# Page:SearleEllipsoid.djvu/4

the surface of the ellipsoid, shall vanish at infinity, and shall satisfy (7). We see at once that if $f(x, y, z)$ satisfies $\nabla^{2}f=0$, then $f(x/\sqrt{\alpha},y,z)$ satisfies (7). Now from electrostatics we know that

$\Phi=\int_{\lambda}^{\infty}\frac{d\lambda}{\sqrt{\left(a^{'2}+\lambda\right)\left(b^{2}+\lambda\right)\left(c^{2}+\lambda\right)}}$,

where $\lambda$ is connected with $x, y, z$ by the relation

$\frac{x^{2}}{a^{'2}+\lambda}+\frac{y^{2}}{b^{2}+\lambda}+\frac{z^{2}}{c^{2}+\lambda}=1$,

satisfies $\nabla^{2}\Phi=0$.

Hence

 $\mathbf{\Psi}=\int_{\lambda}^{\infty}\frac{Ad\lambda}{\sqrt{\left(a^{'2}+\lambda\right)\left(b^{2}+\lambda\right)\left(c^{2}+\lambda\right)}}$, (8)

where $\lambda$ is connected with $x, y, z$ by the relation

 $\frac{x^{2}}{\left(a^{'2}+\lambda\right)}+\frac{y^{2}}{b^{2}+\lambda}+\frac{z^{2}}{c^{2}+\lambda}=1$, (9)

satisfies (7).

Writing $a^2$ for $\alpha a'^2$, (8) and (9) become

 $\mathbf{\Psi}=\int_{\lambda}^{\infty}\frac{Ad\lambda}{\sqrt{\left(a^{2}+\alpha\lambda\right)\left(b^{2}+\lambda\right)\left(c^{2}+\lambda\right)}}$ ; (10)
 $\frac{x^{2}}{a^{2}+\alpha\lambda}+\frac{y^{2}}{b^{2}+\lambda}+\frac{z^{2}}{c^{2}+\lambda}=1$. (11)

This value of $\mathbf{\Psi}$ is constant over the surface of the ellipsoid $a, b, c$, for $\lambda=0$ at all points of this surface; it also vanishes at infinity, and it satisfies (7). It is therefore the value of $\mathbf{\Psi}$ required. To find the constant A we make $\sigma$ have its proper value $q/4\pi bc$ at the end of axis $a$.

Now

$\sigma=\frac{\mathrm{K}}{4\pi}\mathbf{E}_{n} =\frac{\mathrm{K}}{4\pi}\mathbf{E}_{1}$

at the end of the axis.

But by (5)

$\mathbf{E}_{1}=-\frac{d\mathbf{\Psi}}{dx}$.

Again, at $x=a, y=z=0$ we have $d\lambda/dx=2a/\alpha$ and consequently

$\frac{d\mathbf{\Psi}}{dx}=\frac{d\mathbf{\Psi}}{d\lambda}\frac{d\lambda}{dx}=-\frac{A}{abc}\cdot\frac{2a}{\alpha}$.

Hence

$A=\frac{q\alpha}{2\mathrm{K}}$.