Page:SearleEllipsoid.djvu/12

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When h is large the quantity in [ ]

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

vanishing when h=\infty.

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

\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

\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

\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 l^{2}=a^{2}-\alpha b^{2}.

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

\begin{align}
\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{align} (24)

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

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

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

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

If u is small compared with v we have

\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 u^2 /v^2 the magnetic energy is

\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.