# Page:SearleEllipsoid.djvu/3

made by Mr. Oliver Heaviside, F.H.S.,[1] that a distribution of electricity on the surface of a charged body such as to give zero disturbance at all points inside the surface is an equilibrium distribution. Since $\mathbf{F}$ satisfies curl $\mathbf{F} =0$ and $\mathbf{F}$ vanishes inside the surface, it follows that on the outside of the surface $\mathbf{F}$ is perpendicular to the surface. This implies that $\mathbf{\Psi}$ is constant over the surface. But as neither the electric force $\mathbf{E}$ nor the mechanical force experienced by each part of the charged surface (calculated from the Maxwell stress) is normal to the surface, I felt unable to accept the validity of Mr. Heaviside's assumption until I discovered {§ 15} that $\mathbf{F}$ is the mechanical force on an isolated moving unit charge, and that the term $-V\mathbf{GD}$, which appears in the expression for the force experienced by the surface, has no influence in causing convection of electricity from one part of the surface to another. Here $\mathbf{D}$ is the electric displacement, and $\mathbf{G}$ the "magnetic current" $\mu\tfrac{d\mathbf{H}}{dt}$.
Since $\mathbf{\Psi}$ is a true potential for the mechanical force $\mathbf{F}$, I have called $\mathbf{\Psi}$ the "electric convection potential."
When there has been established the boundary condition that $\mathbf{\Psi}$ is constant over the surface, with its consequence that there is zero disturbance within the surface, it is very easy to show that the distribution on an ellipsoid is the same for motion as for rest. Suppose the ellipsoid to have the same distribution as when it is at rest, so that $\sigma =qp/4\pi abc$, where $q$ is the charge, $a, b, c$ the axes of the ellipsoid, and $p$ the perpendicular from the centre upon the tangent-plane at the point. Through any internal point M as vertex draw a slender double cone intercepting two areas N, N' on the surface. Now the electric force due to a moving point-charge is still radial and still varies inversely as the square of the distance, although it alters with change of direction of the radius vector. Thus it follows just as in electrostatics, since $\sigma \propto p$, that the effects at M of N and N' are exactly equal and opposite. The whole surface can be treated in the same manner, and thus it follows that $\mathbf{E} = 0$ at all internal points. Hence $\mathbf{H} = 0$ also. Thus the assumed distribution is in equilibrium and is therefore the actual distribution. Thus the motion has no influence upon the distribution, and this result is true whatever the direction of motion with respect to the axes of the ellipsoid.
In order to find the state of the field near a charged ellipsoid moving with velocity $u$ parallel to the aria of $x$, it is necessary to find a value of $\mathbf{\Psi}$ which shall be constant over