Jackson 1.7 Homework Problem Solution
Dr. Christopher S. Baird
University of Massachusetts Lowell
PROBLEM:
Two long, cylindrical conductors of radii a1 and a2 are parallel and separated by a distance d, which is
large compared with either radius. Show that the capacitance per unit length is given approximately by
−1
( )
d
C≈π ϵ0 ln
a
where a is the geometrical mean of the two radii. Approximately what gauge wire (state diameter in
millimeters) would be necessary to make a two-wire transmission line with a capacitance of 1.2 x 10-11
F/m if the separation of the wires was 0.5 cm? 1.5 cm? 5.0 cm?
SOLUTION:
Place the center of cylinder 1 at the origin and the center of cylinder 2 at x = d. Place a total charge per
unit length -Q/L on cylinder 1 and +Q/L on cylinder 2. We first pretend cylinder 2 is not present and
solve by superposition. Without cylinder 2, the field lines from cylinder 1 are radially outwards. This
symmetry allows us to use Gauss's law in integral form. Draw a cylindrical Gaussian surface centered
on cylinder 1 and enclosing it.
∮ E1⋅n da=
E 1=
qenc
ϵ0
−Q
r̂
2 π r L ϵ0
The second cylinder by itself will create a similar field, only shifted to the right:
E2 =
Q
2
2 π∣x−d ̂i∣ L ϵ0
(x−d ̂i)
So that the total field is:
E=
−Q
Q
( x ̂i+ y ̂j)+
(( x−d ) ̂i+ y ̂j)
2
2
2
2 π( x + y ) L ϵ0
2 π (( x−d ) + y ) L ϵ0
2
The potential is then:
∇ Φ=
Q
Q
( x ̂i+ y ̂j)−
(( x−d ) ̂i+ y ̂j)
2
2
2
2 π(x + y ) L ϵ 0
2 π (( x−d ) + y ) L ϵ0
2
To calculate conductance, we don't need to know the potential everywhere, only the potential
difference between the two conductors. Also, the conductor surfaces are equipotentials, so we only
need to know the potential at one point on the surface to know it everywhere. Let us calculate the
potential difference between the points (x = a1, y = 0) and (x = d - a2, y = 0). Dot the above equation
with ̂i .
dΦ
Q
Q
=
x−
( x−d )
2
2
d x 2 π( x + y ) L ϵ0
2 π((x−d ) 2+ y 2) L ϵ 0
Integrate between our two points:
d Φ=
Q
Q
x−
( x−d ) d x
2
2
2
2 π( x + y )L ϵ 0
2 π(( x−d ) + y ) L ϵ 0
2
d −a2
∫
ΔV =
d −a2
∫
d Φ=
a1
a1
Q
ΔV =
2 π ϵ0 L
[
Q
Q
−
dx
2 π x L ϵ 0 2 π(x−d ) L ϵ 0
[∫
]
[ ( ) ( )]
)]
[(
d −a2
a1
d −a2
1
1
d x− ∫
dx
x
x−d
a
1
ΔV =
d −a 2
−a 2
Q
ln
−ln
2 π ϵ0 L
a1
a 1−d
ΔV =
d −a 1 d −a 2
Q
ln (
)(
)
2 π ϵ0 L
a2
a1
C L=
Q
ΔV
[(
−1
)]
d −a1 d −a 2
C=2 π ϵ0 ln (
)(
)
a2
a1
For d >> a1, a2
−1
[ ( )]
d
C≈π ϵ0 ln
√ a1 a2
−1
( )
C≈π ϵ0 ln
a≈d e
−πϵ 0
C
d
a
where a= √ a1 a 2
]
For a two-wire transmission line with a capacitance of 1.2 x 10-11 F/m, this becomes:
a≈(0.1) d
or in terms of the diameter:
diameter≈( 0.2)d
If the separation of the wires was 0.5 cm, 1.5 cm, or 5.0 cm, the gauge should be:
1 mm, 3 mm, or 10 mm