Calculating the Capacitance — Question 3

Question 3

A coaxial capacitance depends logarithmically on its radius ratio

Long coaxial conductors have radii a and b>a and overlap length L. The gap is vacuum. Neglect end effects and let the inner conductor charge per length be $\lambda$.

1. Derive the capacitance per unit length.

2. Find its leading approximation for b=a+d with d much smaller than a and identify the effective plate area.

Question 3 - Solution

A coaxial capacitance depends logarithmically on its radius ratio

(a) Method and calculation

$$E_r=\lambda/(2\pi\epsilon_0r),\quad V=\int_a^bE_rdr=\frac{\lambda}{2\pi\epsilon_0}\ln(b/a).$$$$C/L=\lambda/V=\frac{2\pi\epsilon_0}{\ln(b/a)}.$$

(b) Interpretation and further calculation

$$\ln(1+d/a)\simeq d/a\Rightarrow C\simeq\frac{2\pi\epsilon_0aL}{d}.$$$$A_{\mathrm{effective}}=2\pi aL.$$

Physical check

The logarithm contains a dimensionless radius ratio; capacitance is proportional to length only while end effects remain negligible.