Fall 2003 3 Em

Consider a metallic wire of length $L$ and radius $a$, which runs parallel to a conducting plane a distance $D$ away; assume that $L \gg D \gg a$. Find an approximate expression for the capacitance of this system.




This problem requires the use of images. The potential anywhere in space is due to the potential of the wire and the potential of the image wire. Using Gauss's Law, for an infinite cylinder with charge plastered on it, we get

\begin{align} E_{\rho_1} 2 \pi \rho_1 L = 4 \pi \lambda L \end{align}


\begin{align} \vec{E} = \frac{2 \lambda}{\rho_1} \hat{\rho_1} \end{align}

which gives a potential of

\begin{align} \phi_1 = 2 \lambda \log \rho_1 \end{align}

For the image wire, we stick charge of the opposite site on, therefore the potential is

\begin{align} \phi_2 = -2 \lambda \log \rho_2 \end{align}

Note that $\rho_1$ and $\rho_2$ refer to the distances we are from each specific wire.

The potential at the conducting plane is 0, which is what we want. The two terms just cancel each other. The potential at the surface of the real wire is approximately

\begin{align} \rho_{wire} = 2 \lambda \log a - 2 \lambda \log 2 D \end{align}

Since $D \gg a$, we can assume the surface of the real wire is a distance $2 D$ away from the image wire (approximating a line charge). Therefore the difference in potential between the plane and the wire is

\begin{align} \Delta \phi = \frac{2 Q}{L} \log \frac{2 D}{a} \end{align}

where $\lambda = \frac{Q}{L}$. Since $Q = C \Delta \phi$, we can solve for $C$ and we get

\begin{align} C = \frac{L}{2} \left ( \log \frac{2 D}{a} \right )^{-1} \end{align}

If we instead tried to calculate the capacitance of the two wire system (real and image) we would have gotten

\begin{align} C = \frac{L}{4} \left ( \log \frac{2 D}{a} \right )^{-1} \end{align}

The way we can see the the capacitance of the plane-wire system is doubled is that it has half the total energy (above the plane and below the plane would have the same amount of energy in the wire-wire system). The charge on the wire is kept constant when switching to the plane-wire system (and difference in potential is halved), so from

\begin{align} U = \frac{Q^2}{2 C} \end{align}

we see that the capacitance must be doubled

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