ 567
19-1 Making Electrostatic Force and
Charge Concepts Quantitative
Knob that can twist
fiber to vary r
Long
distance
French physicist Charles Augustin de Coulomb (1736–1806)
put the concepts of electrostatic force and electric charge on
a quantitative basis. (The historical record seems to indicate
that Cavendish did this earlier, but Coulomb’s results were
the first to be widely disseminated in print.) Coulomb
reported a series of measurements on charged bodies using
a torsion balance apparatus much like the one Cavendish
1 2
used (Figure 8-16) to determine the universal gravitational
constant G. In Coulomb’s apparatus (Figure 19-1), if the fiber
r
is twisted, it exerts a restoring torque proportional to the
angle of twist. The restoring torque it has to exert to keep
the “barbell” in equilibrium is in turn proportional to the electrostatic force that
sphere 1 exerts on sphere 2 when both are charged. It had previously been
suspected, by analogy with gravitational forces, that the electrostatic force
between charged point objects varied inversely as r2. By observing the angle of
twist with the spheres at different separations r, Coulomb showed that this was
indeed the case.
Coulomb could also reduce the charge on either sphere by half to see how
the force varied with charge. He did this by touching the sphere in question with
a conducting sphere identical to it but uncharged. He reasoned that the excess
charges on the charged sphere would by mutual repulsion spread out equally
over both spheres. Then when the introduced sphere is removed, it should take
half the excess charge with it. In this way he found the electrostatic force to be
proportional both to the charge on sphere 1 and the charge on sphere 2. These
relationships are summarized by Coulomb’s law.
Coulomb’s law: The magnitude F 0 F on 1 by 2 0 0 F on 2 by 1 0 of the electrostatic
force that each of two point objects with charges q1 and q2 exerts on the other
at a separation r12 is given by
S
Fk
S
q1 q2
r 212
(19-1)
Note: The magnitude gives the strength of the interaction.
Direction: The forces on the two bodies are directed along the line connecting
them. The force on either body is toward the other body if the two have opposite charges and away if they have like charges.
Note that the signs of the charges are used only in determining the force’s direction, not to find the magnitude of the force.
✦UNITS In SI, the unit of charge, not surprisingly, is called the coulomb (C).
(Its definition follows from ideas that we will treat later on.) The value of the
proportionality constant k could in principle be found by measuring the force that
1 C charges exert on each other at a separation of 1 m. The resulting force would
be 9 109 N (verify for yourself that this is about a million tons!). This tells us
that one coulomb is an extremely large amount of charge. It then follows from
Equation 19-1 that
kF
or
11 m2 2
r 212
19 109 N2
q1 q2
11 C211 C2
k 9 109
N m2
C2
(19-2)
Fiber
Uncharged sphere equal
in weight to sphere 2
keeps “barbell” balanced
Figure 19-1 Coulomb’s torsion
balance. Based on drawing in
Coulomb’s paper in Histoire et
Mémoires de l’Académie Royal des
Sciences, 1785.
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568 ◆ Chapter 19 Electrical Field and Electrical Potential
We cannot physically do the measurement using 1 C charges, but because k is a
constant, we could get the same value from a much smaller measured force
between much smaller charges.
With the humongous coulomb as our unit of charge, the size of nature’s basic
unit of charge, that of a single electron, is just a tiny fraction of a coulomb:
e 1.602 1019 C 1.6 1019 C
(19-3)
(We’ll generally use the more rounded-off value.) The symbol e represents the
magnitude only: The charge of a proton is e; that of an electron is e. Electrostatic forces now join the inventory of forces that we add as vectors when we
S
consider © F in Newton’s second law.
Example 19-1
Vector Addition of Electrostatic Forces
For a guided interactive solution, go to Web Example 19-1 at
www.wiley.com/college/touger
line of action
of Fon A by C
y
B+
A
+
q
–
C at
(2.0m, 2.0m)
x
P
line of action
of Fon A by B
Free body
diagram of A
The charges and masses of the three point objects shown below are
qA 2.4 106 C
qB 1.3 106 C
mA 0.001 kg
Fon A by C
(attractive)
Fon A by B
(repulsive)
mB 0.002 kg
qC 2.0 106 C
mC 0.002 kg
a. If bodies B and C are fixed in their positions, find the total force on A
when it is at the origin.
b. Find the acceleration of A at the instant when it is at this position.
Brief Solution
Choice of approach. a. We must (1) identify the individual forces on A and
their directions in a free-body diagram (see figure), (2) determine the magnitude of each of these forces using Coulomb’s law, and then (3) find the x and
y components and use them to find the resultant vector. In (1), we use the
signs on the charges to determine whether each force is attractive or repulsive
and thus establish its direction along its line of action.
b. To find the acceleration, we use Newton’s second law.
What we know/what we don’t.
To find Fon A by B
To find Fon A by C
0qA 0 2.4 106 C
0qA 0 2.4 106 C
0qB 0 1.3 106 C
0qC 0 2.0 106 C
rAB 2.0 m
Constants
k 9 109
N m2
rAC 212.0 m2 2 12.0 m2 2 2.8 m
Second law applied to A
mA 0.001 kg
total Fon A ?
uF ?
a?
The mathematical solution. a. We find magnitudes by Equation 19-1:
Fon A by B k
9 109
r2AB
7.0 103 N
Fon A by C k
N m2 12.4 10
6
qA qB
qA qC
9 109
r2AC
3
5.5 10
N
2
C
C2 11.3 10
12.0 m2
6
C2
2
6
6
N m2 12.4 10 C2 12.0 10 C2
C2
12.8 m2 2
2
C
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19-2 Electric Fields ◆ 569
In finding the x and y components of these forces, we note that Fon A by B is in
the negative y direction, and from the geometry of triangle APC, u 45°. Thus,
1Fon A by B 2 x 0
1Fon A by B 2 y 7.0 103 N
1Fon A by C 2 x 1Fon A by C 2 cos 45° 15.5 103 N2 10.7072 3.9 103 N
1Fon A by C 2 y 1Fon A by C 2 sin 45° 15.5 103 N2 10.7072 3.9 103 N
Now we add components to get the components of the total force on A:
1Fon A 2 x 1Fon A by B 2 x 1Fon A by C 2 x 0 3.9 103 N 3.9 103 N
1Fon A 2 y 1Fon A by B 2 y 1Fon A by C 2 y 7.0 103 N 3.9 103 N
3.1 103 N
Finally, we convert to a magnitude and direction description:
Fon A 21Fon A 2 2x 1Fon A 2 2y 213.9 103 N2 2 13.1 103 N2 2
5.0 103 N
tan uF 1Fon A 2 y
1Fon A 2 x
3.1 103 N
3.9 103 N
0.79
uF 38 138° below the x axis2.
so
b. By Newton’s second law,
a
Fon A
5 103 N
5 ms2
mA
1 103 kg
◆ Related homework: Problems 19-7, 19-8, 19-9, and 19-11.
19-2
Electric Fields
For an important new perspective on the calculation we did in Example 19-1,
notice that we multiplied by the charge qA in finding each of the forces Fon A by B
and Fon A by C. We used these in turn to find the components of these forces and
then to find the total components 1Fon A 2 x and 1Fon A 2 y and the magnitude Fon A
of the resultant force. So 0qA 0 was built in as a multiplier in finding all of these.
Thus, each of the electrostatic forces on the charge qA, and each of their components, can be viewed as the product of 0qA 0 and “everything else” multiplying
S
it. For example we can write the total force on A 1 F on A 2 very informally as
1Fon A 2 x 0qA 0 a
everything
b
else
x
1Fon A 2 y 0qA 0 a
everything
b
else
y
It then follows that the magnitude Fon A of the total electrostatic force is likewise
the product of 0qA 0 and “everything else.” If qA is at a point P, the “everything
else” multiplying it involves the other charges, their distances from P, and because
we consider directions when we find components, the directions from P to those
charges. In short we can say that the “other stuff” involves the electrical environment of the point P at which qA was placed. To see this reasoning presented
in graphic step-by-step detail, go to WebLink 19-1.
Recall from Chapter 4 that a force is one side of an interaction. A body and
its environment exert equal and opposite electrostatic forces on each other. We
may consider the magnitude of either force to be a measure of the strength of
the interaction. The strength of the interaction is thus the product of a property
of the object itself (its charge) and a term involving the electrical properties
(charges and their configuration) of its environment. This latter term is what we
For WebLink 19-1:
The Electric Field—
The Charge Carrier’s
Environment, go to
www.wiley.com/college/touger
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570 ◆ Chapter 19 Electrical Field and Electrical Potential
call the electrical field, and we denote its magnitude by E:
or
Fon A
Fon A
0qA 0
0qA 0
a
“everything
b
else”
E
(19-4)
strength of
property properties of A’s
electrical
of A
environment
interaction
between A and its
environment
(We use the absolute value of qA because F and E are magnitudes, so they are always positive.)
You can associate the symbol E with electrical environment, or with “everything else.”
Notice that if qA were moved to a different position, its distance and direction from each other charge would change, that is, the “other stuff” contributing
to the field E would change. In essence, qA experiences a different environment—
its surroundings appear different—when it is in a different position. The mathematical way of saying this is: The electric field is a function of position. Changing
position affects each component of the force on A, so in two dimensions we can
write an equation like 19-4 for each direction
Fx on A qAEx
Fy on A qAEy
(19-5x, 19-5y)
Here we use the charge rather than its absolute value because the x and y components of a force can be either positive or negative. We summarize the two
component equations by the vector equation
F on A qA E
S
S
(19-5)
STOP&Think If after doing Example 19-1, you had to find the force on a different
charge (call it qD) placed at the origin, would you have had to do the entire calculation over again using the value of qD? ◆
The above question suggests there is an advantage to factoring the force as
in Equation 19-5. If we did the calculation in Example 19-1 with the factor qA
left out, we could as easily have multiplied by qA or qD at the end. In other
words, we could first have calculated the field at a particular point, and then multiplied by the charge to find the force on any charge carrier placed at that point.
F
From Equation 19-4, the magnitude of the electric field is E 0 qonA A0 , so it is
measured in N/C. This is the number of newtons on each coulomb of charge.
The electric field at a point is the force that will be exerted on each coulomb of charge
placed at that point.
In Equation 19-5, the force vector is the product of the scalar charge and the
vector field. Multiplying a vector by a positive scalar preserves its direction; multiplying it by a negative scalar reverses its direction:
If the charge is positive, the electrostatic force on it is in the same direction as the field;
if the charge is negative, the force is opposite in direction to the field.
Another important point follows from this:
The direction of the electric field at any point is the direction of the total electrostatic
force that a positive charge placed at that point would experience.
✦ELECTRIC FIELD DUE TO A POINT CHARGE For two isolated charges
q1 and q2, Equation 19-4 becomes
Fon 2 by 1 0q2 0 Edue to 1
(19-6)
where Edue to 1 is the field at the point—let’s call it P—where q2 is placed. The
field is due to the environment of charges surrounding P; in this case the only
charge in the environment is q1. Comparing Equation 19-6 with Coulomb’s law
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19-2 Electric Fields ◆ 571
Fon –
A –––––
C
++
Fon +
Fon +
EA
by C
by A
by B
EB
––
+
Fon –
B
++
EC
(a)
Fon +
by C
Fon –
by B
by A
(b)
(c)
0q 0
(Equation 19-1), we see that Edue to 1 k r 2121 . Because the field is the force on each
coulomb (C) placed at P, and because 1 C is a positive amount of charge, the field
at P is in the direction of the force that any positive charge placed there would
experience. Simplifying our notation, we can now summarize.
Electric field at a position a distance r from a point charge q
Magnitude:
Ek
0q 0
r2
(19-7)
Direction: Away from q if q positive,
toward q if q negative.
If the environment of a point P is made up of many point charges, the total
field at P is the vector sum of the fields due to the individual charges. An important first step is to draw a diagram showing the individual fields contributing to
the sum. In the diagram in Figure 19-2a,
S
S
1. E B and E C are directed away from the positive charges on B and C (see rules
S
for direction under Equation 19-7) but E A is toward the negative charge on A;
2. although A and B are the same distance from point P, EA EB because the
magnitude of the charge on A is greater; and
3. although B and C are equally charged, EB EC because B is further from P
(so r in Equation 19-7 is larger).
