The value of $\mathbf {\Psi }$ in terms of $h$ thus becomes

$\mathbf {\Psi } ={\frac {q\alpha }{\mathrm {K} }}\int _{h}^{\infty }{\frac {dh}{h^{2}-l^{2}}}$. |
(17) |

Equation (11) now becomes

${\frac {x^{2}}{h^{2}}}+{\frac {\rho ^{2}\alpha }{h^{2}-l^{2}}}=1$, |
(18) |

so that instead of the cylindrical coordinates $x$ and $\rho (={\sqrt {y^{2}+z^{2}}})$ we, can take $h$ and $\phi$ where

$x=h\cos \phi ,\ \rho ={\frac {\sqrt {h^{2}-l^{2}}}{\sqrt {\alpha }}}\sin \phi$. |
(19) |

From (18) we have in terms of $h$ and $\phi$

${\frac {dh}{dx}}={\frac {(h^{2}-l^{2})\cos \phi }{h^{2}-l^{2}\cos ^{2}\phi }},\ {\frac {dh}{d\rho }}={\frac {h{\sqrt {h^{2}-l^{2}}}\sin \phi {\sqrt {\alpha }}}{h^{2}-l^{2}\cos ^{2}\phi }}$.
Hence

$\mathrm {E} _{1}=-{\frac {d\mathbf {\Psi } }{dh}}\cdot {\frac {dh}{dx}}={\frac {\alpha q\ \cos \phi }{K{\sqrt {h^{2}-l^{2}\cos ^{2}\phi )}}}}$, |
(20) |

$\mathrm {E} _{\rho }=-{\frac {1}{\alpha }}{\frac {d\mathbf {\Psi } }{dh}}\cdot {\frac {dh}{d\rho }}={\frac {qh\ \sin \phi {\sqrt {\alpha }}}{\mathrm {K} {\sqrt {h^{2}-l^{2}}}(h^{2}-l^{2}\cos ^{2}\phi )}}$, |
(21) |

$\mathrm {H} =\mathrm {K} u\mathrm {E} _{\rho }={\frac {quh\ \sin \phi {\sqrt {\alpha }}}{\sqrt {h^{2}-l^{2}(h^{2}-l^{2}\cos ^{2}\phi )}}}$, |
(22) |

I now pass on to calculate the total energy possessed by the ellipsoid when in motion along its axis of figure. In making the calculation I shall suppose that $a^{2}>ab^{2}$, i.e., that $l^{2}$ is positive. The case in which $a^{2}<ab^{2}$ can be deduced by the appropriate mathematical transformation.

I have shown {§ 22} that the total energy, viz. the volume integral of ${\frac {\mathrm {K} \mathbf {E} ^{2}+\mu \mathbf {H} ^{2}}{8\pi }}$, due to the motion of a charge on any surface, is

$\mathrm {W} ={\frac {1}{2}}q\mathbf {\Psi } _{0}+2\mathrm {T}$,
where $\Psi _{0}$ is the value of the convection-potential at the surface of the body, and T is the magnetic part of the energy, viz., the volume integral of $\mu H^{2}/8\pi$.