# Page:SearleEllipsoid.djvu/12

When ${\displaystyle h}$ is large the quantity in [ ]

${\displaystyle =h-l\left({\frac {h^{2}}{l^{2}}}+1\right)\left({\frac {l}{h}}+{\frac {l^{2}}{3h^{3}}}\dots \right)}$

vanishing when ${\displaystyle h=\infty }$.

Thus, making use of ${\displaystyle \mu \mathrm {K} v^{2}=1}$ we have for the magnetic energy

${\displaystyle \mathrm {T} ={\frac {q^{2}u^{2}}{4l\mathrm {K} v^{2}}}\left\{{\frac {a^{2}-l^{2}}{2l^{2}}}\log {\frac {a+l}{a-l}}-{\frac {a}{l}}\right\}}$.

Now by (17) we have at the surface of the ellipsoid

${\displaystyle \mathbf {\Psi } _{0}={\frac {q\alpha }{\mathrm {K} }}\int _{a}^{\infty }{\frac {dh}{h^{2}-l^{2}}}={\frac {q\alpha }{2\mathrm {K} l}}\log {\frac {a+l}{a-l}}}$.

Hence the total electromagnetic energy of the ellipsoid is

 ${\displaystyle \mathrm {W} ={\frac {1}{2}}q\mathbf {\Psi } _{0}+2\mathrm {T} ={\frac {q^{2}}{4\mathrm {K} l}}\left\{\left(1+{\frac {u^{2}a^{2}}{v^{2}l^{2}}}\right)\log {\frac {a+l}{a-l}}-2{\frac {u^{2}a}{v^{2}l}}\right\}}$. (23)

Here we must remember that ${\displaystyle l^{2}=a^{2}-\alpha b^{2}}$.

(A) Energy of Heaviside Ellipsoid. If we put ${\displaystyle a/l}$= S and make S large we have

 {\displaystyle {\begin{aligned}\mathrm {W} &={\frac {q^{2}\mathrm {S} }{2\mathrm {K} a}}\left\{\left(1+{\frac {u^{2}\mathrm {S} ^{2}}{v^{2}}}\right)\left({\frac {1}{\mathrm {S} }}+{\frac {1}{3\mathrm {S} ^{3}}}+\cdot \cdot \cdot \right)-{\frac {u^{2}\mathrm {S} }{v^{2}}}\right\}\\&={\frac {q^{2}}{2\mathrm {K} a}}\left(1+{\frac {1}{3}}{\frac {u^{2}}{v^{2}}}\right)\ \mathrm {when} \ \mathrm {S} =\infty \end{aligned}}} (24)

This corresponds to the Heaviside ellipsoid, for when ${\displaystyle S=\infty }$, ${\displaystyle a^{2}=ab^{2}}$. The energy of the same ellipsoid at rest is

${\displaystyle {\frac {q^{2}{\sqrt {\alpha }}}{2\mathrm {K} a}}\cdot {\frac {v}{u}}\sin ^{-1}{\frac {u}{v}}}$.

(B) Energy of a Sphere. Putting ${\displaystyle b=a}$ we have ${\displaystyle l=au/v}$, and thus

 ${\displaystyle \mathrm {W} ={\frac {q^{2}}{2\mathrm {K} a}}\left({\frac {v}{u}}\log {\frac {v+u}{v-u}}-1\right)}$. (25)

If ${\displaystyle u}$ is small compared with ${\displaystyle v}$ we have

${\displaystyle \mathrm {W} ={\frac {q^{2}}{2\mathrm {K} a}}\left(1+{\frac {2}{3}}{\frac {u^{2}}{v^{2}}}+\cdot \cdot \cdot \right)}$.

It will be found that as far as ${\displaystyle u^{2}/v^{2}}$ the magnetic energy is

${\displaystyle {\frac {q^{2}u^{2}}{3\mathrm {K} av^{2}}}={\frac {\mu q^{2}u^{2}}{3a}}}$

as has been found by Mr. Heaviside.[1] It follows from this

1. 'Electrical Papers' vol. ii. p. 505.