The diagram is much like a free-body diagram but inventories fields rather than
forces. Another important difference is this: There need not be any object at P. The
environment of point P is present whether or not anything is at P to interact with
it. In contrast to the two-tone arrows representing forces in Figures 19-2b and c, the
field arrows are a single color to remind you that fields represent only one contribution to the interaction. If a charged body is then placed at P, it provides the other
contribution, resulting in a force. We then get the free-body diagrams in Figures
19-2b and c, showing the forces on a positively and a negatively charged body at P.
Note that for the negatively charged body, the forces are opposite to the fields.
PROCEDURE 19-1
Finding the Electric Field at a Point P
Due to an Array of Point Charges
1. Draw a field inventory diagram (such as Figure 19-2a) showing the fields
at P due to each individual charge. The directions are away from positive
charges and toward negative charges.
2. Find the distance from P to each charge.
3. Use Equation 19-7 to calculate the magnitude of each field.
4. Find the x and y components of each field and do vector addition to find
the resultant field vector (Procedure 3-4).
Figure 19-2 Field inventory and
free-body diagrams. (a) Field
inventory diagram showing fields at
point P due to surrounding charges.
(b) Free-body diagram of point
object with charge 1 C placed at P.
(c) Free-body diagram of point
object with charge 1 C placed at P.
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572 ◆ Chapter 19 Electrical Field and Electrical Potential
Example 19-2
A Field Approach to Example 19-1
For a guided interactive solution, go to Web Example 19-2 at
www.wiley.com/college/touger
Eat origin
due to C
a. Redo Example 19-1a by finding the field first.
b. Repeat for the case when point object A is an electron.
qF
Ftotal on
–
negative charge
qE
Etotal
Eat origin
due to B
qE =
field makes
with +x axis
qF =
force makes
with +x axis
(a) Field inventory (b) Directions of
diagram
resultant field
and force vectors
Brief Solution
Choice of approach. In following Procedure 19-1, we repeat the calculation
we did for Example 19-1a but leaving out the factor 0qA 0 until the end.
STOP&Think How does the above field inventory diagram compare with the
free-body diagram in Example 19-1? ◆
What we know/what we don’t. See Example 19-1.
The mathematical solution. a. (Compare with Example 19-1) Apply Equation
19-7 at the origin:
Edue to B k
Edue to C k
0qB 0
9 10
0qC 0
9 109
r2AB
r2AC
6
m2 11.3 10 C2
6
N m2 12.0 10 C2
9N
1Edue to B 2 x 0
3
C2
12.0 m2 2
2.9 10 N/C
C2
12.8 m2 2
2.3 103 N/C
1Edue to B 2 y 2.9 103 N/C
1Edue to C 2 x 1Edue to C 2 cos 45° 1.6 103 N/C
1Edue to C 2 y 1Edue to C 2 sin 45° 1.6 103 N/C
As before, we add components to get the component description of the
total field at the origin (the position where A is or will be placed).
Ex 1Edue to B 2 x 1Edue to C 2 x 1.6 103 N/C
Ey 1Edue to B 2 y 1Edue to C 2 y 1.3 103 N/C
and convert to a magnitude and direction description:
E 2E 2x E 2y 2.1 103 N/C
tan uE Ey
Ex
0.81 so u 39
E is in the same direction as the force in Example 19-1 (except for error due to
rounding off). Also agreeing with Example 19-1, the magnitude of the force is
Fon A 0qA 0 E 12.4 106 C212.1 103 N/C2 5.0 103 N
b. Because we calculated the field first, we need only multiply by the different charge ( 0qA 0 e for an electron) placed at the origin:
Fon e e E 11.6 1019 C212.1 103 N/C2 3.4 1016 N
The direction of the force on a negative charge is reversed (changed by 180°;
see part b of figure) from that of the field, so in this case
uF 39° 180° 141
◆ Related homework: Problems 19-13, 19-14, 19-15, and 19-19.
19-3
Fields Due to Continuous Charge Distributions
When the source of the field is not a point object, we may have to consider
fields due to charges spread out more or less continuously over part or all of the
body. We will use qualitative reasoning to draw conclusions about some common distributions of charge.
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19-3 Fields Due to Continuous Charge Distributions ◆ 573
✦CONDUCTORS IN ELECTROSTATIC EQUILIBRIUM In Example 18-2, we
deduced that if a body is a conductor in electrostatic equilibrium, all excess charge
must be on the body’s surface. If there is an electric field in the interior of such a
body, positive charge or conventional current will flow in the direction of the field.
In electrostatic equilibrium, there is no net motion of charge, so we must conclude:
There is no electric field anywhere in the interior of a conductor in electrostatic
equilibrium.
What about at the surface? Suppose a field at a point P on the surface (Figure 19-3) has components parallel and perpendicular to the surface. By Equation
19-5, the electrostatic force Fe on any excess charge at that point would also have
parallel and perpendicular components. But although 1Fe 2 would be opposed
by the constraining forces that keep the charges from leaving the body, 1Fe 2 //
would be unopposed, so charge would move along the surface in the direction
of 1Fe 2 //. But then the conductor would not be in electrostatic equilibrium. We
avoid this contradiction only if 1Fe 2 //, and therefore E//, is zero.
Fe on +
E
E⊥
(Fe )⊥
E
P
P
+
(Fe )
Fconstraining
Figure 19-3 Fields and forces at
the surface of a conductor.
The electric field must be perpendicular to the surface at all points on the surface of a
conductor in electrostatic equilibrium.
Applying Procedure 19-1 in detail to a continuous charge distribution would
require calculus. But we can apply Procedure 19-1 in part, together with symmetry
arguments, to determine the direction of the field due to some simple charge configurations. Fortunately, these are the configurations commonly found in electric circuits.
Example 19-3
The Electric Field along the Axis
of a Uniformly Charged Circular Ring
Find the direction of the electric field along the axis of a circular ring with a
uniformly distributed positive charge (Figure 19-4a).
E1 + E2
E2
E1
E2y
E1y
etc...
P
E2x
+
+
+
+
+
+
+ +
+
+
+
+ +
+
+
(a)
+
+ +
+
+
+
+
E1x
E3 + E4
4
+
E1 + E2
1
+
+
2
+
3
(b)
(c)
Solution
Choice of approach. Mentally we can break up the ring of charge into pieces
so tiny that they are nearly point charges. Guided by symmetry considerations,
we can select representative pieces. We will draw a field inventory diagram
for a point P on the axis (Step 1 of Procedure 19-1), and then sketch the vector addition to determine the direction of the total field.
The reasoning. We can choose pairs of pieces (e.g., 1 and 2, 3 and 4 in Figure 19-4b) that are diametrically across from each other. The field due to each
Figure 19-4 The electric field
along the axis of a uniformly
charged circular ring. See
Example 19-3.
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574 ◆ Chapter 19 Electrical Field and Electrical Potential
EXAMPLE 19-3 continued
piece is directed away from that piece. Figure 19-4b shows that due to the
S
S
symmetry in our choice, the horizontal components of the fields E 1 and E 2
(due to charges 1 and 2) are equal and opposite. Then the resultant of these
two fields is along the axis. The same reasoning applies to each such pair of
charges, so the vector sum of the contributions of all pairs (Figure 19-4c)—the
field due to the entire ring—is likewise along the axis. By considering the surface of a conducting wire to be made up of many such rings, we can build
on this result to find the field along the axis of the wire—see Example 9-4.
◆ Related homework: Problem 19-22.
✦CONDUCTORS NOT IN ELECTROSTATIC EQUILIBRIUM By maintaining
a difference in the surface charge density at its two poles, a battery in a simple
circuit prevents the conducting path from reaching electrostatic equilibrium, so
there continues to be a net motion of charge, that is, a current. Because electrostatic equilibrium is not reached, the interior field does not become zero. In
Example 19-4, we find its direction by extending the approach of Example 19-3.
Example 19-4
Electric Field in a Wire
Connected to a Battery
For a guided interactive solution, go to Web Example 19-4 at
www.wiley.com/college/touger
In Chapter 18 we showed that when a conducting wire is connected between
the terminals of a battery, the battery causes a surface charge density gradient
along the wire.
a. Find the direction of the electric field in the interior of the wire. In your
reasoning, consider a long, straight segment of the wire (Figure 19-5a).
b. Find the direction of the electrostatic force on an electron moving through
the wire, and on a positive charge carrier (such as we consider in a conventional current) moving through the wire.
+
+
+
+
+
+
+
3 +
+
+
4
+
+
+
+
+
–
–
–
–
–
––
–
–
––
–
(a)
ring 1
+
+
+
+
+
+ ring 3
+
E4
region
blown up
in (b)
E1
+
P
P
1 +
2
Figure 19-5 Electric field in a
straight wire. (Summary of main
field ideas in Example 19-4).
region
blown up
in (c)
E2
+
–
–
–
– ring 2
vector sum
E1 + E2 downward
(b)
E3
+
+ ring 4
+
vector sum
E3 + E4 downward
(c)
Brief Solution
Choice of approach. We can find the direction of the field by the same approach
as the previous example. Because a straight wire is cylindrical, we can think
of the charge on its surface as being made up of a stack of loops of charge,
each like the one in Example 19-3.
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19-4 Picturing Electric Fields ◆ 575
The details.
a. First consider the field at a point P on the axis. From Example 19-3, we
know the field contribution from each ring is along the axis at point P.
Consider pairs of rings at equal distances above and below P. If the ring
above is positive and the ring below is negative, as in Figure 19-5b, the
contribution due to each ring is in the positive-to-negative direction, as is
their sum. If both rings are positive, as in Figure 19-5c, the contributions
are in opposite directions, but because the lower ring is less positive, its
field contribution is weaker, so the sum of these two contributions is likewise in the positive-to-negative direction. We can similarly show that the
field due to each pair of rings chosen in this way, and therefore the total
field, is in the direction of decreasing surface charge density.
b. Electrostatic forces are in the same direction as the field on positive charges
and opposite on negative charges. Thus, conventional current flows in the
direction of decreasing surface charge density, but electrons flow oppositely.
◆ Related homework: Problem 19-23.
19-4
Picturing Electric Fields
It is often useful to get an overall picture of the electric field, which we can do
as follows. We can sample the effects of the field by placing a point positive
charge—a “test charge”—at different positions and observing how the magnitude
and direction of the force on it vary from point to point. This is shown in Figure 19-6. The test charge in Figure 19-6a is sampling the field due to the point
charge in the center; the figure shows the force per unit charge on it (the field)
at several different points. Figure 19-6b shows a more detailed mapping, but it is
still representative; if we made the map complete, the arrows would totally blacken
the region and make the map unreadable. So we have to imagine the arrows at
the in-between points; in reality, the field is continuous, changing gradually in
magnitude and direction from one point to the next except right at the point
charge(s) giving rise to the field. The procedure and some of the implications of
the mapping that we get are shown in step-by-step detail in WebLink 19-2.
The mappings are sometimes summarized by drawing a continuous set of
lines, called electric field lines (Figure 19-6c). The field vectors in Figure 19-6b
would be directed along these lines. For fields due to more complicated arrangements of charge, the field lines might be curved. The field vector at each point
would then be directed tangent to the field line.
Because the field lines in Figure 19-6c don’t show the lengths of the field
vectors, they must communicate the field’s strength in another way. Figure 19-6d
suggests how this happens. There, because the field is due to a greater point
+
(a)
For WebLink 19-2:
Mapping an Electric
Field, go to
www.wiley.com/college/touger
+
+
++
–
(b)
(c)
(d )
(e)
Figure 19-6 Electric fields due to point charges. (a) Force per unit charge (field) on a
test charge at several different positions. (b) A more detailed mapping of field vectors at
different positions. (c) Electric field lines for the mapping in b. (d) Electric field lines for
a greater point charge. (e) Electric field lines for a negative point charge.
