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Physics For Scientists And Engineers 6E - part 192

SECTION 25.2 • Potential Differences in a Uniform Electric Field

765

For instance, an electron in the beam of a typical television picture tube may have a
speed of 3.0 % 10

m/s. This corresponds to a kinetic energy of 4.1 % 10

J, which is

equivalent to 2.6 % 10

eV. Such an electron has to be accelerated from rest through a

potential difference of 2.6 kV to reach this speed.

▲

PITFALL PREVENTION

25.3 The Electron Volt

The  electron  volt  is  a  unit  of
energy,  NOT  of  potential.  The
energy  of  any  system  may  be
expressed  in  eV,  but  this  unit  is
most  convenient  for  describing
the  emission  and  absorption  of
visible light from atoms. Energies
of  nuclear  processes  are  often
expressed in MeV.

Quick Quiz 25.1

In Figure 25.1, two points A and B are located within a

region in which there is an electric field. The potential difference $V # V

(a) positive (b) negative (c) zero.

Quick Quiz 25.2

In Figure 25.1, a negative charge is placed at A and then

moved to B. The change in potential energy of the charge–field system for this process
is (a) positive (b) negative (c) zero.

Figure 25.1 (Quick Quiz 25.1)

Two points in an electric field.

25.2 Potential Differences in a Uniform

Electric Field

Equations 25.1 and 25.3 hold in all electric fields, whether uniform or varying, but they
can be simplified for a uniform field. First, consider a uniform electric field directed
along the negative y axis, as shown in Figure 25.2a. Let us calculate the potential differ-
ence between two points A and B separated by a distance

#s# # d, where s is parallel to

the field lines. Equation 25.3 gives

Because E is constant, we can remove it from the integral sign; this gives

(25.6)

The negative sign indicates that the electric potential at point B is lower than at point
A; that is, V

Electric field lines always point in the direction of decreasing

electric potential, as shown in Figure 25.2a.

V # !E

s # !E

# $

V # !

E"d

s # !

cos

0')d

s # !

E ds

(a)

(b)

Figure 25.2 (a) When the electric field E is directed downward, point B is at a lower

electric potential than point A. When a positive test charge moves from point A to

point B, the charge–field system loses electric potential energy. (b) When an object of

mass m moves downward in the direction of the gravitational field g, the object–field

system loses gravitational potential energy.

Potential difference between

two points in a uniform electric

field

Now suppose that a test charge q

moves from A to B. We can calculate the change

in the potential energy of the charge–field system from Equations 25.3 and 25.6:

(25.7)

From this result, we see that if q

is positive, then $U is negative. We conclude that

system consisting of a positive charge and an electric field loses electric poten-
tial energy when the charge moves in the direction of the field. This means that
an electric field does work on a positive charge when the charge moves in the direction
of the electric field. (This is analogous to the work done by the gravitational field on a
falling object, as shown in Figure 25.2b.) If a positive test charge is released from rest
in this electric field, it experiences an electric force q

E in the direction of E

(downward in Fig. 25.2a). Therefore, it accelerates downward, gaining kinetic energy.
As the charged particle gains kinetic energy, the charge–field system loses an
equal amount of potential energy. This should not be surprising—it is simply conser-
vation of energy in an isolated system as introduced in Chapter 8.

If q

is negative, then $U in Equation 25.7 is positive and the situation is reversed:

A system consisting of a negative charge and an electric field gains electric po-
tential energy when the charge moves in the direction of the field. If a negative
charge is released from rest in an electric field, it accelerates in a direction opposite
the direction of the field. In order for the negative charge to move in the direction of
the field, an external agent must apply a force and do positive work on the charge.

Now consider the more general case of a charged particle that moves between A

and B in a uniform electric field such that the vector

s is not parallel to the field lines,

as shown in Figure 25.3. In this case, Equation 25.3 gives

(25.8)

where again we are able to remove

E from the integral because it is constant. The

change in potential energy of the charge–field system is

(25.9)

Finally, we conclude from Equation 25.8 that all points in a plane perpendicular to a

uniform electric field are at the same electric potential. We can see this in Figure 25.3,
where the potential difference V

is equal to the potential difference V

(Prove this to yourself by working out the dot product

E ! s for s

A:B

, where the angle (

between

E and s is arbitrary as shown in Figure 25.3, and the dot product for s

A:C

where ( # 0.) Therefore, V

The name equipotential surface is given to any

surface consisting of a continuous distribution of points having the same electric
potential.

