231
E
LECTRIC
P
OTENTIAL
23.1.
IDENTIFY:
Apply Eq.(23.2) to calculate the work. The electric potential energy of a pair of point charges is given
by Eq.(23.9).
SET UP:
Let the initial position of
2
q
be point
a
and the final position be point
b
, as shown in Figure 23.1.
0.150 m
a
r
=
22
(0.250 m)
(0.250 m)
b
r
=+
0.3536 m
b
r
=
Figure 23.1
EXECUTE:
ab
a
b
WU
U
→
=−
66
92
12
0
1
( 2.40 10 C)( 4.30 10 C)
(8.988 10 N m /C )
40
.
1
5
0
m
a
a
qq
U
r
π
−−
2
+×
−×
==
×
⋅
P
0.6184 J
a
U
0
1
.
3
5
3
6
m
b
b
U
r
2
×
⋅
P
0.2623 J
b
U
0.6184 J
( 0.2623 J)
0.356 J
a
b
U
→
=
−
− −
EVALUATE:
The attractive force on
2
q
is toward the origin, so it does negative work on
q
2
when
q
2
moves to
larger
r
.
23.2.
IDENTIFY:
Apply
.
a
b
U
→
SET UP:
8
5.4 10 J.
a
U
−
×
Solve for
.
b
U
EXECUTE:
8
1.9 10 J
.
a
b
U
−
→
×
=
−
88
8
1
.
91
0
J (5
.
41
J
) 7
.
31
J
.
baa
b
UUW
−
→
=×
−
EVALUATE:
When the electric force does negative work the electrical potential energy increases.
23.3.
IDENTIFY:
The work needed to assemble the nucleus is the sum of the electrical potential energies of the protons
in the nucleus, relative to infinity.
SET UP:
The total potential energy is the scalar sum of all the individual potential energies, where each potential
energy is
00
(1/ 4
)(
/ ).
Uq
q
r
=
P
Each charge is
e
and the charges are equidistant from each other, so the total
potential energy is
222
2
13
.
44
eee
e
U
rrr
r
ππ
⎛⎞
+
=
⎜⎟
⎝⎠
PP
EXECUTE:
Adding the potential energies gives
21
9
2
9
2
2
13
15
0
3
3(1.60 10
C) (9.00 10 N m /C )
3.46 10
J
2.16 MeV
42
.
0
0
1
0
m
e
U
r
−
−
−
××
⋅
=
× =
×
P
EVALUATE:
This is a small amount of energy on a macroscopic scale, but on the scale of atoms, 2 MeV is quite a
lot of energy.
23
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Chapter 23
23.4.
IDENTIFY:
The work required is the change in electrical potential energy. The protons gain speed after being
released because their potential energy is converted into kinetic energy.
(a)
SET UP:
Using the potential energy of a pair of point charges relative to infinity,
00
(1/ 4
)(
/ ).
Uq
q
r
π
=
P
we have
22
21
02 1
1
.
4
ee
WU
U
U
rr
⎛⎞
=Δ
=
−
=
−
⎜⎟
⎝⎠
P
EXECUTE:
Factoring out the
e
2
and substituting numbers gives
()
(
)
2
92
2
1
9
14
15
15
11
9.00 10 N m /C
1.60 10
C
7.68 10
J
3.00 10
m
2.00 10
m
W
−
−
−−
=×⋅
×
−
=
×
××
(b)
SET UP:
The protons have equal momentum, and since they have equal masses, they will have equal speeds
and hence equal kinetic energy.
12
1
.
2
U
K
K
K
mv
mv
Δ= + =
=
=
EXECUTE:
Solving for
v
gives
14
27
7.68 10
J
1.67 10
kg
U
v
m
−
−
Δ×
==
×
= 6.78
×
10
6
m/s
EVALUATE:
The potential energy may seem small (compared to macroscopic energies), but it is enough to give
each proton a speed of nearly 7 million m/s.
23.5.
(a) IDENTIFY:
Use conservation of energy:
other
aa
bb
KUW
KU
++
=+
U
for the pair of point charges is given by Eq.(23.9).
SET UP:
Let point
a
be where
q
2
is 0.800 m from
q
1
and point
b
be where
q
2
is 0.400 m
from
q
1
, as shown in Figure 23.5a.
Figure 23.5a
EXECUTE:
Only the electric force does work, so
other
0
W
=
and
0
1
.
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 Spring '11
 Shaefer
 Charge, Electrostatics, Potential Energy, Electric charge

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