0540T_c19_566-597.qxd 9/24/04 17:21 Page 576
EQA
576 ◆ Chapter 19 Electrical Field and Electrical Potential
equ
qua
ual
al
are
ar
reeaas
as
3 lines
cross one
square cm
(a)
a Density at this surface is 3 lines/cm2
(b) Same totall number of lines crosses each sphere, so
fewer lines cross each cm2 of larger spherical surface.
charge, the lines are more densely packed. The density of field lines visually communicates the strength of the field. Notice, too, that in both Figures 19-6c and
19-6d the lines fan out (become less dense) as they go further out from the
charge. By doing so, they show us that the field gets weaker as the distance r
from a point charge increases. Figure 19-6e shows that the direction of the field
is reversed if the point charge is negative.
In summary, the field line “map” communicates information because they follow a few basic rules:
• Rule 1: Field lines start and end at charge carriers (or else go to infinity). They
are directed away from positive charge and toward negative charge.
• Rule 2: The direction of the field line at any point (or of the tangent at that point
if the field line is curved) is simply the direction of the field vector at that point.
• Rule 3: The density of field lines (how closely packed they are) in a small
region about a point is drawn to be proportional to the magnitude of the field
vector at that point.
Quantitatively, the density of field lines is the number of field lines crossing
each unit of area perpendicular to the lines in a three-dimensional picture (Figure 19-7a). The following line of reasoning demonstrates that with this definition,
2
the depicted field due to a point charge (Figure 19-7b) is proportional to 1 r , as
Coulomb’s law requires it to be. In preparation for our argument, the figure shows
imaginary spheres of different radii r centered at the point charge; these provide
the surface areas that we picture the field lines crossing.
Argument Statement in Words
Equivalent Mathematical Statement
1. The same number of field lines crosses all the
spheres in Figure 19-7b.
Number of
field lines c1 (a constant)
2. The number of field lines crossing each unit
area on a particular sphere of radius r is the
total number of lines divided by the sphere’s
surface area 4pr2.
Number of
c1
lines/unit area 4pr2
3. The number of field lines per unit area that
we draw is proportional to the field.
c2E 4. Then the field due to the point charge is
inversely proportional to r2.
Ea
c1
4pr2
1c2 is another constant2
c1
1
b
4pc2 r2
a constant
that we call k
b
Figure 19-7 The density of field
lines indicates the magnitude of
the field. (a) The density at this
2
surface is 3 linescm . (b) The same
total number of lines crosses each
sphere, so fewer lines cross each
cm2 of the larger spherical surface.
0540T_c19_566-597.qxd 9/24/04 15:54 Page 577
EQA
19-4 Picturing Electric Fields ◆ 577
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
(a) Charge
configuration
(b) Field near
surface
(c) Field far from
surface (zoom out)
For a more intuitive approach to this
at the end of the chapter.
1
r2
(d ) Lines are continuous
between near and far
limits
Figure 19-8 Electric field lines
due to a charged conducting
plate.
dependence, work through Problem 19-1
✦GAUSS’S LAW Doing Problem 6-55 shows that the irradiance due to a point
light source is proportional to r12. The irradiance is also called the energy flux
density. In this terminology there is the same total flow or flux of energy per
second across each sphere, rather than the same “total flow” of field lines. Then
we can speak of the electric field as the electric flux density. A total electric
flux, proportional to the total number of lines, can be obtained for the point
charge by multiplying the field at any distance r by the spherical surface to which
kq
2
it is everywhere perpendicular, giving us 1 r 2 2 14pr 2 4pkq (a constant times q).
For WebLink 19-3:
Thus, the total flux across any of the imaginary spheres enclosing the charge is
Electric Field Due
proportional to the charge enclosed. By deforming the spheres, we could show
to a Charged Conducting
that the total flux across an enclosing surface of any shape is proportional to the
Plate, go to
total amount of charge enclosed, however the charge is configured. This generwww.wiley.com/college/touger
alization is called Gauss’s law, after the great German mathematician Karl
Friedrich Gauss (1777–1855). It is useful for finding the electric fields due to various continuous arrangements of charge,
but its detailed mathematical treatment is beyond the scope
A
of this book.
B
A charged conductor is not a point object. To get a picture of the field due to such an object (Figure 19-8a), we
+
+
use the fact that close to a conducting object, the field must
be perpendicular to the surface (Figure 19-8b). At a great disC
tance, though, the object looks like a point object, so the
field at this distance must look like the field due to a point
object (Figure 19-8c). Because field lines always start and end
at charge carriers or else go to infinity, there can be no breaks
(a) The two fields add vectorially at each point.
in the lines between the near and far regions, so we can fill
in the connecting parts (Figure 19-8d). For a more detailed
graphic presentation of this reasoning, go to WebLink 19-3.
In general, we can think of an extended charged object
A
as though it were built up of point charges. Each point charge
B
has a field like the fields in Figure 19-6. We now consider
how these fields combine to give us a total field.
+
+
✦SUPERPOSITION OF FIELDS The electric field lines due
to an arrangement of charges communicate information in
the same way (Rules 1–3) as the electric field lines due to a
point charge. But now the rules tell us how to read the direction and comparative magnitude of the total electric field
from the field lines. From the field lines for each of two point
charges, we can find the two fields at each point in space
and then combine them by vector addition (Figure 19-9a).
C
(b) The field lines are a map of the total field
Figure 19-9 Superposition of fields due to two point
charges. (a) The two fields add by vector addition at each
point. (b) The field lines are a map of the total field.
0540T_c19_566-597.qxd 9/24/04 15:54 Page 578
EQA
578 ◆ Chapter 19 Electrical Field and Electrical Potential
+
+
–
+
–
+
–
+
–
+
–
+
–
+
–
–
+
–
–
+
–
–
+
–
–
+
–
–
+
–
E due to +
+
A
Total E
+
+
+
(a) Charge configuration
E due to –
B
(b) Superposition of
fields between plates
D
+
–
–
(approx. zero)
Figure 19-10 Electric field of a
charged parallel plate capacitor.
+
–
+
–
+
–
+
–
+
–
+
–
+
–
+
–
+
–
(d ) Superposition of fields
near edges
E due
to +
Total E ≈ 0
(c) Superposition of
fields outside
C
+
E due
to –
(approx. zero)
(e) Overall picture
The resulting electric field lines for the combination of charges map information
about the resultant of the two field vectors at each point (Figure 19-9b). This
point-by point vector addition is called a superposition of the two fields. To
see how we apply this reasoning to obtain the result in Figure 19-10, work
through Example 19-5.
Example 19-5
Electric Field
of a Parallel Plate Capacitor
For a guided interactive solution, go to Web Example 19-5 at
www.wiley.com/college/touger
Draw the electric field lines for a charged parallel plate capacitor, in which the
distance between the plates is very small compared to the dimensions of either
plate.
Brief Solution
Choice of approach.
1. Draw a charge configuration diagram.
2. Consider the superposition of fields in three separate regions: a. between
the plates, b. to the left and right of both plates, and c. around the edges.
The details.
1. As in Example 18-3, the charge distributes over the inner faces of the two
plates, forming two opposite planes of charge (Figure 19-10a).
2. A typical point A between the plates (away from the edges) is close
enough to each plane of charge so that they both look effectively infinite
from A (Figure 19-10b). The uniform fields due to the two planes are
both away from positive and toward negative charge, as is the resultant
field.
At a typical point B a short distance to either side of both plates, the planes
of charge still look more or less infinite (Figure 19-10c). But now their fields,
still away from positive and toward negative and still equal in magnitude, are
0540T_c19_566-597.qxd 9/24/04 15:54 Page 579
EQA
19-4 Picturing Electric Fields ◆ 579
opposite in direction, for a total field of zero. At much larger distances from
both plates, the fields due to the two planes of charge are very nearly (but
not exactly) equal and opposite, so the total field is nearly (but not exactly)
zero.
At points near the edges, such as C and D in Figure 19-10d, the fields are
not equal and opposite, and the total field varies in direction from point to
point. This effect is called fringing.
Figure 19-10e shows the overall field lines resulting from these superpositions.
◆ Related homework: Problems 19-27, 19-28, and 19-29.
Terminal Speed, Electrophoresis, and DNA Testing
An object falling through a fluid (such as air) in a gravitational field is subject
to a drag force that increases with speed. When the magnitude of the drag
force becomes as great as the gravitational force, the total force on the object
becomes zero and the object stops accelerating. The speed it has reached at
this stage is called the terminal speed. If the magnitude of the drag force is
proportional to the speed 1Fd zv, where the proportionality constant or drag
coefficient z, the Greek letter zeta, depends on the object’s size and shape),
mg
then at terminal speed vt, zvt mg, and vt z .
In the same way, when a uniform electric field is applied to ions in solution
(also a fluid), the electric force on the ions is opposed by a drag force. As in
air, the drag coefficient will be greater for ions with larger cross-sections, and
those objects will reach smaller terminal speeds. In this case, the electrical
qE
force equals the drag force at terminal speed: zvt qE, and vt z . Because
these terminal speeds are typically reached over very small distances, each
type of ion travels through most of the solution at its terminal speed—a
different terminal speed for each one. The ions thus separate out. The process
of separating out ions in solution by applying an electric field is called
electrophoresis.
If the ions are permitted to pass through a porous medium, such as a gel,
procedures such as staining will show how far the different ions have traveled
through the gel. In a given time interval, ions of different sizes, with different
terminal speeds, travel different characteristic distances (see Problem 19-72).
The resulting separation, and the density of each ion that shows up on staining, makes possible analysis of which ions are present and to what extent.
(See photo.) This process is called gel electrophoresis.
DNA fragments typically occur in an ionized state in certain kinds of solutions and can therefore be analyzed by gel electrophoresis. Performing gel electrophoresis on DNA, however, is complicated by the fact although the drag
coefficient z increases with fragment size, so does the total charge q. This is
because there is a charge on each of the base pairs making up the rungs of
the DNA ladder, so the total charge depends on how many pairs are strung
together to make up the total length of the fragment. To the extent that both
q
z and q are proportional to length, the ratio z doesn’t change with length, and
qE
the terminal speed vt z remains the same for fragments of all lengths in a
field E. In this case, a gel with smaller pores is used, so that larger fragments
are likely to become trapped over a shorter distance. Gel electrophoresis is a
standard procedure for DNA analysis. It is used for such purposes as establishing who is the father of a child, or identifying blood samples in legal cases
such as rape or murder trials.
DNA testing. The photo shows the
separation of DNA components by
gel electrophoresis.
0540T_c19_566-597.qxd 9/24/04 15:54 Page 580
EQA
580 ◆ Chapter 19 Electrical Field and Electrical Potential
19-5 Electrical and Gravitational Systems: Similarities
and Differences; Electrical Potential Energy
Figure 19-11 Small, irregularly
shaped shepherd moons keep
the material in Saturn’s rings
moving in formation.
Gravitational and electrostatic interactions are due to different inherent properties
of the particles that constitute matter: gravitational mass and electric charge. But
mathematically, they obey similar inverse square laws (Coulomb’s law and Newton’s
universal law of gravitation).
Just as the total electrostatic force on a charged body is the vector sum of
the forces exerted on it by all other charged bodies, the total gravitational force
on a body is the sum of the gravitational forces exerted on it by all other masses.
For example, the sun and Earth both exert significant gravitational forces on the
moon, and the paths of the chunks of debris that make up the rings of Saturn
are affected by several significant gravitational forces. These include the forces
exerted by Saturn, by the rest of the debris in the ring, and quite significantly by
small moons that rotate with the rings and help keep the debris in ring-shaped
configurations (Figure 19-11). Because they “herd” the debris in this way, they
are called shepherd moons.
