WiReD Capacitance

Electricity and Magnetism Level 2

consider $2$ long and parallel oppositely charged thick wires of radius $d$ and with their central axes separated by a distance $D$ .Obtain an expression for the capacitance per unit length of this pair of wires.

If the capacitance per unit length $C'$ can be written as

$C'$ $=$ $\alpha$ $π$ $\epsilon_0$ $/$ $ln$ $($ $[$ $D$ $-$ $\beta$ $d$ $]$ $/$ $\gamma$ $d$ $)$

What is $\alpha$ $+$ $\beta$ $+$ $\gamma$ ?

Details and Assumptions :

$1)$ $\alpha$ , $\beta$ , $\gamma$ are integers which may or may not be equal to one another,but are in their simplest form.

$2)$ $ln(x)$ refers to natural logarithm of $x$ $($ $log_e (x)$ $)$ where $e$ is euler 's number.

$3)$ $\epsilon_0$ refers to the vacuum permittivity constant.

The answer is 3.

1 solution

Steven Chase
Oct 9, 2017

Suppose each wire has an area charge density $\sigma$ . Applying Gauss's law to such a conductor yields the following (considering a certain length of conductor ( $l$ )):

$E \, A = \frac{q_{enc}}{\epsilon_0} \\ E \, 2 \pi r l = \frac{2 \pi \, d \, l \, \sigma}{\epsilon_0} \\ E = \frac{d \, \sigma}{r \epsilon_0}$

Find the voltage from one conductor surface to the other conductor surface, due to a single conductor. Then double it to get the total voltage.

$V_{single} = \frac{d \, \sigma}{\epsilon_0} \int_d^{D-d} \frac{1}{r} dr = \frac{d \, \sigma}{\epsilon_0} ln \Big(\frac{D-d}{d} \Big) \\ V = 2 V_{single} = \frac{2 d \, \sigma}{\epsilon_0} ln \Big(\frac{D-d}{d} \Big)$

Find the charge associated with length $l$ (one conductor only, as per standard convention):

$Q = 2 \pi \, d \, l \, \sigma$

Calculate the capacitance:

$Q = CV \\ 2 \pi \, d \, l \, \sigma = C \frac{2 d \, \sigma}{\epsilon_0} ln \Big(\frac{D-d}{d} \Big) \\ C = \frac{\pi \, l \, \epsilon_0}{ln \Big(\frac{D-d}{d} \Big)}$

Divide by $l$ to get the capacitance per unit length:

$C' = \frac{\pi \, \epsilon_0}{ln \Big(\frac{D-d}{d} \Big)}$

WiReD Capacitance

The answer is 3.

1 solution

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