CHAPTER
25
Electrostatic Energy and Capacitance
1* ·
Three point charges are on the
x
axis:
q
1
at the origin,
q
2
at
x
= 3 m, and
q
3
at
x
= 6 m. Find the electrostatic
potential energy for (
a
)
q
1
=
q
2
=
q
3
= 2
μ
C, (
b
)
q
1
=
q
2
= 2
μ
C and
q
3
= –2
μ
C, and (
c
)
q
1
=
q
3
= 2
μ
C and
q
2
= –2
μ
C.
1. List
r
1,2
,
r
1,3
, and
r
2,3
(
a
), (
b
), (
c
) Use Equ. 25-1
r
1,2
= 3 m,
r
1,3
= 6 m,
r
2,3
= 3 m
(
a
)
U
= 8.99
×
10
9
×
4
×
10
–12
(5/6) J = 30.0 mJ
(
b
)
U
= –5.99 mJ
(
c
)
U
= –18.0 mJ
2
·
Point charges
q
1
,
q
2
, and
q
3
are at the corners of an equilateral triangle of side 2.5 m. Find the electrostatic
potential energy of this charge distribution if (
a
)
q
1
=
q
2
=
q
3
= 4.2
μ
C, (
b
)
q
1
=
q
2
= 4.2
μ
C and
q
3
= –4.2
μ
C,
(
c
)
q
1
=
q
2
= –4.2
μ
C and
q
3
= +4.2
μ
C.
(
a
), (
b
), (
c
) Use Equ. 25-1; note that all distances
between charges are the same
(
a
)
U
= 8.99
×
10
9
×
4.2
2
×
10
–12
(3/2.5) J = 0.19 J
(
b
)
U
= 0.0634 J;
(
c
)
U
= –0.0634 J
3
·
What is the electrostatic potential energy of an isolated spherical conductor of radius 10 cm that is charged
to 2 kV?
U
=
1/2
QV
;
V
=
kQ
/
r
;
U
=
1/2
rV
2
/
k
U
= 0.05
×
4
×
10
6
/8.99
×
10
9
= 22.3
μ
J
4
··
Four point charges of magnitude 2
μ
C are at the corners of a square of side 4 m. Find the electrostatic
potential energy if (
a
) all the charges are negative, (
b
) three of the charges are positive and one is negative, and
(
c
) two are positive and two are negative.
1. Determine the distances between charges
(
a
), (
b
), (
c
) Use Equ. 25-1
r
1,2
=
r
1,3
=
r
2,4
=
r
3,4
= 4 m;
r
1,4
=
r
2,3
= 5.657 m
(
a
)
U
= 48.7 mJ; (
b
)
U
= 0; (
c
)
U
= –12.7 mJ or
–23.2 mJ, depending on the configuration
5* ··
Four charges are at the corners of a square centered at the origin as follows:
q
at (–
a
, +
a
); 2
q
at (
a
,
a
); –3
q
at (
a
, –
a
); and 6
q
at (–
a
, –
a
). A fifth charge +
q
is placed at the origin and released from rest. Find its speed
when it is a great distance from the origin.
1. Find
V
(0)
2.
1/2
mv
2
=
qV
(0)
V
(0) = (
kq/a
2
)(1 + 2 – 3 + 6) = 6
kq
/
a
2
v
=
a
k/m
2
6
q
6
··
Four identical particles each with charge
Q
are at the corners of a square of side
L
. The particles are