Suppose we find the force on one chunk of debris, and moments later
a chunk with different mass is at the same point. Rather than recalculate the
total force from scratch, we can be guided by what we did for electrostatic
forces. We can calculate the contribution to the total force at a point from
all the surrounding bodies—the force per unit of mass—and then multiply it
by the mass of whatever we place at that point. Analogous to Equations
19-4 and 19-7, we can again write equations we used to describe gravitational
forces:
Fon A
strength of
gravitational
interaction
between A and its
environment
or
where S
g has magnitude
(a) Far
Earth
(b) Near
Figure 19-12 Gravitational fields
far from and near Earth’s surface.
mA
g
property
of A
property
of A’s
environment
(5-1, like 19-4)
F on A mA S
g
S
gG
m
r2
(8-8 or 8-9, like 19-7)
Continuing the analogy, we call S
g the gravitational field. Near Earth’s surface
S
g has the roughly constant magnitude g G RmE2 9.8 m/s2. It is the gravitational acceleration of a freely falling body. We could treat this value as constant over trajectories near Earth’s surface small enough so we can regard Earth
as flat. How does this compare with electric fields? Viewed from far away,
Earth looks like a point object, and its gravitational field lines (Figure 19-12a)
look like the electric field lines due to a negative point charge. STOP&Think
Why not a positive point charge? ◆ Viewed close up, Earth looks flat, and
the field lines (Figure 19-12b)—everywhere perpendicular to the surface—
look like those near a plane of charge (Figure 19-8b). The lines are parallel,
not spreading out or converging, so the density of lines, indicating the magnitude of the field, doesn’t change. The field is essentially constant in a flat
Earth approximation. (Continuing to treat gravitational field lines as we did
electrical, we could deduce Gauss’s law for gravitational fields: The total
gravitational flux across an enclosing surface is proportional to the mass
enclosed.)
By considering the work done to move bodies in opposition to gravitational
forces within a system, we were able to derive expressions for gravitational
potential energy when g is constant (e.g., over short distances from a planet’s
surface)
PEgrav mgy
(6-7)
0540T_c19_566-597.qxd 9/24/04 17:21 Page 581
EQA
19-5 Electrical and Gravitational Systems: Similarities and Differences; Electrical Potential Energy ◆ 581
+y
+ + + + + +
+
m
0
+
–
M
(a)
m1m2
PEgrav G
r
–
–
–
(c)
–
0
–
–
–
–
–
+
–
+
–
–
+y
PEelec qEy
(19-8)
PEelec k
q1q2
r
+y
0
By rotating the page, you can turn
Figure19-13c into either of these pictures
Figure 19-14 The y direction is
chosen so that the field is in the
negative y direction. Compare
with Figure 19-13c.
(19-9)
Equations 6-7 and 19-8 have the same sign because we chose our coordinate
system so that the field direction is the same in both cases. Thus, in all parts of
Figure 19-13, the hand increases the PE of the body by lifting it. Note that Figure
19-13c is really the same as Figure 19-14a or b: In each case the hand is pulling
the positive charge away from the plane of negative charge that is attracting it; in
each case we pick our coordinates so that the field is in the y direction.
In the r-dependent situations described by Equations 8-10 and 19-9, however,
we are stuck with keeping the object that is the source of the field at the center 1r 02. The hand in Figure 19-15 does positive work either by pulling two
mutually attractive positive masses (masses are always positive) apart or by pushing two mutually repulsive positive charges together. In the gravitational case PE
increases as r increases, but in the electrical case PE increases as r decreases, so
the signs in Equations 8-10 and 19-9 must be opposite.
Example 19-6
–
–
(if we take the direction of the electric field to be the negative y
S
direction, as we did for the gravitational field g )
When we must consider E’s variation with r:
–
(8-10)
In either case, we can think of the gravitational PE as the
energy stored by moving one body in a system against the
gravitational field due to the rest of the system.
Purely by analogy (using the correspondences m 4 q,
g 4 E, and G 4 k), we can obtain expressions for the electrical PE (the energy stored by moving one charged body
in a system against the electric field due to the rest of the
system):
When E is constant:
–
(b)
and when we must consider g’s variation with r
S
–
Figure 19-13 Positive work is
done by lifting either a mass
or a positive charge against a
field directed in the negative
y direction.
Energy of a Proton–Electron System
For a guided interactive solution, go to Web Example 19-6 at
www.wiley.com/college/touger
A proton and an electron are a distance r apart.
a. Compare the electric and gravitational PEs of the proton–electron system.
b. If the distance r is increased, does the electrical PE increase or decrease?
Does the gravitational PE increase or decrease?
Brief Solution
Restating the question. The most meaningful way to compare two quantities of
PEelec
?
very different sizes is by a ratio. In a. we ask, PEgrav
E
+
g
m
+
M
(a)
(b)
Figure 19-15 The hand does
positive work (r-dependent PE
increases) in both cases.
0540T_c19_566-597.qxd 9/24/04 15:54 Page 582
EQA
582 ◆ Chapter 19 Electrical Field and Electrical Potential
EXAMPLE 19-6 continued
Choice of approach. We can treat the proton and electron as point objects, so
q q
m m
the two PEs are PEelec k 1r 2 (Equation 19-9) and PEgrav G 1r 2 (Equation 8-8). The masses and charges are known constants.
What we know/what we don’t.
r?
Known
constants:
mp 1.67 1027 kg
PEelec
?
PEgrav
me 9.11 1031 kg
G 6.67 1011
e
qp e 1.6 1019 C
qe e 1.6 1019 C
k 9 109
N m2
kg2
N m2
C2
The mathematical solution. a. By Equations 19-9 and 8-8,
PEelec
PEgrav
q1q2
kq1q2
r
PEgrav m1m2
Gm1m2
G
r
k
a9 109
N m2
b 11.6 1019 C211.6 1019 C2
C2
N m2
27
a6.67 1011
kg219.11 1031 kg2
2 b 11.67 10
kg
2.30 1039 1unitless2
PEelec is vastly greater than PEgrav; PEgrav is negligibly small in comparison.
b. Both PEs are negative, PEelec because qe is negative and PEgrav because it
has a sign. As r increases, the absolute value of each PE decreases. Then
both PEs become less negative, that is, both increase.
◆ Related homework: Problems 19-31, 19-32, 19-33, 19-36, and 19-37.
Example 19-6 shows that gravitational interactions are insignificantly weak
compared to electric interactions and are only evident between very massive
bodies. Thus, on an atomic scale electrostatic forces are the binding forces
that hold matter together, but on an astronomical scale it is mutual gravitational attraction that causes great clouds of hot gas to condense into stars and
planets.
Like the gravitational force, the electrostatic force is conservative, so Equations 6-10 (conservation of mechanical energy) and 6-16 (the work-energy theorem) remain valid under the same conditions as before when the potential energy
is partly or totally electrical.
Example 19-7
Repelling Klingon Space Rats
A Klingon spacecraft hovers 18 m from the vast flat surface of a space station.
Klingon attack rats, each with a mass of 2.00 kg, can propel themselves across
the gap at speeds of up to 12 m/s. But space station engineers have figured
out that in the process of propelling themselves, the rats acquire a positive
4
charge of 5.4 10 C. They therefore devise a way of charging up the plane
surface of the station so that it will have a uniform electric field. Find the magnitude and direction of the minimum electric field required to repel all rats.
0540T_c19_566-597.qxd 9/24/04 15:55 Page 583
EQA
19-6 Potential and Potential Differences ◆ 583
Solution
Choice of approach. We apply conservation of mechanical energy
(Equation 6-10) to the case when the PE is purely electrical and is due
to a uniform field. (Remember that for other arrangements of charge,
like a point charge, the field is not uniform and we cannot use Equation 19-8.) We desire a final state in which the rat’s velocity is reduced
to zero. We first sketch the situation (Figure 19-16) to show the necessary direction of the field so that the force on the positively charged
rat opposes its motion. We choose our coordinates so that, consistent
with Equation 19-8, the positive y direction is opposite to the electric
field.
18 m
O
++
+
++
+
+
+
+ ++
+ +
+
What we know/what we don’t.
yA 0
State B
vA 12 m/s
PEA qEyA KEA +
+
E
State A
yB 18 m vB 0
2
1
2 mvA
PEB qEyB KEB 0
E?
+
+
+
+
+
+
+
+
+
+
+
+
+
CM
+
Constants
m 2.00 kg
q 5.4 104 C
+y
Plane
Figure 19-16 A sketch of
the physical situation for
Example 19-7.
E?
The mathematical solution.
1Total Energy2 A 1Total Energy2 B
PEA KEA PEB KEB
or
(6-10)
qEyA 12 mv2A qEyB 0
becomes
Solving for E,
E
2
12.00 kg2112 m/s2 2
mvA
1.5 104 N/C
2q 1yB yA 2
215.4 104 C2 118 m 02
◆ Related homework: Problems 19-34, 19-35, 19-38, and 19-39.
19-6
Potential and Potential Differences
Just as the electrostatic force on any body may be treated as the product of two
contributions, one from the body itself (its charge Q) and the other from the electrical environment (the field), the electric PE may likewise be separated into two
contributions:
PEelec Q a
PEelec
b
Q
body’s
contribution
environment’s
contribution
(19-10)
Suppose the body is small enough for us to consider Q a point charge. As the
electric field is a function of position giving the force on each unit of charge placed
PE
at that position, so the quantity Qelec, which is called the potential, is a function
of position giving the potential energy of each unit of charge at that position. The
units of potential are units of energy (joules) divided by units of charge (coulombs)
and are therefore J/C.
It follows from Equations 19-8 and 19-9 that
potential
QEy
PEelec
Ey
Q
Q
(19-11)
(in a region of constant field in the y direction, that is, near an effectively infinite plane of
charge or close enough to a uniformly charged surface so that it appears flat)
➥ We will shortly introduce another
name for J/C, but not yet, because
we want the units to serve as a
reminder that we are talking about
PE per unit charge.
0540T_c19_566-597.qxd 9/24/04 15:55 Page 584
EQA
584 ◆ Chapter 19 Electrical Field and Electrical Potential
and
potential
PEelec
Q
kQq
kq
r
r
Q
(19-12)
(at a great enough distance r from a body with charge q so that
the body may be regarded as a point object)
Example 19-8
+y
y2
Change of Potential Experienced
by an Electron
An electron is released at rest at a point where the potential is 3.0 J/C. The
point is in a region of uniform electric field of magnitude 75 N/C. Set in motion
by the field, the electron reaches a point 0.020 m from its starting point. What
is the potential at this point?
Felec
e–
∆y = 0.020 m
y1
E
Solution
S
Choice of approach. We sketch the situation with E in the negative y direction
(Figure 19-17) so that we can apply Equation 19-11. Because the force on a
negative charge is opposite to the field, the electron moves ¢y 0.020 m in
the positive y direction. Because PEQelec Ey,
Figure 19-17 A sketch of
the physical situation for
Example 19-8.
¢a
PEelec
b E ¢y
Q
(E constant)
(19-13)
What we know/what we don’t.
¢y 0.020 m
a
PEelec
b 3.0 J/C
Q 1
E 75 N/C
a
PEelec
b ?
Q 2
The mathematical solution.
¢a
a
PEelec
b E ¢y 175 N/C210.020 m2 1.5 J/C
Q
11 J 1 N m2
PEelec
PEelec
PEelec
b a
b ¢a
b 3.0 J/C 1.5 J/C 4.5 J/C
Q 2
Q 1
Q
Making sense of the results. To lose potential energy, negatively charged bodies must move from lower to higher potential. Symbolically, ¢1 PEQelec 2 must be
positive for ¢PEelec Q¢1 PEQelec 2 to be negative when Q is negative.
◆ Related homework: Problem 19-43.
In Chapter 6, we pointed out that it is changes in potential energy rather than
PE values at a single point that have physical meaning. Likewise, the change in
potential energy per unit charge—the change in potential from one point to
another—is what has physical significance, not its value at either point. That’s
why we have not assigned potential its own symbol, but we now choose VAB as
our symbol for the potential difference between two points A and B:
VAB a
PEelec
PEelec
¢PEelec
b a
b Q B
Q A
Q
(19-14)
Units of potential difference, like potential, are J/C. A more common name for
these units, especially in the context of electric circuits, is volts (V):
1 volt 1
joule
coulomb
11 V 1 J/C2
(19-15)
0540T_c19_566-597.qxd 9/24/04 18:26 Page 585
EQA
19-6 Potential and Potential Differences ◆ 585
Thus, voltage is a common expression for potential difference and motivates our
choice of symbol.
Another unit of energy, often used when applying these ideas on an atomic
scale, is the electron volt (eV). Because an amount of potential difference multiplied by an amount of charge gives us an amount of energy, the electron volt
is defined as 1 volt multiplied by the magnitude of the charge of one electron.
