Page:SearleEllipsoid.djvu/7

 moving charges. When they move parallel to each other with the speed of light the force between them vanishes.

If the ellipsoid is more oblate than Heaviside's the limiting internal surface of ellipsoidal form, whose action is the same as that of the ellipsoid, is a disk of radius $$\sqrt{b^{2}-a^{2}/\alpha}$$, the axis of the disk coinciding with the axis of $$x$$.

The form of the lines of the electric force $$\mathbf{E}$$ due to an ellipsoid of revolution is easily found. Putting $$\rho^{2}$$ for $$y^{2}+z^{2}$$, the equilibrium surfaces are given by

Now the mechanical force $$\mathbf{F}$$ is normal to this surface, and therefore

$\frac{\mathbf{F}_{\rho}}{\mathbf{F}_{1}}=\frac{\rho(a^{2}+\alpha\lambda)}{x(b^{2}+\lambda)}$,|undefined

where

$\mathbf{F}_{\rho}^{2}=\mathbf{F}_{2}^{2}+\mathbf{F}_{3}^{2}$

But by (5),

$\mathbf{E}_{1}=\mathbf{F}_{1}$ and $\mathbf{E}_{\rho}=\mathbf{F}_{\rho}/\alpha$ ;

so that

Now consider the conic

The tangent of the angle which the geometrical tangent makes with the axis of $$x$$ is

But if the point $$x, \rho$$ lies on both (13) and (15), it follows that

$-\frac{x(b^{2}+v)}{\rho(a^{2}+\alpha v)}=\frac{1}{\alpha}\frac{\rho(a^{2}+\alpha\lambda)}{x(b^{2}+\lambda)}$.

Hence by (14) and (16) the electric force is always tangential to the conic (15). But this conic has exactly the same equation as the equilibrium surfaces. Thus the single equation (13) represents both the equilibrium surfaces and the lines of electric force.

If any point $$x, \rho$$ be taken, there are two values of $$\lambda$$ which will satisfy (13) considered as a quadratic in $$\lambda$$. One value