Electric Potential
23-29
(c) S
ET
U
P
:
part (a):
0
0
ln
ln 1
.
4
4
Q
x
a
Q
a
V
a
x
a
x
π
π
+
⎛
⎞
⎛
⎞
=
=
+
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
P
P
From Appendix B,
2
ln(1
)
/ 2 ...,
u
u
u
+
=
−
so
2
2
ln(1
/
)
/
/2
a x
a x
a
x
+
=
−
and this becomes
/
a x
when
x
is large.
E
XECUTE
:
Thus
0
0
.
4
4
Q
a
Q
V
a
x
a
π
π
⎛
⎞
→
=
⎜
⎟
⎝
⎠
P
P
For large
x,
V
becomes the potential of a point charge.
part (b):
2
2
2
2
0
0
ln
ln
1
.
4
4
a
a
y
Q
Q
a
a
V
a
y
a
y
y
π
π
⎡
⎤
⎛
⎞
⎛
⎞
+
+
⎢
⎥
⎜
⎟
⎜
⎟
=
=
+
+
⎜
⎟
⎜
⎟
⎢
⎥
⎝
⎠
⎝
⎠
⎣
⎦
P
P
From Appendix B,
2
2
2
2
1/ 2
2
2
1
/
(1
/
)
1
/2
a
y
a
y
a
y
+
=
+
=
+
+
…
Thus
2
2
2
2
/
1
/
1
/
/ 2
1
/
.
a
y
a
y
a
y
a
y
a
y
+
+
→
+
+
+
→
+
…
And then using ln(1
)
u
u
+
≈
gives
0
0
0
ln(1
/
)
.
4
4
4
Q
Q
a
Q
V
a
y
a
a
y
y
π
π
π
⎛
⎞
→
+
→
=
⎜
⎟
⎝
⎠
P
P
P
E
VALUATE
:
For large
y, V
becomes the potential of a point charge.
23.80.
I
DENTIFY
:
The potential at the surface of a uniformly charged sphere is
kQ
V
R
=
.
S
ET
U
P
:
For a sphere,
3
4
3
V
R
π
=
. When the raindrops merge, the total charge and volume is conserved.
E
XECUTE
:
(a)
12
4
( 1.20
10
C)
16.6 V
6.50
10
m
kQ
k
V
R
−
−
−
×
=
=
= −
×
.
(b)
The volume doubles, so the radius increases by the cube root of two:
4
3
new
2
8.19
10
m
R
R
−
=
=
×
and the new
charge is
12
new
2
2.40
10
C.
Q
Q
−
=
= −
×
The new potential is
12
new
new
4
new
( 2.40
10
C)
26.4 V
8.19
10
m
kQ
k
V
R
−
−
−
×
=
=
= −
×
.
E
VALUATE
:
The charge doubles but the radius also increases and the potential at the surface increases by only a
factor of
2 / 3
1/ 3
2
2
2
=
.
23.81.
(a) I
DENTIFY
and
S
ET
U
P
:
The potential at the surface of a charged conducting sphere is given by Example 23.8:
0
1
.
4
q
V
R
π
=
P
For spheres
A
and
B
this gives
0
4
A
A
A
Q
V
R
π
=
P
and
0
.
4
B
B
B
Q
V
R
π
=
P
E
XECUTE
:
A
B
V
V
=
gives
0
0
/4
/4
A
A
B
B
Q
R
Q
R
π
π
=
P
P
and
/
/
.
B
A
B
A
Q
Q
R
R
=
And then
3
A
B
R
R
=
implies
/
1/3.
B
A
Q
Q
=
(b) I
DENTIFY
and
S
ET
U
P
:
The electric field at the surface of a charged conducting sphere is given in
Example 22.5:
2
0
1
.
4
q
E
R
π
=
P
E
XECUTE
:
For spheres
A
and
B
this gives
2
0
4
A
A
A
Q
E
R
π
=
P
and
2
0
.
4
B
B
B
Q
E
R
π
=
P
2
2
2
0
2
0
4
/
(
/
)
(1/3)(3)
3.
4
B
B
A
B
A
A
B
A
B
A
Q
E
R
Q
Q
R
R
E
R
Q
π
π
⎛
⎞
⎛
⎞
=
=
=
=
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠⎝
⎠
P
P
E
VALUATE
:
The sphere with the larger radius needs more net charge to produce the same potential. We can write
/
E
V
R
=
for a sphere, so with equal potentials the sphere with the smaller
R
has the larger
V.
23.82.
I
DENTIFY
:
Apply conservation of energy,
a
a
b
b
K
U
K
U
+
=
+
.
S
ET
U
P
:
Assume the particles initially are far apart, so
0
a
U
=
, The alpha particle has zero speed at the distance of
closest approach, so
0
b
K
=
.
19
1 eV
1.60
10
J
−
=
×
. The alpha particle has charge
2
e
+
and the lead nucleus has
charge
82
e
+
.

This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*