Chapter 22 Practice Pr solns:
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.
W
q V
μ
= ∆
=
=
(
Note:
Since only magnitudes are needed in this problem, we omitted the subscripts
A
and
B
.)
17.
D
EVELOP
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

 .
U
q V
∆
=
∆
E
VALUATE
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
K
e
V


∆
= 
∆
=
=
×
=
×
A
SSESS
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
U
q V
∆
=
=
∆
hence
AB
V
∆
=
45 J/15 mC
3 kV.
=
(Since the work required to move the charge from
A
to
B
is positive,
and
B
A
AB
V
V
V
∆
is positive.)
19.
I
NTERPRET
This problem is about conversion of units.
D
EVELOP
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).
=
⋅
E
VALUATE
Combining the two expressions gives
1 V
1 J/C
1 N m/C.
=
=
⋅
It follows that
1 V/m
1 N/C.
=
A
SSESS
These are the units for the electric field strength.
20.
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 Summer '08
 Hoston,Amahd,Bakanowski
 Physics, Charge, Energy, Potential Energy, Work, Electric charge

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