Chapter 16
Electrical Energy and
Capacitance
Problem Solutions
16.1
(a)
The work done is
(
)
cos
cos
W
F s
qE
s
θ
θ
=
⋅
=
⋅
, or
(
)
(
)
(
)
19
2
19
1.60
10
C
200 N C
2.00
10
m
cos0
6.40
10
J
W



=
×
×
° =
×
(b)
The change in the electrical potential energy is
19
6.40
10
J
e
PE
W

Δ
= 
= 
×
(c)
The change in the electrical potential is
19
19
6.40
10
J
4.00 V
1.60
10
C
e
PE
V
q

Δ

×
Δ
=
=
=

×
16.3
The work done by the agent moving the charge out of the cell is
(
)
(
)
(
)
19
3
20
J
1.60
10
C
90
10
1.4
10
J
C
input
field
e
W
W
PE
q
V



= 
=  Δ
= +
Δ
=
×
+
×
=
×
16.5
6
2
25 000 J C
1.7
10 N C
1.5
10
m
V
E
d

Δ
=
=
=
×
×
16.8
From conservation of energy,
(
)
(
)
2
2
1
0
or
2
f
f
q
V
mv
q
V
v
m
Δ

=
Δ
=
(a)
For the proton,
(
)
(
)
19
5
27
2
1.60
10
C
120 V
1.52
10 m s
1.67
10
kg
f
v


×

=
=
×
×
35
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
36
CHAPTER 16
(b)
For the electron,
(
)
(
)
19
6
31
2
1.60
10
C
120 V
6.49
10 m s
9.11
10
kg
f
v



×
+
=
=
×
×
16.12
1
2
1
2
1
2
e
q
q
V
V
V
k
r
r
=
+
=
+
where
1
0.60 m
0
0.60 m
r
=

=
, and
2
0.60 m
0.30 m
0.30 m.
r
=

=
Thus,
2
9
9
9
2
2
N
m
3.0
10
C
6.0
10
C
8.99
10
2.2
10
V
C
0.60 m
0.30 m
V


⋅
×
×
=
×
+
=
×
16.15
(a)
2
9
9
9
2
N
m
5.00
10
C
3.00
10
C
8.99
10
103 V
C
0.175 m
0.175 m
e
i
i
i
k q
V
r


=
⋅
×
×
=
×

=
∑
(b)
(
) (
)
2
12
9
9
2
7
9
2
5.00
10
C
3.00
10
C
N m
8.99
10
3.85
10
J
C
0.350 m
e
i
k q q
PE
r



=
×

×
⋅
=
×
=

×
The negative sign means that
positive work must be done
to separate the charges
(that is, bring them up to a state of zero potential energy).
This is the end of the preview.
Sign up
to
access the rest of the document.
 Capacitance, Energy, Potential Energy, Work, electrical energy

Click to edit the document details