SME
University Physics II: Electricity & Magnetism · Problem type
The capacitance of a coaxial cable
Use this for “A coaxial cable has a 0.8 mm inner wire inside a 3 mm shield. Find the capacitance of 10 m of it.”
Table
Radius ratio b ÷ a (n)
Capacitance (C)
(pF)
This is an example. Type your own values to start a new problem.
C (
)
=
2πε₀
L (m)
ln
b (
)
a (
)
n
Radius ratio b ÷ a
calculated
Formulas
n
=
b
a
4
=
2
0.5
C
=
2π
×
8.85
×
L
ln(
n
)
40.1114
=
2π
×
8.85
×
1
ln(4)
Assumptions
•
The cable is much longer than b, so E points straight out from the inner conductor and the ends don’t matter.
•
Gauss’s law on a cylinder between the conductors gives E = λ ÷ (2πε₀r); adding E dr from a to b gives V = λ ln(b ÷ a) ÷ (2πε₀).
•
C = Q ÷ V with Q = λL, so C = 2πε₀L ÷ ln(b ÷ a), with ε₀ = 8.85 × 10⁻¹² F/m = 8.85 pF/m and air in the gap.
•
Only the ratio b ÷ a matters, not the sizes: doubling both radii leaves C the same.
Step-by-step
WE KNOW
a = 0.5 mm, b = 2 mm, L = 1 m
FIND
radius ratio b ÷ a (n), capacitance (C)
Step 1 · Find radius ratio b ÷ a (n)
Use n = b ÷ a
How many times wider the outer conductor is than the inner one.
n
=
2
0.5
n
=
4
Step 2 · Find capacitance (C)
Use C = 2π × 8.85 × L ÷ ln(n)
Charge
over
voltage:
the
charge
per
length
λ
cancels
from
both,
leaving
2πε₀
L
over
ln
n
;
ε₀
=
8.85
pF/m
gives
C
in
pF.
C
=
2π
×
8.85
×
1
ln(4)
C
=
40.1114
pF
Check
Put the numbers back in. Both sides should match.
4
=
2
0.5
✓
40.1114
=
2π
×
8.85
×
1
ln(4)
✓
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Potential and capacitance
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