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)}

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

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