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

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

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

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

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

Hence

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

satisfies (7).

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

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} $|undefined

at the end of the axis.

But by (5)

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

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

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

Hence

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