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.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
 Summer '08
 Hoston,Amahd,Bakanowski
 Physics, Charge, Energy, Potential Energy, Work, Electric charge

Click to edit the document details