Page:SearleEllipsoid.djvu/4

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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}}.