Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/210

Then, by Green’s theorem, the potential energy of $E$ on the electrified surface is equal to that of the electrified surface on $E$, or

$\iint V\sigma dS=\sum V_{i}dE,$

(50)

where the first integration is to be extended over every element $dS$ of the surface of the sphere, and the summation $\sum$ is to be extended to every part $dE$ of which the electrified system $E$ is composed.

But the same potential function $V_{i}$ may be produced by means of a combination of $2^{i}$ electrified points in the manner already described. Let us therefore find the potential energy of $E$ on such a compound point.

If $M_{0}$ is the charge of a single point of degree zero, then $M_{0}V$ is the potential energy of $V$ on that point.

If there are two such points, a positive and a negative one, at the positive and negative ends of a line $h_{i}$, then the potential energy of $E$ on the double point will be

$-M_{0}V+M_{0}\left(V+h_{1}{\frac {dV}{dh_{1}}}+{\frac {1}{2}}h^{2}{\frac {d^{2}V}{dh_{1}^{2}}}+\mathrm {etc} .\right);$

and when $M_{0}$ increases and $h_{1}$ diminishes indefinitely, but so that

$M_{0}h_{1}=M_{1},$

the value of the potential energy will be for a point of the first degree

$M_{1}{\frac {dV}{dh_{1}}}.$

Similarly for a point of degree $i$ the potential energy with respect to $E$ will be

$M_{1}{\frac {d^{i}V}{dh_{1}dh_{2}\dots dh_{i}}}$

This is the value of the potential energy of $E$ upon the singular point of degree $i$. That of the singular point on $E$ is $\sum V_{i}dE$ and, by Green’s theorem, these are equal. Hence, by equation (50),

$\iint V\sigma \ dS=M_{i}{\frac {d^{i}V}{dh_{1}\dots dh_{i}}}.$

If $\sigma =CY_{i}$ where $C$ is a constant quantity, then, by equations (49) and (14),

$M_{i}={\frac {4\pi C}{|{\underline {i}}}}{\frac {a^{i+2}}{2i+1}}$

(51)

Hence, if $V$ is any potential function whatever which satisfies Laplace’s equation within the spherical surface of radius $a$, then the