CHAPTER 26 ELECTROSTATIC ENERGY AND CAPACITORS
Section 261: Energy of a Charge Distribution
Problem
1.
Three point charges, each of
+
q
, are moved from infinity to the vertices of an equilateral triangle of side
l
. How much
work is required?
Solution
The sentence preceding Example 261 allows us to rewrite Equation 261 (for the electrostatic energy of a distribution of
point charges) as
W
kq q
r
i
j
ij
= ∑
pairs
=
. For three equal charges (three different pairs) at the corners of an equilateral triangle
(
r
ij
=
l
for each pair)
W
kq
=
3
2
=
l
.
Problem
2.
Repeat the preceding problem for the case of two charges
+
q
and one

q
.
Solution
When two pairs have opposite charges (
),
q q
q
i
j
= 
2
and one pair has equal charges (
),
q q
q
i
j
=
2
the electrostatic energy is
W
k
q
q
q
kq
=


+
= 
(
)
.
2
2
2
2
=
=
l
l
Problem
3.
Four 50
C
m
charges are brought from far apart onto a line where they are spaced at 2.0cm intervals. How much work
does it take to assemble this charge distribution?
Solution
Number the charges
q
i
i
=
=
50
1
2
3
4
C
m
,
,
,
,
, as they are spaced along the line at
a
=
2 cm intervals. There are six
pairs, so
W
kq q
r
k q q
a
q q
a
q q
a
q q
a
q q
a
q q
a
kq
a
i
j
ij
= ∑
=
+
+
+
+
+
=
+
+
+
+
+
=
pairs
=
=
=
=
=
=
=
=
(
)
(
)(
)
1
2
1
3
1
4
2
3
2
4
3
4
2
1
2
1
3
1
2
2
3
2
1
1
1
13
3
13 9
10
50
3
2
4 88
2
9
2
kq
a
=
=
=
×
×
=
(
/ )(
)
(
)
.
.
m F
C
cm
kJ
m
(See solution to Problem 1.)
Problem 3 Solution.
Problem
4.
Repeat Example 261 for the case when the negative charge is

q
rather than

q
=
2.
Solution
If the negative charge in Example 261 is

q
W
,
2
and
W
3
are unchanged, but
W
k
q
a
q
a
q
a
4
2
2
2
2
=



(
).
=
=
=
Therefore,
W
W
W
W
=
+
+
=
2
3
4
0. (In this case, the work needed to assemble the positive charges equals the energy gained adding
the negative charge.)
Problem
5.
Suppose two of the charges in Problem 1 are held in place, while the third is allowed to move freely. If this third charge
has mass
m
, what will be its speed when it’s far from the other two charges?
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CHAPTER 26
Solution
With one charge removed to infinity, the potential energy is reduced to that of just one pair of charges,
W
kq
f
=
2
=
l
. The
initial potential energy was
W
kq
i
=
3
2
=
l
(see Problem 1), so the kinetic energy of the charge at infinity (from the
conservation of energy) is
K
W
W
kq
i
f
=

=
2
2
=
l
. Thus,
v
=
=
=
=
2
0
K m
q
m
pe
l
.
Problem
6.
To a very crude approximation, a water molecule consists of a negatively charged oxygen atom and two “bare” protons,
as shown in Fig. 2625. Calculate the electrostatic energy of this configuration, which is therefore the magnitude of the
energy released in forming this molecule from widely separated atoms. Your answer is an overestimate because
electrons are actually “shared” among the three atoms, spending more time near the oxygen.
Solution
The electrostatic potential energy of the water molecule (in this approximation) is
U
W
kq q
r
i
j
ij
=
= ∑
pairs
=
(generalization
of Equation 261). The two oxygenhydrogen pairs have separation
a
=

10
10
m, while the hydrogenhydrogen pair has
separation 2
37 5
159
a
a
cos
.
.
.
° =
Therefore,
U
k e
e
a
ke
a
ke
a
=

+
= 
= 
×
2
2
337
337 9
10
2
2
9
( )(
)
.
.
.
(
)
=
=
=
( .
)
.
.
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 Spring '08
 Dougherty
 Electric charge, Energy density

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