1 eV 11.6 1019 C2 11 V2 1.6 1019 J
(19-16)
We must emphasize that voltage is a difference; it does not tell us the PE per
unit charge at one point but how much it changes between two different points.
For example, the voltage of a battery is the increase (a positive difference) in
energy that each unit of charge gets in passing from one terminal to the other.
Because the manufacturers don’t know what will be connected to a battery or
for how long, the useful thing to tell you is how much nonelectrostatic work the
battery does on each unit of charge (and how much energy it thus transfers to
each) passing between its endpoints.
Case 19-1
◆
Electric and Gravitational Circuits: Exploring an Analogy
In Figure 19-18, the escalator does the same amount of
nongravitational work on each unit of a population of
mice (we may take
our unit of mass in
this case to be one
standard
mouse)
and so increases the
gravitational PE of
each unit of mass by
the same amount,
no matter how
Escalator
many mice ride the
escalator in all.
In the gravitational circuit, there
is a value of the Enough
Slide
coefficient of fric- friction to
tion between mouse prevent
speeding up
and slide for which
the
amount
of
energy dissipated
exactly equals the
(a)
a Gravitational circuit
gravitational PE lost, so that the mice move at constant
speed rather than gaining or losing KE. Somewhat analogously,
because
Electrons
there is resistance in
the wire, the electrons travel at an
Negative
pole
unchanging
drift
velocity. But like all
analogies, this one
has its limitations. If
the coefficient of
Battery
friction is increased,
the mice will be
slowed down, ultiPositive
mately causing a
pole
back-up of mice at
the top of the escalator because the
escalator keeps carConducting
wire
rying the mice at the
same
speed
as
before. But in the
(b) Electrical curcuit
Figure 19-18 Gravitational and electrical circuits compared.
(a) Gravitational circuit
(b) Electrical circuit
Similarities
Each mouse gains PE when the electromechanical mechanisms
of the escalator do work to move the mice in opposition to
gravitational forces.
The energy gained is dissipated because of friction along the
path.
We can characterize the escalator by how much PE in joules it
provides to each unit of mass (mouse or kg).
Each electron gains PE when the chemical mechanisms of
the battery do work to move the electrons in opposition to
electrostatic forces.
The energy gained is dissipated because of resistance along
the path.
We can characterize the battery by how much PE in joules it
provides to each unit of charge (electron or coulomb).
11 JC 1 volt2
Differences
The current or flow of mice consists of a few mice doing complete laps of the circuit.
The net flow of electrons is the result of vast numbers of electrons shifting very small distances at very slow drift velocities.
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586 ◆ Chapter 19 Electrical Field and Electrical Potential
Case 19-1
◆
Electric and Gravitational Circuits: Exploring an Analogy (continued )
electrical circuit, if electrons are slowed down by greater
resistance and begin to back up, they repel the electrons coming up behind them, slowing them down as
well. This feedback continues throughout the circuit so
the electrons travel the entire circuit at a slower speed
(resulting in a smaller current).
There are other differences. Individual mice may
board the escalator at the bottom and do repeated laps
of the gravitational circuit in a short time. In the electric circuit, in contrast, all parts of the circuit contribute
electrons to the flow. These electrons actually drift
extremely slowly through the circuit, typically traveling
on the order of 10 cm in an hour (we’ll do the calculation in the next chapter). But there are so many of
19
them that, for example, under ordinary conditions 10
of them will pass any point on the filament of a 75 W
bulb each second. Remember also that the individual
electrons do not travel at constant speed; they are
accelerated by the field and slowed down by collisions.
The drift velocity is an average.
To further clarify the connection between voltage and energy concepts, work
through Example 19-9.
Example 19-9
A Photoelectron
For a guided interactive solution, go to Web Example 19-9 at
www.wiley.com/college/touger
Light striking a metal surface may under certain conditions provide some surface electrons with sufficient energy to break away from the surface. This effect
is called the photoelectric effect, and the resulting emission of electrons is called
photoemission. Suppose that when light strikes one of two parallel plates, elec5
trons are emitted with a maximum velocity of 4.8 10 m/s. What voltage
must be maintained between the plates if none of the emitted electrons are to
reach the opposite plate?
Brief Solution
Choice of approach. Because voltages are PE differences per unit charge, we
focus on energy. We can apply conservation of mechanical energy: The loss in
KE as the electron is brought to a stop must equal the gain in PE. The force on
a negative charge is opposite the field, so the field is toward the opposite plate.
What we know/what we don’t.
State A
State B
Constants
vA 4.8 10 m/s
vB 0
me 9.11 1031 kg
PEA ? KEA 12 mv2A
PEB ? KEB 0
5
VAB 19
q e 1.6 10
C
PEB PEA
?
q
The mathematical solution.
PEA KEA PEB KEB
becomes
Then
V AB PEB PEA
q
2
1
2 mvA
2
1
2 mvA
PEB PEA
10.5219.11 10
q
31
(6-10)
kg214.8 105m/s2 2
1.6 1019 C
0.66 V
Making sense of the result. The decrease in potential, when multiplied by a
negative charge, will yield an increase in PE.
◆ Related homework: Problems 19-46 and 19-48.
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19-7 Batteries ◆ 587
✦POTENTIAL DIFFERENCE AND SURFACE CHARGE DENSITY
GRADIENTS In Chapter 18 we established that a current flows in a conductor if the surface charge density changes along the conductor. In Example 19-4
we saw that the surface charge density gradient along the conductor results in
an electric field directed from higher to lower surface charge density along the
conductor’s axis. The net field at any interior point depends on differences
between the surface charge upstream and downstream from the point. So where
the density is changing uniformly (the gradient is constant), the field is likewise
constant (except near the ends of the region of uniform change). Where the field
is constant, the voltage or potential difference between any two points along the
conductor is given by Equation 19-13, which we can rewrite as
VAB E¢y
1 ¢y yB yA 2
(19-17)
If the surface charge densities at points A and B differ, there will be a field along
the path between the points. If there is a field, there will be a potential difference
or voltage between the points. The voltage, then, is a measure of the surface
charge density difference between A and B. We can think of it as a measure of
the battery’s ability to “pump” charge to maintain this difference (somewhat as
an ordinary air pump maintains a difference between the volume density of molecules at opposite ends of the pump). Whatever we said of the surface charge
density gradient in Chapter 18 also holds true for voltage. For example, if there
is a nonzero voltage (now replacing the words surface charge density gradient)
between two points on a conductor, there will be a movement of charge (a current) between the points.
y
Again, a gravitational analogy is useful. In Figure 19-19
yA
we retain the convention that the electric and gravita∆Vab and
tional fields are both in the negative y direction. Then the
∆PEgrav
= g∆y
m
gravitational and electrical PE of a body both decrease in
Unit
of
Unit of
yB
+
are both negative
this direction. The change in PE per unit of mass is
charge as body falls toward
mass
¢PEgrav
mg¢y
1 kg
m g¢y; It is a decrease when the object is
“ground” or Earth
∆y = yB – yA
m
falling and ¢y is negative. We usually choose our coordiPEgrav
0
nates so that m 0 when y 0. In the same way in the
“Ground” ( PE
Earth
PE
q = 0)
electrical case, we choose our coordinates so that qelec 0
when y 0. Like ground level in the gravitational case, this Figure 19-19 Change in PE when going from higher to
lower potential in electrical and gravitational systems.
is electrical “ground.”
Analogies can also help in thinking about circuits. For example, as is true for
gravitational PE, the change in electric PE is the same for any path between the
same two points (recall the discussion of Figures 6-17 and 6-18). So it must also
hold true for the voltage, the change in electric PE per unit charge. In electric
circuits, then, we know that any wires or devices connected between the same
two points are subject to the same voltage. We will use this idea extensively in
Chapter 21.
19-7
Batteries
How do batteries produce potential differences? Batteries increase electric PE by
separating opposite charges that are ordinarily drawn toward each other. Where
does this energy come from? STOP&Think Do batteries create the energy? ◆
Recall from Section 6-1 (Figures 6-6 and 6-7 or WebLink 6-2) that chemical
reactions can release energy that had been stored as chemical energy. What occurs
is an energy conversion, not a violation of energy conservation. Common reactions in batteries include the dissolving or ionization to differing degrees of two
different metals immersed in a solution, which is often acidic. In such a case the
two pieces of metal, such as a piece of copper and a piece of zinc, form the
poles of the battery. As the positive ions go into solution, their abandoned outer
electrons remain behind on the metals. Because the two metals ionize to different
0540T_c19_566-597.qxd 9/24/04 17:21 Page 588
EQA
588 ◆ Chapter 19 Electrical Field and Electrical Potential
Milliammeter
0
Zinc
Figure 19-20 Two examples of
batteries. (a) “Pickle power”: The
pickle’s juice is an acid solution. The
deflection of the meter indicates a
small current between the dissimilar
metal poles of this battery. (b) A
12.6-V lead-acid storage battery for
an automobile.
Copper
(b)
(a)
degrees, the net negative charges left behind on the pieces of metal are different.
Typically only one metal (like the copper) dissolves significantly in an acid solution. Random (thermal) motion carries some of the positive ions away from the
dissolving pole. Then if a wire is connected between the two poles, mutual repul➥A note on language: Originally a sion will cause the electrons left behind to flow to the other pole (Figure 19-20).
battery of cells referred to a number Ions from the first pole that make it to this one will bind to its surplus electrons,
of cells used in combination. Today thus adhering to and “plating” the second pole. (Copper plating is done in this
we commonly speak even of a sin- way.)
gle cell as a battery.
In some batteries, the regions near the two poles are separated by a mechanism like a porous barrier, which permits larger ions to pass less readily than
Flow of e–’s through external wire
smaller ions. If the positive and negative ions are of substantially different
sizes, the differential flow of the two kinds of ions through the barrier will
entional curr
v
n
give rise to a difference in charge density across the barrier. In the same
ent
Co
way, the fact that the membranes of living cells are not equally permeable
to different ions can give rise to potential differences across cell membranes.
Figure 19-21 shows typical movements of charge carriers and the
resulting conventional currents for a type of battery called a galvanic or
voltaic cell. Note that although the charge carriers in the external wire are
Flow of – ions
electrons, the carriers within the battery may be ions. Although
More e–’s
Conventional current
Fewer e–’s
the current may be equal everywhere in the circuit, the carrileft behind
left behind
Flow of + ions
by ionization
ers comprising the current can be different in different parts of
by ionization
the circuit.
Metal A dissolves
Porous
Metal B dissolves
The increase in PE from one pole to the other can be
more (more + ions
membrane less (fewer + ions
equated
to an amount of work, commonly described as work
into solution)
or material into solution
due
to
non-electrostatic
forces. Actually, electrostatic forces are
Figure 19-21 Principle features in
indirectly
involved
because
they
govern
chemical
reactions, but no electrostatic
the functioning of a galvanic or
forces
due
to
external
application
of
an
electric
field
are involved.
voltaic cell.
✦SUM M ARY✦
The electrostatic force between point charges q1 and q2 is
described by Coulomb’s law.
body A is placed at that position,
Fon A
Coulomb’s law: The magnitude
Fk
q1 q2
(19-1)
r 212
1k 9 109 N m2C2 2
The direction of the force on either body is toward the other
body if the two have opposite charges, and away if they have
like charges.
strength of
electrical
interaction
between A and its
environment
0qA 0
E
property
of A
(19-4)
properties of
environment
of point where
A is placed
Electric field at a position a distance r from a point charge q
Magnitude:
Ek
0q 0
r2
(19-7)
S
The force per unit charge, or electric field E , is defined
S
by F on A qA E and is a function of position. If a charged
S
Direction: away from q if q positive, toward q if q negative
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EQA
Summary ◆ 589
Because electric fields are vectors, finding the electric field at
a point P due to an array of point charges (Procedure 19-1)
involves vector addition.
For a conductor in electrostatic equilibrium,
• there can be no electric field at any point in its interior;
PEelec k
and
• Rule 1: The direction of the field line at any point (or of
the tangent at that point if the field line is curved) is simply the direction of the total field vector at that point.
