PHY 5346
HW Set 4 Solutions – Kimel
1. 2.8 The system is pictured below
a) Using the known potential for a line charge, the two line charges above give the potential
φr⃗ =
1
2π 0
λ ln r
′
r = V, a constant. Let us define V
′ = 4π 0V
Then the above equation can be written
r ′
r
2
= e
V′
λ or r ′2 = r2e
V′
λ
Writing r ′2 =
r⃗ − R⃗
2
, the above can be written
r⃗ + ẑ
R
e
V′
λ − 1
2
=
R2e
V′
λ
e
V′
λ − 12
The equation is that of a circle whose center is at −ẑ
R
e
V′
λ −1
, and whose radius is a =
Re
V′
2λ
e
V′
λ −1
b) The geometry of the system is shown in the fugure.
Note that
d = R + d1 + d2
with
d1 =
R
e
Va
′
λ −1
, d2 =
R
e
−Vb
′
λ −1
and
a =
Re
Va
′
2λ
e
Va
′
λ − 1
, b =
Re
−Vb
′
2λ
e
−Vb
′
λ − 1
Forming
d2 − a2 − b2 = R + R
e
Va
′
λ −1
+
R
e
−Vb
′
λ −1
2
−
Re
Va
′
2λ
e
Va
′
λ − 1
2
−
Re
−Vb
′
2λ
e
−Vb
′
λ − 1
2
or
d2 − a2 − b2 =
R2 e
Va
′ −Vb
′
λ + 1
e
Va
′
λ − 1e
−Vb
′
λ − 1
Thus we can write
d2 − a2 − b2
2ab
=
e
Va
′ −Vb
′
λ + 1
2e
Va
′
2λ e
−Vb
′
2λ
= e
Va
′ −Vb
′
2λ + e
− Va
′ −Vb
′
2λ
2
= cosh
Va
′ − Vb
′
2λ
or
Va − Vb
λ
=
1
2π 0
cosh−1 d
2 − a2 − b2
2ab
Capacitance/unit length = C
L
=
Q/L
Va − Vb
=
λ
Va − Vb
=
2π 0
cosh−1
d2−a2−b2
2ab
c) Suppose a2 << d2, and b2 << d2, and a ′ = ab , then
C
L
=
2π 0
cosh−1
d2−a2−b2
2a ′2
=
2π 0
cosh−1
d2 1−a2+b2/d2
2a ′2
cosh−1
d21 − a2 + b2/d2
2a ′2
=
2π 0L
C
d21 − a2 + b2/d2
2a ′2
= e
2π0L
C
2
+ negligible terms if 2π 0L
C
>> 1
or
ln
d21 − a2 + b2/d2
a ′2
=
2π 0L
C
or
C
L
=
2π 0
ln
d2 1−a2+b2/d2
a ′2
Let us defind α2 = a2 + b2/d2, then
C
L
=
2π 0
ln
d2 1−α2
a ′2
= 2π 0
ln d
2
a ′2
+ 2π
0
ln2 d
2
a ′2
α2 + Oα4
The first term of this result agree with problem 1.7, and the second term gives the appropriate
correction asked for.
d) In this case, we must take the opposite sign for d2 − a2 − b2, since a2 + b2 > d2. Thus
C
L
=
2π 0
cosh−1
a2+b2−d2
2a ′2
If we use the identiy, lnx + x2 − 1 = cosh−1x, G.&R., p. 50., then for d = 0
C
L
=
2π 0
ln
a2+b2
2ab
+ a
2−b2
2ab
=
2π 0
ln
a
b
in agreement with problem 1.6.