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 and the function ω0 itself will satisfy the condition

Comparing this with (2), we see that in corresponding points

and consequently

In virtue of what has been remarked at the end of § 4, the components of the electric force in the system S will therefore be

Parallel to OX we have the same electric force in S and S0, but in a direction perpendicular to OX the electric force in S will be $$\frac{1}{k}$$ times the electric force in S0.

By means of this result every electrostatic problem for a moving system may be reduced to a similar problem for a system at rest; only the dimensions in the direction of translation must be slightly different in the two systems. If, e.g., we wish to determine in what way innumerable ions will distribute themselves over a moving conductor C, we have to solve the same problem for a conductor C0, having no translation. It is easy to show that if the dimensions of C0 and C differ from each other in the way that has been indicated, the electric force in one case will be perpendicular to the surface of C, as soon as, in the other case, the force $$\mathfrak{E}_{0}$$ is normal to the surface of C0.

Since

exceeds unity only by a quantity of the second order — if we call $$\frac{\mathfrak{p}_{x}}{V}$$ of the first order — the influence of the earth's yearly motion on electrostatic phenomena will likewise be of the second order.