Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/116

Let the force at a distance $r$ from a point at which a quantity $e$ of electricity is concentrated be $R$, where $R$ is some function of $r$. All central forces which are functions of the distance admit of a potential, let us write ${\tfrac {f(r)}{r}}$ for the potential function due to a unit of electricity at a distance $r$.

Let the radius of the spherical shell be $a$, and let the surface-density be $\sigma$. Let $P$ be any point within the shell at a distance $p$ from the centre. Take the radius through $P$ as the axis of spherical coordinates, and let $r$ be the distance from $P$ to an element $dS$ of the shell. Then the potential at $P$ is

$V=\iint \sigma \,{\frac {f(r)}{r}}dS$,

$V=\int _{0}^{2\pi }\int _{0}^{\pi }\sigma \,{\frac {f(r)}{r}}\,a^{2}\,sin\theta \,d\theta \,d\phi$.

Now

$r^{2}=a^{2}-2ap\,cos\theta +p^{2}$,

{{{2}}}

$V=2\pi \sigma {\frac {a}{p}}\int _{a-p}^{a+p}f(r)dr$;

and V must be constant for all values of $p$ less than $a$.

Multiplying both sides by $p$ and differentiating with respect to $p$,

$V=2\pi \sigma a{f(a+p)+f(a-p)}\,\!$.

Differentiating again with respect to $p$,

$0=f'(a+p)-f'(a-p)\,\!$

, Since a and p are independent,

$f'(r)=C\,\!$, a constant.

Hence

$f(r)=Cr+C'\,\!$

and the potential function is

${\frac {f(r)}{r}}=C+{\frac {C'}{r}}$

The force at distance $r$ is got by differentiating this expression with respect to $r$, and changing the sign, so that

$R={\frac {C'}{r^{2}}}$

or the force is inversely as the square of the distance, and this therefore is the only law of force which satisfies the condition that the potential within a uniform spherical shell is constant^[1]. Now

1. See Pratt's Mechanical Philosophy, p. 144.