The equipotential surfaces of a uniform electric field consist of a family of parallel

planes that are all perpendicular to the field. Equipotential surfaces for fields with
other symmetries are described in later sections.

U # q

V # !q

E"s

V # !

E"d

s # !

E"s

U # q

V # !q

766

C H A P T E R 2 5 • Electric Potential

Figure 25.3 A uniform electric

field directed along the positive

x axis. Point B is at a lower electric

potential than point A. Points

B and C are at the same electric

potential.

9 V

8 V

7 V

6 V

Figure 25.4 (Quick Quiz 25.3)

Four equipotential surfaces.

Quick Quiz 25.3

The labeled points in Figure 25.4 are on a series of equipo-

tential surfaces associated with an electric field. Rank (from greatest to least) the work
done by the electric field on a positively charged particle that moves from A to B ; from
B to C ; from C to D ; from D to E.

Quick Quiz 25.4

For the equipotential surfaces in Figure 25.4, what is

the approximate direction of the electric field? (a) Out of the page (b) Into the page
(c) Toward the right edge of the page (d) Toward the left edge of the page (e) Toward
the top of the page (f) Toward the bottom of the page.

Change in potential energy

when a charged particle is

moved in a uniform electric field

SECTION 25.2 • Potential Differences in a Uniform Electric Field

767

Example 25.1 The Electric Field Between Two Parallel Plates of Opposite Charge

uniform. (This assumption is reasonable if the plate separa-
tion is small relative to the plate dimensions and if we do not
consider locations near the plate edges.) Find the magnitude
of the electric field between the plates.

Solution The  electric  field  is  directed  from  the  positive
plate (A) to the negative one (B), and the positive plate is at
a  higher  electric  potential  than  the  negative  plate  is.  The
potential  difference  between  the  plates  must  equal  the
potential  difference  between  the  battery  terminals.  We  can
understand this by noting that all points on a conductor in
equilibrium are at the same electric potential

; no potential

difference exists between a terminal and any portion of the
plate to which it is connected. Therefore, the magnitude of
the electric field between the plates is, from Equation 25.6,

The configuration of plates in Figure 25.5 is called a

parallel-plate capacitor, and is examined in greater detail in
Chapter 26.

4.0 % 10

V/m

E #

# V

12 V

0.30 % 10

A battery produces a specified potential difference $V
between conductors attached to the battery terminals. A 12-V
battery is connected between two parallel plates, as shown in
Figure 25.5. The separation between the plates is d # 0.30 cm,
and we assume the electric field between the plates to be

2.8 % 10

m/s

√

2(1.6 % 10

C)(!4.0 % 10

1.67 % 10

v #

√

(2e

)

(

0) ) e

V # 0

$K ) $U # 0

–

∆V = 12 V

Figure 25.5 (Example 25.1) A 12-V battery connected to two

parallel plates. The electric field between the plates has a mag-

nitude given by the potential difference $V divided by the plate

separation d.

Example 25.2 Motion of a Proton in a Uniform Electric Field

Solution The charge–field system is isolated, so the mech-
anical energy of the system is conserved:

A proton is released from rest in a uniform electric field that
has a magnitude of 8.0 % 10

V/m (Fig. 25.6). The proton

undergoes a displacement of 0.50 m in the direction of

(A)

Find the change in electric potential between points A

and B.

Solution Because the positively charged proton moves in
the direction of the field, we expect it to move to a position
of lower electric potential. From Equation 25.6, we have

–

= 0

Figure 25.6 (Example 25.2) A proton accelerates from A to B

in the direction of the electric field.

What If?

What if the situation is exactly the same as that

shown in Figure 25.6, but no proton is present? Could both
parts (A) and (B) of this example still be answered?

The electric field vanishes within a conductor in electrostatic equilibrium; thus, the path integral

between any two points in the conductor must be zero. A more complete discussion of this point is
given in Section 25.6.

