Page:Motion of Electrification through a Dielectric.djvu/10

 disturbance the difference between V and the electrostatic difference of potential is great.

But it is worth noticing, as a rather remarkable circumstance, that when we derive the system (32) by elementary considerations, viz. by extending the diffusion-system by the addition of the of inertia and leakage-current, we apparently as a matter of course take V to mean the same as in the diffusion-system. The resulting equations are correct, and yet the assumption is certainly wrong. The true way appears to be that given by me in the paper last referred to, by considering the line-integral of electric force in a closed curve. [vol. II., p. 187. Also p. 87]. We cannot, indeed, make a separation of the electric force of inertia from $$\scriptstyle{-\nabla\Psi}$$ without some assumption, though the former is quite definite when the latter is suitably defined. But, and this is the really important matter, it would be in the highest degree inconvenient, and lead to much complication and some confusion, to split V into two components, in other words, to bring in Ψ and A.

In thus running down Ψ, I am by no means forgetful of its utility in other cases. But it has perhaps been greatly misused. The clearest course to pursue appears to me to invariably make E and H the primary objects of attention, and only use potentials when they naturally suggest themselves as labour-saving appliances.

Special Tests. The Connecting Equations.

13. Returning to the solutions (29), the following are the special tests of their accuracy. Let E1 and E2 be the z and h components of E. Then, by (11) and (13), with the special meaning assumed by p, we have

{{centre|$$\left.\begin{array}{rcl} \scriptstyle{\frac{1}{h}\frac{d}{dh}hH} & \scriptstyle{=} & \scriptstyle{-cu\frac{dE_{1}}{dz},}\\ \scriptstyle{-\frac{dH}{dz}=-cu\frac{dE_{2}}{dz},\quad} & \scriptstyle{\mathrm{or}} & \scriptstyle{\quad H=cuE_{2}}\\ \scriptstyle{\frac{dE_{1}}{dh}-\frac{dE_{2}}{dz}=-\mu_{0}u\frac{dH}{dz},\quad} & \scriptstyle{\mathrm{or}} & \scriptstyle{\quad\frac{dE_{1}}{dh}=\left(1-\frac{u^{2}}{v^{2}}\right)\frac{dE_{2}}{dz}.}\end{array}\right\}\scriptstyle{\cdots\text{(33)}}$$}}

In addition to satisfying these equations, the displacement outward through any spherical surface centred at the charge may be verified to be q; this completes the test of the accuracy of (29).

But (33) are not limited to the case of a single point-charge, being true outside the electrification when there is symmetry with respect to the z-axis, and the electrification is all moving parallel to it at speed u.

When u=v, E1=0, and E2=E=μvH, so that we reduce to

outside the electrification. Thus, if the electrification is on the axis of z, we have

differing from (31) only in that q, the linear density, may be any function of z.