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22.1
ELECTRIC POTENTIAL
22
EXERCISES
Section 22.1 Electric Potential Difference
16.
The potential difference and the work per unit charge, done by an external agent, are equal in magnitude, so
(50 C)(12 V)
600 J.
WqV
μ
=Δ =
=
(
Note:
Since only magnitudes are needed in this problem, we omitted the
subscripts
A
and
B
.)
17.
INTERPRET
This problem is about the energy gained by an electron as it moves through a potential difference
.
V
Δ
DEVELOP
We assume that the electron is initially at rest. When released from the negative plate, it moves toward
the positive plate, and the kinetic energy gained is

Uq
V
.
Δ =Δ
EVALUATE
As the electron moves from the negative side to the positive side (i.e., against the direction of the
electric field), the
kinetic
energy it gains is
19
17
(
)

120 eV
(1.6 10
C)(120 V)
1.92 10
J
Ke
V
−−
Δ=−Δ=
= ×
=
×
ASSESS
Moving a negative charge through a positive potential difference is like going downhill; potential energy
decreases. However, the kinetic energy of the electron is increased.
18.
The work done by an external agent equals the potential energy change,
45 J
,
AB
AB
V
Δ
==
Δ
hence
AB
V
Δ
=
45 J/15 mC
3 kV.
=
(Since the work required to move the charge from
A
to
B
is positive,
and
B
AA
B
V
>Δ
VV
is
positive.)
19.
INTERPRET
This problem is about conversion of units.
DEVELOP
By definition,1 volt
1 joule/coulomb
=
(1 V
1 J/C).
=
On the other hand,1
joule
1 newtonmeter
=
(1 J
1 N m).
=⋅
EVALUATE
Combining the two expressions gives1 V
1 J/C
1 N m/C.
=
It follows that1V
/m 1N/C.
=
ASSESS
These are the units for the electric field strength.
20.
For
in the direction of a uniform electric field, Equation 22.2b gives
.
r
Δ
K

(650 N/C)(1.4 m)
910 V
VE
r
Δ =Δ=
=
K
(See note in solution to Exercise 16. Since
,
dV
E dr
=
−⋅
G
K
the potential always decreases in the direction of the
electric field.)
21.
INTERPRET
This problem is about the work done by the battery to move the charge from the positive terminal to
the negative terminal.
DEVELOP
The work done by the battery is equal to the kinetic energy gained by the charge, and is equal to
Wq
V
=Δ
.
EVALUATE
Substituting the values given, we have
(3.1 C) (9.0 V)
27.9 J
=Δ=
=
ASSESS
The charged particle gains kinetic energy as it moves toward the negative plate (in the direction of the
electric field). The battery is needed to maintain the potential difference between the plates.
22.
The energy gained is
(see Example 22.1). The proton and singlyionized helium atom have charge
e
, so they
gain100
while the
qV
Δ
19
17
eV
(1.6 10
C)(100 V)
1.6 10
J,
=×
=
×
particle
α
has charge 2
e
and gains twice this energy.
Section 22.2 Calculating Potential Difference
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View Full Document22.2
Chapter 22
23.
INTERPRET
In this problem we are given a uniform electric field and asked to calculate the potential difference
between two points.
DEVELOP
For a uniform field, the potential difference between two points
a
and
b
is given by Equation 22.1b:
ab
b
a
VV
V
E
r
Δ
=−=
−
⋅
Δ
G
G
where
is a vector from
a
to
b
.
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 Fall '10
 Pheong

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