HRW Solution Chapter 25.pdf

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Halliday
♦
Resnick
♦Walker
FUNDAMENTALS OF PHYSICS
SIXTH EDITION
Selected Solutions
Chapter 25
25.17
25.23
25.33
25.41
17. (a) The electric potential
V
at the surface of the drop, the charge
q
on the drop, and the radius
R
of
the drop are related by
V
=
q/4πε
0
R.
Thus
q
(8.99
×
10
9
N
·
m
2
/C
)(30
×
10
−12
C)
R
=
=
= 5.4
×
10
−4
m
.
4πε
0
V
500 V
2
(b) After the drops combine the total volume is twice the volume of an original drop, so the radius
R
of the combined drop is given by (R )
3
= 2R
3
and
R
= 2
1/3
R.
The charge is twice the charge of
original drop:
q
= 2q. Thus,
V
=
1
q
1
2q
= 2
2/3
V
= 2
2/3
(500 V)
≈
790 V
.
=
1/3
R
4πε
0
R
4πε
0
2
23. A positive charge
q
is a distance
r
−
d
from
P
, another positive charge
q
is a distance
r
from
P
, and a
negative charge
−q
is a distance
r
+
d
from
P
. Sum the individual electric potentials created at
P
to
find the total:
q
1
1
1
V
=
+
−
.
4πε
0
r
−
d r
r
+
d
We use the binomial theorem to approximate 1/(r
−
d)
for
r
much larger than
d:
1
d
1
= (r
−
d)
−1
≈
(r)
−1
−
(r)
−2
(−d) = +
2
.
r
−
d
r
r
Similarly,
1
1
d
≈ −
2
.
r
+
d
r
r
Only the first two terms of each expansion were retained. Thus,
V
≈
q
1 1
d
d
2d
1
q
1 2d
q
+
2
+
−
+
2
=
+
2
=
1+
4πε
0
r
r
r
r
r
4πε
0
r
r
4πε
0
r
r
.
33. (a) The charge on
√
every part of the ring is the same distance from any point
P
on the axis. This
distance is
r
=
z
2
+
R
2
, where
R
is the radius of the ring and
z
is the distance from the center of
the ring to
P
. The electric potential at
P
is
V
=
1
4πε
0
dq
1
=
r
4πε
0
√
1
dq
1
√
=
4πε
0
z
2
+
R
2
z
2
+
R
2
dq
=
1
q
√
.
4πε
0
z
2
+
R
2
(b) The electric field is along the axis and its component is given by
E
=
=
This agrees with Eq. 23-16.
−
∂V
q ∂
2
=
−
(z +
R
2
)
−1/2
∂z
4πε
0
∂z
z
1
q
q
.
(z
2
+
R
2
)
−3/2
(2z) =
4πε
0
2
4πε
0
(z
2
+
R
2
)
3/2
41. The particle with charge
−q
has both potential and kinetic energy, and both of these change when the
radius of the orbit is changed. We first find an expression for the total energy in terms of the orbit radius
r. Q
provides the centripetal force required for
−q
to move in uniform circular motion. The magnitude
of the force is
F
=
Qq/4πε
0
r
2
. The acceleration of
−q
is
v
2
/r,
where
v
is its speed. Newton’s second
law yields
mv
2
Qq
Qq
=
=⇒
mv
2
=
,
4πε
0
r
2
r
4πε
0
r
and the kinetic energy is
K
=
total energy is
1
2
2
mv
=
Qq/8πε
0
r.
The potential energy is
U
=
−Qq/4πε
0
r,
and the
Qq
Qq
Qq
−
=
−
.
8πε
0
r
4πε
0
r
8πε
0
r
E
=
K
+
U
=
When the orbit radius is
r
1
the energy is
E
1
=
−Qq/8πε
0
r
1
and when it is
r
2
the energy is
E
2
=
−Qq/8πε
0
r
2
. The difference
E
2
−
E
1
is the work
W
done by an external agent to change the radius:
W
=
E
2
−
E
1
=
−
Qq
8πε
0
1
1
−
r
2
r
1
=
Qq
8πε
0
1
1
−
r
1
r
2
.

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