• Rule 2: The density of field lines (how closely packed they
are) in a small region about a point is drawn to be proportional to the magnitude of the total field vector at that
point.
• Rule 3: Field lines start and end at charge carriers (or else
go to infinity). They are directed away from positive charge
and towards negative charge.
Because Coulomb’s law and Newton’s universal law of
gravitation have similar forms, we can treat many aspects of
electrical systems like gravitational systems (and vice versa—
we can treat S
g as a gravitational field, map gravitational fields,
etc.). Using the correspondences m 4 q, g 4 E, and G 4 k,
we obtain
PEelec qEy
(19-8)
S
if E constant in the negative y direction
(19-9)
r
for point charges a distance r apart
The electric potential energy per unit charge is called the
electric potential or simply the potential:
• at all points on its surface, the electric field must be perpendicular to the surface.
But if a conducting wire is not in electrostatic equilibrium, the
interior field is in the direction of decreasing surface charge
density (the direction of the negative gradient).
Electric field lines communicate information because
they follow a few basic rules:
q1q2
Potential
PEelec
Ey
Q
(19-11)
(in a region of constant field in the y direction, that is, near
an effectively infinite plane of charge or close enough to a
uniformly charged surface so that it appears flat)
and
Potential
kq
PEelec
r
Q
(19-12)
(at a great enough distance r from a body with charge q so
that the body may be regarded as a point object)
As is true of PE, only differences in potential have physical meaning. The potential difference or voltage between
two points A and B is
VAB a
PEelec
PEelec
¢PEelec
b a
b Q B
Q A
Q
Units: 1 volt 1
11 V 1 JC2
joule
coulomb
(19-14)
(19-15)
The electron volt (eV), a unit of energy, is then defined as
1 eV 11.6 1019 C211 V2 1.6 1019 J
(19-16)
In a constant field (taken to be in the negative y direction)
VAB E¢y
(19-17)
1 ¢y yB yA 2
From this it follows that the electric field is the negative
gradient of the potential. The table below summarizes the conditions in a conductor.











Conditions
Rate of change (in
direction of decrease)
Response of positive charge carriers
to conditions (negative do the opposite)
Negative density gradient
Movement of positive charge in direction of
negative gradient
Conventional current in direction of field
Total change
Underlying picture
Formal concept
Surface charge density
difference
Potential difference
Negative potential
gradient field
Batteries establish potential differences by a combination of
chemical means and mechanisms resulting in unequal flow of
positive and negative ions.
The table below contains some important things to
remember about the quantities we introduced in this chapter.
Vector:
adds by vector addition
Total amount
S
Electrostatic force F
Scalar:
adds by scalar addition
Electric potential energy PEelec
Work done moving charge a distance
¢s in one dimension: ¢W F¢s
Potential difference between endpoints
Amount per unit of charge
S
Electric field E
Electric potential
PEelec
q
of ¢s: Vab ¢W
F¢s
E¢s
q
q
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590 ◆ Chapter 19 Electrical Field and Electrical Potential
✦ Q U A L I TAT I V E A N D Q U A N T I TAT I V E P R O B L E M S ✦
Hands-On Activities and Discussion Questions
The questions and activities in this group are particularly suitable for
in-class use.
Board B
Board A
19-1. Discussion Question.
Exploring an analogy: In Figure 19-22a, the paint from the
painter’s spray can spread out
to board A in a cone-shaped
r
region. If board A is removed,
the cone will continue to
2r
spread to board B, twice as far
from the nozzle of the can.
a. When board A is removed,
(a)
a
how will the total amount of
paint reaching board B each second compare to the total amount
of paint that previously reached
B
A
board A each second?
b. How will the area that gets
r
painted on board B compare
2r
with the area that gets painted on
area B
board A? Express as a ratio area A.
(b)
c. If each board got sprayed for
1 s, how will the thickness of Figure 19-22 Problem
the paint on the two boards (the 19-1
amount of paint on each square
B
centimeter of board) compare? Express as a ratio thickness
thickness A .
d. Now compare the spray paint situation to the situation of
electric field lines directed outward from a positive point
charge in Figure 19-22b. The field lines cross two imaginary
spheres, both centered at the point charge. To which quantities in the paint situation can each of the following be
compared: the areas of the two spheres? the total number
of field lines crossing each sphere? the magnitude of the
electric field? Briefly explain each.
e. How does the magnitude of the electric field compare at
B
the surfaces of the two spheres? Express as a ratio EE at
at A ,
and briefly explain your reasoning.
19-2. Discussion Question.
a. Other than in superconductors, can there ever be a current
between two points A and B if the potential difference VAB
is zero?
b. Figure 18-8 shows the anode and cathode in a cathode ray
tube as two parallel plates, with a hole in the negative cathode that permits electrons to pass through. Assume the
plates are effectively infinite in size compared to the width
of the gap between them. The electron is accelerated from
the negative cathode to the positive anode. Why isn’t the
electron slowed down by attraction to the anode once it
moves to the right of it?
c. Is there a potential difference between the anode and the
screen?
d. Is there an electric field to the right of the anode? What
does Equation 19-16 tell you about your answer to c?
e. Is there a current between the anode and the screen?
f. Are your answers to c and e consistent with your answer
to a? If not, make any necessary corrections.
Review and Practice
Section 19-1 Making Electrostatic Force and Charge
Concepts Quantitative
19-3. Two point charges exert an attractive force of 5.0 N on
each other. How much force would they exert on each other
if the distance between them were
a. ten times as great?
b. one-tenth as great?
c. How would your answers to a and b be altered, if at all,
if the forces the charges exert on each other were repulsive
rather than attractive?
19-4. How many electrons must be brought together to constitute a total charge (ignoring sign) of one coulomb?
19-5.
a. Find the electrostatic force that two electrons exert on each
other when they are 1 mm apart.
b. Compare this with the gravitational force they exert on
each other at the same separation 1me 9.11 1031 kg2.
Felec
Express your comparison as a ratio Fgrav
.
19-6. What must be the charge on each of two identical bodies if they are to exert a repulsive force of 1 N on each other
at a separation of 5 cm?
19-7. SSM WWW A point object with a charge of 1.00 108 C
is placed at the origin.
SSM Solution is in the Student Solutions Manual
a. Where must a proton be positioned to experience a force
of 1.44 1015 N directed to the left?
b. Where must an electron be positioned to experience a force
of 1.44 1015 N directed to the left?
19-8. Three charged point objects A, B, and C are all
placed along the x axis. A is at the origin, and B is at
x 3 cm. A and C have charges of 1.0 106 C; B’s charge
is 1.6 106 C. Find the magnitude and direction of the total
electrostatic force on C if C is placed 1 cm a. to the left
of B. b. to the right of B. c. to the left of A. d. to the right
of A.
19-9. Suppose that in Problem 9-8, object C has a mass of
5.0 1021 kg. What is its acceleration when it is positioned
1 cm to the left of A?
19-10. Charged point objects A and B are placed along the
x axis at x 0.04 m and x 0.04 m, respectively. The
magnitude of the charge on A is double that of B. Where on
the x axis will a positively charged point object experience
zero total force if a. both A and B are positive? b. A is positive
and B is negative?
19-11. Repeat Problem 19-8 for the case in which A and B
remain where they were but C is placed along the y axis at
y 4 cm.
WWW Solution is at http://www.wiley.com/college/touger
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EQA
Qualitative and Quantitative Problems ◆ 591
19-12. At a certain instant, a proton approaching another pro5
2
ton has an acceleration of 4.0 10 ms . What is its acceleration at a later distance when it is only half as far away?
Section 19-2 Electric Fields
19-13. Two charged point objects A and B are placed along
the x axis. A is at the origin and B is at x 3 cm. A’s charge
6
6
is 1.0 10 C, and B’s charge is 1.6 10 C. Find the magnitude and direction of the total electric field due to these
charges at a point 1 cm a. to the left of B. b. to the right of B.
c. to the left of A. d. to the right of A.
19-14. Use the results of Problem 19-13 to find the force on
a point object C placed 1 cm to the right of B if C has a
charge of a. 1.0 106 C. b. 2.0 106 C.
19-15. SSM Consider charges A and B in Problem 19-13.
a. Find the magnitude and direction of the total electric field
due to these charges at the point 1x 0, y 4 cm2. b. Find
the force on an electron placed at this point.
19-16. Find the magnitude and direction of the electric field
at a certain point if a force of 8.0 1011 N to the right is
exerted on a. a sodium ion placed at that point. b. a chloride
ion placed at that point.
19-17. Suppose an oil droplet in the Millikan experiment has
two excess electrons on it. If the total non-electrostatic force
on the droplet is 4.0 1015 N, what electric field (magnitude
and direction) must be applied if the droplet is to descend at
constant velocity?
19-18. A 0.0050-kg ball of metal foil is suspended from an
insulating thread. When a charged rod is held at a fixed distance directly to the left of the ball, the thread hangs at an
angle of 15° to the right of vertical. Calculate the electrostatic
force that the rod exerts on the ball.
19-19. SSM WWW For each of the following questions, either
answer it or tell why it cannot be answered.
a. A point object has a charge of 2 1018 C. Find the magnitude of the electric field at a position 1 m from this object
if there is nothing at that position.
b. Repeat a for the case when there is an electron at the position in question.
c. Two point objects, each with a charge of 0.003 C, are positioned 0.40 m apart. Find the magnitude of the electric field.
19-20. Figure 19-23 shows the positions of two point objects
with equal positive charges Q. Four points in the vicinity of
these objects are labeled A, B, C, and D.
a. Rank these points in order according to the magnitude of
the total electric field at each point. Rank them from least
to greatest, making sure to indicate any equalities.
b. Repeat a for the case when the charge at x 5 cm is negative instead of positive.
c. Suppose now that the charge at x 1 cm is negative and
the charge at x 5 cm is positive. If you listed the points
y (in cm)
in rank order as in a and b, how would the order compare
to the orderings you listed for those two situations?
19-21. A singly charged positive ion is located at the origin
and a singly charged negative ion is located on the x axis at
x 6.00 103 m.
a. Find the electric field (magnitude and direction) at the point
y 1.44 102 m on the y axis.
b. Find the instantaneous acceleration (magnitude and direction) of an electron as it passes through this point.
Section 19-3 Fields Due
to Continuous Charge Distributions
19-22. Consider a square loop of an imaginary material having the peculiar property that any charge placed on it remains
where it is put. Suppose that a certain amount of positive
charge is uniformly distributed over two adjacent sides of this
square and an equal magnitude of negative charge is uniformly
distributed over the other two sides. By symmetry arguments,
determine the direction of the electric field at the center of the
square.
19-23. Figure 19-24 shows part of a closed circuit with the
wires connected to the terminals of the battery greatly
enlarged. The diagram focuses on the surface charge density
on four ring-shaped regions of the wire surfaces, although
there is charge distributed over the remainder of the wire surfaces as well. Ring B has a greater density of positive charge
than ring A; ring C has a greater density of negative charge
than ring D. Point P1 is midway between rings A and B; point
P2 is midway between rings C and D.
a. What is the direction of the electric field at point P1 due to
S
ring A alone? (Call this E A.)
b. What is the direction of the electric field at point P1 due to
S
ring B alone? (Call this E B.2
c. At point P1, what is the direction of the total electric field
S
S
E A E B?
d. The rest of the surface charge distribution on the wire to the
battery’s left can be divided into rings of charge similar to A
and B. What is the direction of the total field at point P1 due
to any one pair of rings? What is the direction of the total
field at point P1 due to all such pairs of rings combined?
e. What is the direction of the electric field at point P2 due to
S
ring C alone? (Call this E C.)
f. What is the direction of the electric field at point P2 due to
S
ring D alone? (Call this E D.)
g. At point P2, what is the direction of the total electric field
S
S
E C E D?
h. The rest of the surface charge distribution on the wire to the
battery’s right can be divided into rings of charge similar to
C and D. What is the direction of the total field at point P2
due to any one pair of rings? What is the direction of the
total field at point P2 due to all such pairs of rings combined?
i. Write a sentence to describe the direction of the electric
field along the entire closed loop of the circuit, external to
the battery.
