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PROBLEM SOLUTIONS
16.1
(a)
Because the electron has a negative charge, it experiences a force in the direction opposite
to the ﬁ
eld and, when released from rest, will move in the negative
x
direction. The work
done on the electron by the ﬁ
eld is
WF x q
E x
xx
=
( ) = ( ) =− ×
( )( )
−
−
ΔΔ
1 60
10
375
3
19
..
C
N
C
20
10
1 92
10
21
8
×
( )
=×
−−
m
J
.
(b)
The change in the electric potential energy is the negative of the work done on the particle
by the ﬁ
eld. Thus,
Δ
PE
W
=−
= −
×
−
192 10
18
.
J
continued on next page
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Chapter 16
(c)
Since the Coulomb force is a conservative force, conservation of energy gives
∆∆
KE
PE
+=
0, or
KE
m
PE
PE
fe
f
=−
=
−
v
2
20
, and
v
f
e
PE
m
=
−
( )
=
−−
×
( )
×
−
−
2
2
1 92 10
911 10
18
31
∆
.
.
J
kg
=
×
−
205 10
6
.
m s in the
direction
x
16.2
(a)
The change in the electric potential energy is the negative of the work done on the particle
by the F
eld. Thus,
∆
∆
PE
W
qE
x
qE
y
qx
xy
( ) + ( )
⎡
⎣
⎤
⎦
( )
+×
0
5 40
10
.
−
−
( )
+
( )
−×
( )
⎡
⎣
⎤
⎦
=+
×
62
327
32 0
10
5 65 10
C
N C
m
..
−
4
J
(b)
The change in the electrical potential is the change in electric potential energy per unit
charge, or
∆
∆
V
PE
q
==
×
−
−
565 10
10
105
4
.
J
+5.40
C
V
6
16.3
The work done by the agent moving the charge out of the cell is
WW
P
E
q
V
e
input
field
C
=− −
( )=+ ( )
=×
−
160 10
19
.
( )
⎛
⎝
⎜
⎞
⎠
⎟
90
10
1 4
10
32
0
J
C
J
.
16.4
PE
q
V
q V
V
ef
i
=
( )
( )
, so
q
PE
VV
e
fi
=
−
=
×
−
−
∆
192 10
320 10
17
.
.
J
+60.0 J C
19
C
16.5
E
V
d
×
−
∆
25 000
15 10
17 10
2
6
JC
m
NC
.
.
16.6
Since potential difference is work per unit charge
∆
V
W
q
=
, the work done is
WqV
=
( )
( )
+
( )
∆
3 6
10
12
4 3 10
56
C
J C
J
16.7
(a)
E
V
d
×
−
∆
600
533 10
113 10
3
5
m
.
.
(b)
Fq
E
×
( )
×
( )
1 60 10
1 13 10
1 80 10
19
5
14
.
CN
C
N
(c)
WF
s
=⋅
( )
−
( ) ×
⎡
cos
.
θ
180 10
533 290
10
14
3
N
m
⎣
⎤
⎦
−
cos
.
0
4 38 10
17
°
J
Electrical Energy and Capacitance
61
16.8
(a)
Using conservation of energy,
ΔΔ
KE
PE
+=
0, with
KE
f
=
0 since the particle is
“stopped,” we have
PE
KE
m
i
=−
−
⎛
⎝
⎜
⎞
⎠
⎟
=+
×
(
−
0
1
2
1
2
911 10
23
1
e
kg
v
.
)
×
( )
×
−
2 85 10
3 70 10
7
2
16
..
ms
J
The required stopping potential is then
Δ
Δ
V
PE
q
==
+×
−×
×
−
−
370 10
160 10
231 10
16
19
.
.
.
J
C
3
231
V
kV
.
(b)
Being more massive than electrons, protons traveling at the same initial speed will have
more initial kinetic energy and require a greater magnitude stopping potent
ial .
(c)
Since
Δ
V
PE
q
KE
q
m
q
stopping
−
=
−
v
2
2
, the ratio of the stopping potential for a proton to
that for an electron having the same initial speed is
Δ
Δ
V
V
me
mm
i
i
p
e
p
e
pe
=
−+
−−
v
v
2
2
2
2
()
16.9
(a)
Use conservation of energy
KE
PE
PE
KE
PE
PE
se
f
i
++
( )
=++
( )
or
Δ
KE
PE
PE
( ) + ( ) + ( ) =
0
Δ
KE
=
0 since the block is at rest at both beginning and end.
Δ
PE
kx
s
( )
1
2
0
2
max
, where
x
max
is the maximum stretch of the spring.
Δ
PE
W
QE x
e
( )=− =− ( )
max
Thus,
0
1
2
0
2
+−
( ) =
kx
QE x
max
max
, giving
x
QE
k
max
CV
m
78.0 N m
×
( ) ×
( )
−
2
2 350 10
486 10
64
=
×
=
−
436 10
436
2
mc
m
(b)
At equilibrium,
Σ
FFF
k
x Q
E
=+=
−
+ =
00
,
or
eq
Therefore,
x
QE
k
x
eq
cm
=
1
2
218
max
.
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This note was uploaded on 04/23/2010 for the course DS SAD taught by Professor Asd during the Spring '10 term at A.T. Still University.
 Spring '10
 ASD

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