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1
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print-out
should
have 22 questions.
Multiple-choice
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:

gilvin (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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