(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Two long, cylindrical conductors of radii [itex]a_1[/itex] and [itex]a_2[/itex] are parallel and separated by a distance [itex]d[/itex], which is large compared with either radius. Show that the capacitance per unit length is given approximately by

[tex]

C=\pi\epsilon_0 \left(\ln\frac{d}{a}\right)^{-1}

[/tex]

where [itex]a[/itex] is the geometric 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 [itex]1.2\times10^{-11}[/itex] F/m if the separation of the wires was 0.5 cm? 1.5 cm? 5.0 cm?

2. Relevant equations

[tex]

C=\frac{Q}{\phi}

[/tex]

[tex]

\phi=-\int \vec{E}\cdot d\vec{l}

[/tex]

3. The attempt at a solution

Honestly, I'm pretty lost on this... I'm not sure how to construct this from scratch, my only work is from backwards, and I feel shady about it...

Here it is:

We know that [itex]C=Q/\phi[/itex] and we are given the capacitance per unit length [itex]C'=C/l[/itex], so I did this

[tex]

\phi=\frac{q}{C' l}=\frac{\lambda}{\pi\epsilon_0}[\ln(d)-\ln(\sqrt{a_1 a_2})]

[/tex]

with [itex]\lambda[/itex] the charge per unit length, which can be written as

[tex]

\phi=-\int_{d}^{\sqrt{a_1 a_2}} \frac{\lambda}{\pi\epsilon_0 }\frac{1}{r}dr

[/tex]

Then the electric field would be

[tex]

\vec{E}=\frac{\lambda}{\pi\epsilon_0 r}\hat{r}

[/tex]

Now if I were to continue with this (and I'm not committed to this :) ), how could I manage to write this as a superposition of two cylindrical electric fields from two different conductors with equal and opposite charges...?

Can someone give me a hand here?

Thanks in advance,

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# Homework Help: Capacitance of Parallel cylinders

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