# Page:SearleEllipsoid.djvu/4

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

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

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

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

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

Hence

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

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

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

satisfies (7).

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

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

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

Now

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

at the end of the axis.

But by (5)

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

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

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

Hence

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