Interactive

4.0 % 10

V # !Ed # !(8.0 % 10

V/m)(0.50 m) #

(B)

Find the change in potential energy of the proton–field

system for this displacement.

Solution Using Equation 25.3,

The negative sign means the potential energy of the system
decreases as the proton moves in the direction of the elec-
tric field. As the proton accelerates in the direction of the
field, it gains kinetic energy and at the same time the system
loses electric potential energy.

(C)

Find the speed of the proton after completing the

0.50 m displacement in the electric field.

6.4 % 10

# (1.6 % 10

C)(!4.0 % 10

U # q

V # e

25.3 Electric Potential and Potential Energy

Due to Point Charges

In Section 23.4 we discussed the fact that an isolated positive point charge q produces
an electric field that is directed radially outward from the charge. To find the electric
potential at a point located a distance r from the charge, we begin with the general
expression for potential difference:

where A and B are the two arbitrary points shown in Figure 25.7. At any point in
space, the electric field due to the point charge is

(Eq. 23.9), where is

a unit vector directed from the charge toward the point. The quantity

E " ds can be

expressed as

Because the magnitude of is 1, the dot product " d

s # ds cos (, where ( is the angle

between and d

s. Furthermore, ds cos ( is the projection of ds onto r; thus, ds cos ( #

dr. That is, any displacement d

s along the path from point A to point B produces a

change dr in the magnitude of

r, the position vector of the point relative to the charge

creating the field. Making these substitutions, we find that

E " ds # (k

q/r

)dr; hence,

the expression for the potential difference becomes

(25.10)

This equation shows us that the integral of

E " ds is independent of the path between

points A and B. Multiplying by a charge q

that moves between points A and B, we see

that the integral of q

E " ds is also independent of path. This latter integral is the

work done by the electric force, which tells us that the electric force is conservative
(see Section 8.3). We define a field that is related to a conservative force as a

conserv-

ative field. Thus, Equation 25.10 tells us that the electric field of a fixed point charge
is conservative. Furthermore, Equation 25.10 expresses the important result that the
potential difference between any two points A and B in a field created by a point
charge depends only on the radial coordinates r

and r

. It is customary to choose the

reference of electric potential for a point charge to be V # 0 at r

# *

. With this

reference choice, the electric potential created by a point charge at any distance r
from the charge is

(25.11)

V # k

q
r

# !

rˆ

E"d

s # k

rˆ"d

rˆ

E #

rˆ/r

# !

E"d

768

C H A P T E R 2 5 • Electric Potential

charge.  Part  (B)  of  the  example  would  be  meaningless  if
the  proton  is  not  present.  A  change  in  potential  energy  is
related  to  a  change  in  the  charge–field  system.  In  the
absence of the proton, the system of the electric field alone
does not change.

Answer Part (A) of the example would remain exactly the
same  because  the  potential  difference  between  points  A
and  B is  established  by  the  source  charges  in  the  parallel
plates.  The  potential  difference  does  not  depend  on  the
presence  of  the  proton,  which  plays  the  role  of  a  test

At the Interactive Worked Example link at http://www.pse6.com, you can predict and observe the speed of the proton as it
arrives at the negative plate for random values of the electric field.

Figure 25.7 The potential differ-

ence between points A and B due to

a point charge q depends only on the

initial and final radial coordinates r

and r

. The two dashed circles rep-

resent intersections of spherical

equipotential surfaces with the page.

▲

PITFALL PREVENTION

25.4 Similar Equation

Warning

Do  not  confuse  Equation  25.11
for  the  electric  potential  of  a
point  charge  with  Equation  23.9
for  the  electric  field  of  a  point
charge.  Potential  is  proportional
to  1/r,  while  the  field  is  propor-
tional  to  1/r

. The effect of a

charge on the space surrounding
it  can  be  described  in  two  ways.
The charge sets up a vector elec-
tric  field  E,  which  is  related  to
the  force  experienced  by  a  test
charge placed in the field. It also
sets up a scalar potential V, which
is related to the potential energy
of  the  two-charge  system  when  a
test charge is placed in the field.

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