B
1
A
0
+Q
+
1
C
2
Figure 19-23 Problem 19-20
3
D
4
B
A
+
+
+
+Q
+
5
x (in cm)
+
+
+
P1
++
++ ++
+
+
+
+
++ ++
++
+
–
Battery
Figure 19-24 Problems 19-23 and 19-64
–
––
–
––
C
––
––
D
––
–
–
––
–
–
–
P2
–
–
–
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EQA
592 ◆ Chapter 19 Electrical Field and Electrical Potential
Section 19-4 Picturing Electric Fields
19-24. Figure 19-25 shows
the electric field lines due
to two point objects 1 and
2 having equal positive
charges. If points A and B
are equidistant from object
1, sketch the electric field
vectors at these two points.
1 cm
C
A
+
1 cm
Object 1
1m
C
1m
C
B
Object 2
+
1m
(Adapted from P. Shaffer, G. Fran-
Figure 19-27 Problem 19-29
cis, and L. C. McDermott, paper
abstracted in AAPT Announcer, 22,
no. 2, p. 51 [1992].)
Figure 19-25 Problems 19-24
and 19-25
19-25. Figure 19-25 shows the electric field lines due to two
point objects 1 and 2 having equal positive charges. Points A
and B are equidistant from object 1.
a. Is the magnitude of the field at A less than, equal to, or
greater than the field at B? Briefly explain.
b. If another positively charged point object is placed at point
A and released from rest, will the object pass to the left
of, to the right of, or directly through point C? Briefly
explain.
19-26. Two tiny spheres with equal and opposite charges are
separated by a distance much greater than their diameters.
Carefully sketch the electric field lines representing the total
field due to these charges.
19-27.
a. Carefully sketch the electric field lines due to a thin straight
rod of length L with a charge Q distributed uniformly
along it.
b. Repeat a for a similar rod with a charge 2Q distributed
uniformly along it. What are the significant differences?
19-28. Figure 19-26 shows a metal cylinder in both front and
top views. A negative charge is placed on this cylinder and is
allowed to reach electrostatic equilibrium.
a. Sketch what the electric field due to this cylinder looks like
in the front view.
b. Sketch what the field looks like in the top view.
y
y
z
x
Front view
Section 19-5 Electrical and Gravitational Systems:
Similarities and Differences; Electrical Potential Energy
19-30.
a. In Figure 19-18b, in what part(s) of the electric circuit (if
any) does the conventional current of positive charges flow
oppositely to the direction of the electric field?
b. In Figure 19-18a, in what part(s) of the gravitational circuit
(if any) does mass flow oppositely to the direction of the
gravitational field?
c. Is it gravitational forces or other forces that causes mass to
flow opposite to the gravitational field in b ? Briefly explain.
d. Is it electrostatic (Coulomb) forces or other forces that
causes conventional current to flow opposite to the electric
field in a ? Briefly explain.
19-31. Two electrons are a small distance apart. When we
rearrange this system to increase its electrical potential energy,
does the gravitational potential energy of the system increase,
decrease, or remain the same? Briefly explain why.
19-32. When the distance between a proton and an electron
is doubled, the force on the electron changes in magnitude
from F1 to F2, and the potential energy of the two-particle system changes from PE1 to PE2. The final value of each quantity is a certain fraction its initial value. When we compare the
PE
two fractions, do we find that PE21 is less than, equal to, or
F2
greater than F1? Briefly explain.
19-33. An atom of the most common isotope of hydrogen consists of a single proton and an electron. Suppose the electron
starts out at a distance of 5.29 1011 m from the proton, and
after absorbing radiated energy it ends up at four times that
distance. By how much (in J) does the potential energy of this
atom change?
y
x
in each case. On which of the two test charges does the force
change by a larger percent? Briefly explain.
x
Top view
Figure 19-26 Problem 19-28
19-29. Figure 19-27 shows two conducting objects A and B.
The two objects are charged so that the fields at their surfaces
have equal magnitude. Suppose a small test charge A is placed
at the center of object A and then moved 1 cm to the right.
An identical test charge B is placed at the center of object B
and then likewise moved 1 cm to the right. C marks the center
19-34. The electric field
due to the uniform
plane of charge in Figure 19-28 has a magnitude of 400 N/C. At a
particular
instant,
a
charged particle is in the
position shown and has
a speed of 800 m/s
toward the plate.
a. If the charged particle
is a singly charged
positive ion and the
–
–
–
–
–
–
–
–
–
–
–
–
E
v = 800 m/s
+
7.5 × 10–4 m
Figure 19-28 Problem 19-34
0540T_c19_566-597.qxd 9/24/04 15:55 Page 593
EQA
Qualitative and Quantitative Problems ◆ 593
plane’s charge is negative, find the change in potential
energy as the particle travels from its present position to the
point where it strikes the plane.
b. Find the change in kinetic energy.
c. If the ion has a mass of 2.0 1026 kg, calculate the speed
with which it strikes the plane.
19-35. SSM WWW
a. Calculate the speed with which the ion strikes the plane in
Problem 19-34 if everything is the same except that the
plane’s charge is positive.
b. Calculate the speed with which the ion would strike the
positively charged plane if the ion were negative.
19-36. Write a problem involving gravitational rather than electrostatic forces and fields that is completely analogous to Problem 19-34.
19-37.
a. Find the change in the electrical potential energy of a system of two electrons during an interval in which the distance between them decreases from 3.0 1010 m to
2.0 1010 m.
b. Find the change in the gravitational potential energy of the
system over this same time interval.
19-38. The electric field between the two
oppositely charged plates in Figure 19-29
has a magnitude of 1500 N/C.
a. If a proton is released from point P in
the figure, which plate would it strike
and at what speed?
b. If an electron is released from point P,
which plate would it strike and at what
speed?
19-39. Two protons are instantaneously
at rest at a distance of 5.0 1010 m
from each other. If the protons are never
affected by any objects except each other,
what maximum speed will each of the
protons reach? Hint: How do the speeds
of the two protons compare at any
instant? Why?
6.0 × 10–4 m
+
+
+
–
+
+
+
+
+
+
+
+
+
P
–
–
–
–
–
–
–
–
–
–
–
–
2.0 × 10–4 m
Figure 19-29
Problem 19-38
Section 19-6 Potential and Potential Differences
19-40. In Figure 19-30, the charge Q on the object in the center is 1.00 106 C. Calculate the change in potential in going
B
A
D
C
Q
0.10 0.20 0.30
r (in m)
a. from A to B. b. from C to D. c. from A to D. Briefly tell
what relationship there is between this last answer and your
answers to a and b.
19-41. SSM In Figure 19-30, the charge Q on the object in the
center is 1.00 106 C. In each of the following cases, calculate the change of potential energy and put a sign on each
answer to indicate whether the change is an increase 12 or
a decrease 12. (Hint: Think about whether you can use any
of your results from Problem 19-40.)
a. A proton moves outward from A to B.
b. A proton moves outward from C to D.
c. An electron moves outward from C to D.
d. An electron moves inward from D to A.
19-42. The electric field due to a charged point object has a
magnitude of 22.5 N/C at a distance of 0.020 m from the object.
a. Find the potential due to this object at the same distance.
b. At what distance would the electric field be doubled? At
what distance would the potential be doubled?
19-43. Repeat Example 19-8 for the case in which a proton is
released at the given point and reaches a point 0.020 m from
its starting point.
19-44. When we calculate the electric field by E qF in SI
units, we obtain E in N/C. When we calculate the field from
Vab E¢y in SI units, we obtain E in V/m. Show that N/C
and V/m are really the same units.
19-45. An electron in a hydrogen atom gains 10.2 eV of energy
in going from its lowest energy level to its first excited state.
a. How much is this in joules?
b. How many electrons in a sample of hydrogen gas could be
raised to their first excited states by an input of 1 J of energy?
19-46. How much potential difference must be established
between two parallel plates if an electron leaving one of the
two plates with negligible speed is to acquire a speed of
8.5 104 ms on reaching the opposite plate?
19-47. The potential difference between two parallel plates is
12 V. Find the magnitude of the electric field between the
plates if the plates are 0.0025 m apart.
19-48. If the electric field in Example 19-8 is produced by two
equally but oppositely charged plates separated by a distance
of 0.10 m, what potential difference must be applied between
the plates?
Section 19-7 Batteries
19-49. A student watches the instructor make a battery using
two metal electrodes and an acidic solution. The student later
tries to explain it to a friend who missed class. “If I take these
two pieces of copper, I can use them as my electrodes. Then,
when I insert them into opposite ends of this acid bath, I will
be able to measure a potential difference between the ends of
the electrodes that stick out of the bath.” Will there be a potential difference for the set-up this student is describing? Briefly
explain.
19-50. How does the random motion of molecules (thermal
motion) contribute to the working of a battery?
Figure 19-30 Problems 19-40 and 19-41
19-51. The cell membrane of a living organism is not equally
permeable to different kinds of ions. How might this fact give
rise to a potential difference across the cell membrane?
0540T_c19_566-597.qxd 9/24/04 17:21 Page 594
EQA
594 ◆ Chapter 19 Electrical Field and Electrical Potential
Going Further
The questions and problems in this group are not organized by sec- ••b. If instead, object A passes through the origin with a velocity of 5.0 m/s to the left, estimate the direction in which it
tion heading, so you must determine for yourself which ideas apply.
will be moving 1.0 103 s later.
Some of them will be more challenging than the Review and Practice
questions and problems (especially those marked with a • or ••).
19-57. The “leaves” of a certain electroscope are two very small metallic spheres
19-52. A positive ion is positioned at the origin. An electron
of mass 3.0 103 kg suspended from 0.20
approaches it from the right. As the electron draws closer,
m lengths of fine conducting wire. When
a. does its speed increase, decrease, or remain the same?
the spheres hang as shown in Figure 19-35,
b. does its acceleration increase, decrease, or remain the same?
what is the magnitude of the charge on
each of the spheres (assuming the two
19-53. SSM Two spheres of equal size have charges 2 mC
5.8°
5.8°
spheres are equally charged)?
and 6 mC. Which of the diagrams in Figure 19-31 most accurately shows the electrical forces that the two spheres exert on
19-58. A charged sphere (mass 6.0 each other?
4
10 kg) suspended by a thread hangs at
an angle of 3.6° to the left of the vertical Figure 19-35
(a)
when in the presence of a 650 N/C electric Problem 19-57
+2
+6
field directed to the right. Calculate the
(b)
charge on the sphere.
(c)
(d )
+2
+6
+2
+6
+2
+6
19-59. A parrot is on a dry wooden perch in a metal birdcage.
The cage is struck by lightning. The parrot is unharmed. Why?
19-60. Redraw Figure 19-5c for the case where both rings are
negative but the lower ring is more negative. Sketch the field
contribution due to each ring and the resultant field due to
the combined effect of the two rings.
Figure 19-31 Problem 19-53
y
19-54. Figure 19-32
E
shows five points in
a uniform electric
B
E
field directed toward
the right. Rank these
points according to
C
A
D
each of the following, listing them in
order from least to
greatest and indicat- Figure 19-32 Problem 19-54
ing any equalities:
a. the magnitude of the electric field at each point.
b. the electric potential at each point.
x
E
19-55. SSM Figure 19-33
+
+
shows five points in the
y
+
vicinity of an infinite
E
+
B
plane of uniformly dis+
+
tributed positive charge.
C
D
A
+
Rank these points accord+
ing to each of the fol+
lowing, listing them in
+
order from least to greatest and indicating any Figure 19-33 Problem 19-55
equalities:
a. the magnitude of the electric field at
y
each point.
b. the electric potential at each point.
19-56.
a. In Figure 19-34, all three charges have
equal magnitude. If object A is instantaneously at rest when it is at the origin, in what direction will it move
away from that point?
