Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/289

 then $$x$$ and $$y$$ will be conjugate with respect to $$\phi$$ and $$\psi$$, and $$\phi$$ and $$\psi$$ will be conjugate with respect to $$x$$ and $$y$$.

Now let $$x$$ and $$y$$ be rectangular coordinates, and let $$k\psi$$ be the potential, then $$k\phi$$ will be conjugate to $$k\psi$$, $$k$$ being any constant.

Let us put $$\psi=\pi$$, then $$y=A\pi$$, $$x=A\left(\phi-e^{\phi}\right)$$.

If $$\phi$$ varies from $$-\infty$$ to 0, and then from 0 to $$+\infty$$, $$x$$ varies from $$-\infty$$ to $$-A$$ and from $$-A$$ to $$-\infty$$. Hence the equipotential surface for which $$b=\pi A$$ is a plane parallel to $$x$$ at a distance $$b=\pi A$$ from the origin, and extending from $$-\infty$$ to $$x=-A$$.

Let us consider a portion of this plane, extending from

$x=-(A+a)$ to $x=-A$ and from $z=0$ to $z=c$,

let us suppose its distance from the plane of $$xz$$ to be $$y=B=A\pi$$, and its potential to be $$V=k\psi=k\pi$$.

The charge of electricity on any portion of this part of the plane is found by ascertaining the values of $$\phi$$ at its extremities.

If these are $$\phi_{1}$$ and $$\phi_{2}$$, the quantity of electricity is

$\frac{1}{4\pi}ck\left(\phi_{2}-\phi_{1}\right)$

We have therefore to determine $$\phi$$ from the equation

$x=-(A+a)=A\left(\phi-e^{\phi}\right)$

$$\phi$$ will have a negative value $$\phi_{1}$$ and a positive value $$\phi_{2}$$ at the edge of the plane, where $$x=-A,\ \phi=0$$.

Hence the charge on the negative side is $$-ck\phi_{1}$$, and that on the positive side is $$ck\phi_{2}$$.

If we suppose that $$a$$ is large compared with $$A$$,

$\begin{array}{l} \phi_{1}=-\frac{a}{A}-1+e^{-\frac{a}{A}-1+e^{-\frac{a}{A}-+\mathrm{etc.}}}\\ \\ \phi_{e}=\log\left\{ \frac{a}{A}+1+\log\left(\frac{a}{A}+1+\mathrm{etc.}\right)\right\} \end{array}$|undefined

If we neglect the exponential terms in $$\phi_{1}$$ we shall find that the charge on the negative surface exceeds that which it would have if the superficial density had been uniform and equal to that at a distance from the boundary, by a quantity equal to the charge on a strip of breadth $$A=\frac{b}{\pi}$$ with the uniform superficial density.

The total capacity of the part of the plane considered is

$C=\frac{c}{4\pi^{2}}\left(\phi_{2}-\phi_{1}\right)$|undefined