Page:SearleEllipsoid.djvu/6

 approximate to Heaviside ellipsoids as $$\lambda$$ is made very great. The value of $$\mathbf{\Psi}$$ at the surface $$\lambda$$ is $$\tfrac{q\sqrt{\alpha}}{\sqrt{\lambda}}$$.

Putting $$c= b$$ so that we have an ellipsoid of revolution, the axis of revolution being the axis of $$x$$, we see by taking $$\lambda=-b^2$$ that a uniformly-charged line of length $$2\sqrt{a^{2}-b^{2}\alpha}$$ lying along the axis of $$x$$ produces exactly the same effect as the ellipsoid $$a, b, b$$. It may therefore be called its "image." When $$b=a$$ this length becomes $$2au/v$$. Thus, when a charged sphere is at rest it produces the same effect as a point-charge at its centre. When the sphere is in motion it produces the same effect as a uniformly-charged line whose length bears to the diameter of the sphere the same ratio as the velocity of the sphere bears to the velocity of light. When $$u=v$$, so that the sphere moves with the velocity of light, the line becomes the diameter of the sphere; and the same is true for an ellipsoid. Since when $$u=v$$ each element of the charged line produces a disturbance which is confined to the plane through the element perpendicular to the direction of motion {(46)}, it follows that the disturbance is entirely confined between the planes $$x=\pm a$$. Between them the electric force is radial to the axis of $$x$$ and has exactly the same value, viz. $$q/aK\rho$$, as if the line had been of infinite length and had had the same line-density $$q/2a$$. Here $$\rho$$ stands for $$\left\{ y^{2}+z^{2}\right\} ^{\frac{1}{2}}$$. The magnetic force is by (3) $$qu/a\rho$$. Hence the field between the planes $$x=\pm a$$ is independent of $$x$$. There are therefore no displacement-currents except in the two bounding-planes. There is an outward radial current in the front plane and an inward current in the back plane, the total amount of current in each case being $$qu$$, equal in amount to the convection-current carried by the ellipsoid.

It appears, however, that at the velocity of light any distribution on any surface is in equilibrium. For the value of $$\mathbf{\Psi}$$ at any point near a moving point-charge is {(43)}

$\mathbf{\Psi}=\frac{q\sqrt{\alpha}}{\mathrm{K}\sqrt{x^{2}/\alpha+y^{2}+z^{2}}}$,|undefined

and this vanishes when $$u=v$$ (so that $$\alpha =0$$), even when $$x=0$$. Thus the value of $$\mathbf{\Psi}$$ for a point-charge vanishes, and the value of $$\mathbf{\Psi}$$ for any distribution being derivable from that for a point-charge by integration, it follows that $$\mathbf{\Psi}$$ has the constant value zero everywhere. Hence the charge is in equilibrium however it may be distributed. The same result follows from the expression {§ 19} for the force between two