••19-61.
a. A uniformly charged circular ring with radius
R has a total charge
R
of Q. The ring is positioned perpendicular to
x
the x-axis with its center at the origin (Figure
19-36). Show that at any Figure 19-36 Problem 19-61
position x along the positive x-axis, the magnitude of the electric field is
x
kQx
1x2 R 2 2 3/2
State any assumptions you make in your reasoning.
b. Find the approximate value of E when x is so large that R
is negligibly small in comparison. Can you suggest a reason why you might have expected this value?
x
•19-62. A point object with a small positive charge is placed
between equally and oppositely charged plates. Assume the
point object’s charge has a negligible effect on the configuration of charge on the plates. Which of the sketches in Figure 19-37 most accurately shows possible electric field lines
due to this arrangement of charge? Assume the plate dimensions are large compared to the distance between them, so
that the plates extend beyond the top and bottom of the figure. Note: Only a few field lines are shown in each sketch.
B+
A
+
–
C
Figure 19-34
Problem 19-56
x
+
+
+
+
+
+
+
+
(a)
–
–
–
–
–
–
–
+
+
+
+
+
+
+
+
(b)
Figure 19-37 Problem 19-62
–
–
–
–
–
–
–
+
+
+
+
+
+
+
+
(c)
–
–
–
–
–
–
–
+
+
+
+
+
+
+
+
(d )
–
–
–
–
–
–
–
0540T_c19_566-597.qxd 9/24/04 15:55 Page 595
EQA
Qualitative and Quantitative Problems ◆ 595
You might try including some of the others to see whether the
sketch makes sense.
19-63. Student A says, “When I connect a bulb by two wires
to the terminals of a battery, the electric field that causes
charge to flow through the filament of the bulb is due entirely
to the concentrations of positive and negative charge at the
terminals of the battery. There are no concentrations of charge
on the wire itself; they are not needed to cause the current. I
picture it like this (Figure 19-38a).”
Student B says, “I don’t agree. I think the battery will
cause a change in surface charge density along the length of
the wire, like this (Figure 19-38b). Most of the change in surface charge density takes place over the length of the filament,
because that’s where most of the resistance is.”
The two circuits in Figure 19-38c provide us with a way
of seeing which student’s model is better. The circuits are identical, except that in circuit I, the bulb is very near one terminal of the battery, whereas in circuit II the bulb is fairly distant from either terminal.
a. How should the electric field in the bulb filament compare
in the two circuits if we assume that student A’s model is
correct? How should the brightness of the bulb then compare in the two circuits? Briefly explain.
b. How should the electric field in the bulb filament compare
in the two circuits if we assume that student B’s model is
correct? How should the brightness of the bulb then compare in the two circuits? Briefly explain.
c. In real circuits, does the bulb brightness depend on how
far it is from the battery? (Check this experimentally if you’re
not sure.) Which conclusion agrees with what we actually
observe, the one based on student A’s model or the one
based on student B’s model? (Problem based on comments received
in private communication from Bruce Sherwood.)
Bulb
+++
++
++
++
Bulb
+ +
+ +
+ +
+ +
+ +
––––
––
–
––––
Battery
+ +
+ +
+
+
Almost all of
resistance is
in filament
– – –
– – –
–
–
–
– –
– –
–
–
–
–
–
Battery
19-65. A certain wire has a right-angle bend in it. Along each
straight segment, away from the bend, the surface charge density gradient is constant, so the field is constant along the axis
of the wire. Roughly sketch the configuration of charge that is
required in the region of the bend for the field lines to turn
the corner in this region.
19-66. An electron is given an initial veloc+
ity parallel to a flat plate with a positive
+
charge distributed uniformly over its surface
+
(Figure 19-39). The distance d in the figure
+
e– vo
is very small compared to the dimensions of
+
the plate.
d
+
a. Sketch the trajectory of the electron after it
is given its initial velocity. What is the
+
shape of this trajectory? Why?
Figure 19-39
b. Describe a situation involving only gravita- Problem
tional forces that is analogous to this situ- 19-66
ation. Tell what feature of the gravitational
situation corresponds to each feature of this situation. Does
the moving object in the situation you describe follow the
same shape trajectory as the electron?
19-67.
a. Make a list of the correspondences between features of the
gravitational and electric circuits in Figure 19-18.
b. What aspects of the electric circuit do not have corresponding features in the gravitational circuit?
19-68. How can an electron be made to follow a parabolic
trajectory in the presence of a negatively charged plate? Draw
a sketch to clarify what you describe. Specify any conditions
that you think are required.
19-69. Just as we can speak of non-electrostatic forces exerted
on oppositely charged charge carriers in a battery, we can
speak of the non-electrostatic force per unit of charge as a
non-electrostatic field. Estimate the magnitude of the nonelectrostatic field inside a standard 1.5 V D-cell battery and
give its direction (toward which pole).
•19-70. A simple model of a hydrogen atom consists of an
electron of charge e and mass me held in a circular orbit of
radius r by a proton of charge e and mass mp. Show that
2
(a) Student A’s model
Circuit I
(b) Student B’s model
Circuit II
(c) Testing the models:
Figure 19-38 Problem 19-63
19-64.
a. In Figure 19-24, what is the direction of the electric field
associated with electrostatic force inside the battery?
b. In what direction does conventional current flow through
the battery?
c. Under what circumstances, if any, can the electric field associated with electrostatic forces and the conventional current
be in opposite directions?
e
the total energy E of this arrangement is E k 2r
.
19-71. The DNA molecule, often described as a double helix,
can be pictured as a twisted ladder. The two long strands that
wind around each other are connected by bonds between pairs
of bases, one base on each strand. These are the “rungs” of
the ladder. People studying DNA often look at DNA fragments,
which are themselves basically chains of these base pairs. DNA
fragments in a neutral pH solution are negatively charged;
there will be two excess electrons per base pair. If a 0.1 micron
11.0 107 m2 DNA fragment has a charge of 9.0 1017 C,
how many base pairs are there along this fragment?
19-72. Hemoglobin is the protein molecule found in red
blood cells. Under suitable conditions, the ratio of charge q to
drag coefficient z for hemoglobin molecules in solution is
q
10 C m
z 2.0 10
N s . An electric field of 1000 N/C is applied
to the solution in an electrophoresis apparatus.
a. What terminal speed is attained by the hemoglobin molecules?
b. About how far will the hemoglobin molecules travel in a day?
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596 ◆ Chapter 19 Electrical Field and Electrical Potential
19-74. An electric field of 150 N/C is maintained between two
equally and oppositely charged plates with a distance of
1.2 102 m between them. The region between the plates is
evacuated, and the negative plate is sufficiently heated so that
it occasionally emits electrons.
a. If an electron is emitted from the negative plate at negligible speed, with what speed does it reach the positive
plate?
b. How long does it take the electron to reach the positive plate?
19-75. SSM A potential difference of 120 V is maintained
between two equally and oppositely charged plates. The
region between the plates is evacuated, and the negative
plate is sufficiently heated so that it occasionally emits electrons. Answer whichever of the following questions it is possible to answer. If a question cannot be answered, explain
why not.
a. If an electron is emitted from the negative plate at negligible speed, with what speed does it reach the positive
plate?
b. How long does it take the electron to reach the positive plate?
19-76. Two lengths of identical resistance wire are each connected between the terminals of a 1.5 V D-cell battery. (The
wires are said to be connected in parallel.) Length A is 25 cm
long, and length B is 50 cm long.
a. How does the voltage between the ends of length A compare to the voltage between the ends of length B?
b. How does the electric field in length A compare to the electric field in length B?
c. How does the resistance of length A compare to the resistance of length B?
d. How does the resistivity of length A compare to the resistivity of length B?
e. How does the current through length A compare to the current through length B?
19-77. Two lengths of identical resistance wire are each connected between the terminals of a 9.0 V battery. Length A is
0.30 m long, and length B is 0.90 m long. Calculate the electric field in each wire.
19-78. We have spoken of setting up a uniform charge density gradient along the surface of a wire. Each of the choices
below shows the surface charge density (in C/m2) along the
surface of a wire. Which one shows a uniform charge density
gradient?
a. 2
2
2
2
0
2
2
2
1
4
1
8
1
16
2
b. 8
4
2
1
1
2
c. 6
5
4
3
2
1
0
1
2
d. 2
2
2
2
2
2
2
2
2
19-79. We have spoken of setting up a uniform charge density gradient along the surface of a wire. In each of the graphs
in Figure 19-40, surface charge density is plotted against distance along the wire. Which one of these graphs shows a uniform, nonzero charge density gradient?
Surface charge density plotted vertically,
distance along wire plotted horizontally
0
0
0
0
B
A
C
D
Figure 19-40 Problem 19-79
19-80.
a. From point B in Figure 19-9, in what direction would you
travel to experience the potential increasing most rapidly
with position?
b. The direction you found in a is defined to be the direction
of the potential gradient at point B. How does the direction of the electric field at B compare with the direction of
the potential gradient?
19-81. A proton and an electron are a distance r apart. If the
PEelec
distance r is increased, does the ratio PEgrav
increase, decrease,
or remain the same? Briefly explain.
19-82. Figure 19-41 shows an infinite plane of uniformly distributed negative charge. The arrows in the figure show four
possible paths that can be traveled in the vicinity of the plane.
Each path starts at the tail of the arrow and ends at the head
of the arrow. Rank these paths according to each of the following, listing them in order from least to greatest and indicating any equalities:
a. the change in potential from the beginning to the end of
the path.
b. the change in electrical potential energy that would occur if
an electron traveled from the beginning to the end of the
path.
y
A
B
C
D
–––––––––––––––––––––––––––
Figure 19-41 Problem 19-82
19-83. Gravitational and electrical systems compared. The electrical system in Figure 19-42a consists of two wires connected
between the terminals of a battery. Wire 2 is twice as long as
wire 1. In the gravitational system in Figure 19-42b, an object
at point P can reach the ground by falling vertically downward
+ –
Wire 1 (L)
P
Path 1 (L)
19-73. An electron passes a point where the electric field is
directed toward the right and has a magnitude of 300 N/C. If
the electron interacts only with the sources of this field, find
the magnitude and direction of the electron’s acceleration at the
instant it passes through this point.
Pat
h
2(
2L
)
Wire 2 (2L)
Ground
(a) Electrical system
(b) Gravitational system
Figure 19-42 Problem 19-83
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EQA
Qualitative and Quantitative Problems ◆ 597
(path 1) or sliding down a frictionless ramp (path 2). Path 2 is
twice as long as path 1.
a. In the electrical system, suppose an electron travels from
the negative to the positive pole of the battery. Compare
the loss of potential energy when it travels by wire 1 and
when it travels by wire 2.
b. In the electrical system, compare the change in potential
between the two terminals of the battery along wire 1 and
along wire 2.
c. In the electrical system, compare the electrical field along
wire 1 to the electrical field along wire 2.
d. In the gravitational system, suppose an object descends
from point P to the ground. Compare the loss of potential
energy when it descends by path 1 and when it descends
by path 2.
e. Compare the accelerations that the object would experience
along the two paths.
f. In the gravitational system, compare the gravitational field
along path 1 to the gravitational field along path 2.
g. Do the two systems behave analogously?
P r o b l e m s o n We b L i n k s
19-84. In WebLink 19-1, if we replace charge carrier A with a
charge carrier with twice as much charge,
a. will the magnitude of the electric field at the origin increase,
decrease, or remain the same?
b. will the magnitude of the force on the charge carrier at the
origin increase, decrease, or remain the same?
19-85. In WebLink 19-1, if we replace charge carrier A with a
charge carrier with a charge of opposite sign, will the electric
field at the origin change direction? Briefly explain.
19-86. Suppose that when a sphere of radius 1 m is centered
at the positive point charge in WebLink 19-2, 720 field lines
in all cross the surface of the sphere. If this sphere is now
replaced by a sphere of radius 2 m centered at the same point
charge, how many field lines in all cross the surface of the
new sphere?
19-87. Suppose that when a sphere of radius 1 m is centered
at the positive point charge in WebLink 19-2, 60 field lines
cross each square unit of the sphere’s surface. If this sphere is
now replaced by a sphere of radius 2 m centered at the same
point charge, how many field lines cross each square unit of
the new sphere’s surface?
19-88. In WebLink 19-3, does the electric field increase,
decrease, or remain the same as you go from a. point A to
point B? b. point B to a point very far to the right of point B?
c. point A to a point in the interior of the charged plate?