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line marked $\lambda =0$ is the "image". The semi-length of the line and the radius of the disk are each taken as unity.

I have attempted to find the lines ot the mechanical force F, these being everywhere perpendicular to the equilibrium surfaces. But the process involved an impracticable integration, and thus led to no result.

I will now write down the values of $\mathbf {E}$ and $\mathbf {H}$ at any point near the ellipsoid of revolution with axes $a,b,b$. Instead of $\lambda$ it will be convenient to take as the parameter of any one of the equilibrium surfaces its $x$ axis and to denote this by $h$. Thus

$h^{2}{=}a^{2}+\alpha \lambda$ ;

and consequently if we put $l^{2}$ for $a^{2}-\alpha b^{2}$, so that $l$ is the semi-length of the line which is the "image" of the ellipsoid, we have

$b^{2}+\lambda {=}{\frac {h^{2}-l^{2}}{\alpha }}$.