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To make my question more clear, I will use an example to illustrate it.

Consider an infinite conducting plane, with a long conducting cylinder of radius $R$ at a distance $h$ away from it, with the cylinder axis parallel to the plane. I have used conformal transformations to show that the electrostatic potential between the plates is $$u=\cosh^{-1}{\Big(\frac{h}{R}}\Big).$$ The capitance per unit length is then $$\frac{C}{L}=\frac{Q}{L}\frac{1}{\cosh^{-1}{\big(\frac{h}{R}\big)}}.$$ The next part is something that occurs in most complex capacitance problems involving circular/cylindrical cases and is something that I and others often struggle to understand. How do I obtain the following relation? $$\frac{Q}{L}=2\pi\epsilon_0$$ This implies that the capacitance per unit length becomes $$\frac{C}{L}=\frac{2\pi\epsilon_0}{\cosh^{-1}{\big(\frac{h}{R}\big)}},$$ which for this particular case is the correct answer.

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  • $\begingroup$ You seem to have a mistake very early on. If the potential is a constant that does not depend on any charges in the system (as you have written here) then the capacitance is not a constant due to the geometry and material of the arrangement (which it should be), but instead depends on the charge (as you have written; but which it should not). Once you fix that, the factor of $2\pi$ will presumably arise from an application of Gauss's law to a cylindrical geometry of some kind. $\endgroup$ – dmckee May 16 '16 at 0:30
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What you have forgotten is that $u$ is not a potential rather it is a number with no units.

I do not know how you derived your expression for $u$ but here is a way without a lot of the intermediate steps.

The arrangement you have described is similar to that of two parallel line charges with separation $2a$ and charge per unit length $\pm \lambda$ as shown in the diagram below.

enter image description here

The similarity is because the equipotentials for this arrangement are circles and in the diagram centred at $x = \pm h$.
Your conducting cylinder of radius $R$ is the right hand equipotential surface at a potential $V$ with the $y$-axis being the $V=0$ equipotential which is your plane conductor.

Using Gauss the electric field due to a line of charge is $E(r) = \dfrac {\lambda}{2 \pi \epsilon_o}\frac 1 r$ and so the potential $V(r) = - \dfrac {\lambda}{2 \pi \epsilon_o} \ln r + \text{constant}$.

So the potential at point $A$ is $V= \dfrac {\lambda}{2 \pi \epsilon_o} \ln \left (\dfrac r R \right)$

You proceed to let $r = R e^u$ which gives $u = \dfrac {2 \pi \epsilon_o V}{\lambda}$.
In terms of finding the capacitance of the arrangement per unit length you need to find $\dfrac \lambda v = \dfrac {2 \pi \epsilon_o}{u}$ and there is your missing $2 \pi \epsilon_o$.

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