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gilvin (jg47854)  8: The Electric Potential  meyers  (21235)
1
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printout
should
have 22 questions.
Multiplechoice
questions
may
continue
on
the next
column or
page
 find
all choices
before answering.
so the magnitude of the potential difference is
V
= E d =
d
Δ
σ
ϵ
0
This homework is due Tuesday, March 1, at
(
midnight Tucson time.
=
1.1 × 10
−10
C/m
2
) (0.011 m)
8.85419 × 10
−12
C
2
/N · m
2
001
10.0 points
To recharge a 12 V battery, a battery charger
must move
2.4 × 10
5
C of charge from the
negative terminal to the positive terminal.
How
much
work
is
done
by
the
battery
charger?
Correct answer: 2.88 × 10
6
J.
Explanation:
Given :
q = 2.4 × 10
5
C
and
V
= 12 V .
The potential difference is
V
= W
q
,
W
= q V
= (2.4 × 10
5
C) (12 V)
=
2.88 × 10
6
J
.
002
10.0 points
Two flat conductors are placed with their in
ner faces separated by 11 mm.
If the surface charge density on one of the
inner faces is 110 pC/m
2
and the other inner
face −110 pC/m
2
, what is the magnitude of
the electric potential difference between the
two conductors?
Correct answer: 0.136658 V.
Explanation:
Let:
ϵ
0
= 8.85419 × 10
−12
C
2
/N · m
2
,
σ = 110 pC/m
2
= 1.1 × 10
−10
C/m
2
,
and d
= 11 mm = 0.011 m .
The electric field between two flat conduc
tors is
=
0.136658 V
.
003
10.0 points
A voltmeter indicates that the difference in
potential between two plates is
53
V.
The
plates are 0.36 m apart.
What electric field intensity exists between
them?
Correct answer: 147.222 N/C.
Explanation:
Let :
V
= 53 V
and
d = 0.36 m .
The potential difference is
V
= E d
E = V
d
=
53 V
0.36 m
=
147.222 N/C
.
004
10.0 points
Two parallel conducting plates are connected
to
a
constant
voltage
source.
The
magni
tude of the electric field between the plates is
2022 N/C.
If the voltage is quadrupled and the dis
tance between the plates is reduced to
1
5
the
original distance,
what is the magnitude of
the new electric field?
Correct answer: 40440 N/C.
E =
σ
ϵ
0
Explanation:
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View Full Documentgilvin (jg47854)  8: The Electric Potential  meyers  (21235)
2
Q is the total
Let :
E = 2022 N/C ,
charge inside
R
the sphere.
V
′
= 4 V ,
and
d
′
= 1
5 d .
The electric field between two parallel con
ducting plates is E
=
−V
d , where V
is the
p
Using Gauss’s Law, you can show that for
ρ r
voltage between the plates, and d is the dis
tance between the plates, so the new electric
field has a magnitude of
(
)
r < R the electric field is given by E =
,
3
ϵ
0
where ρ is the charge density.
The charge density ρ is
E
′
= −V
′
d
′
= 20 E
= − 4 V
= 20
−V
d
d
1. ρ =
Q
5
π R
3
.
Q
= 20 (2022 N/C)
=
40440 N/C
.
005
10.0 points
When you touch a friend after walking across a
rug on a dry day, you typically draw a spark of
about 2 mm.
2. ρ =
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 Spring '11
 doc
 Electric Potential, Magnetism, Work

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