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 $$\therefore T$$, the kinetic energy due to the electrification

Hence, if m be the mass of the sphere, the whole kinetic energy

or the effect of the electrification is the same as if the mass of the sphere were increased by $$\tfrac{4}{15}\tfrac{\mu e^{2}}{a}$$, or, if V be the potential of the sphere, by $$\tfrac{4}{15}\mu K^{2}V^{2}a$$.

To form some idea of what the increase of mass could amount to in the most favourable case, let us suppose the earth electrified to the highest potential possible without discharge, and calculate the consequent increase in mass. According to Dr. Macfarlane's experiments, published in the Philosophical Magazine for December 1880, the electric force in air at ordinary temperatures and pressures must not exceed $$3\times10^{12}$$ (electromagnetic system of units). The electric force just outside the sphere is V/a; hence the greatest possible value of V is $$3\times10^{12}a$$, where a is the radius of the earth. Putting this value for V, μ=1, $$K=\tfrac{1}{9\cdot10^{20}}$$, $$a=6\cdot4\times10^{8}$$, we get for the corresponding value of the increase of mass $$7\times10^{8}$$ grms., or about 650 tons, a mass which is quite insignificant when compared with the mass of the earth.

For spheres of different sizes, the greatest increase in mass varies as the cube of the radius; hence the ratio of this increase to the whole mass of the sphere is constant for all spheres of the same material; for spheres of different materials the ratio varies inversely as the density of the material.

If the body moves so that its velocities parallel to the axes of x, y, z respectively are p, q, r, then it is evident that the effect of the electrification will be equivalent to an increase of $$\tfrac{4}{15}\mu K^{2}V^{2}a\left(p^{2}+q^{2}+r^{3}\right)$$ in the mass of the sphere.

§ 3. To find the magnetic force produced by the moving sphere at any point in the field. By equations (2) we have, for points outside the sphere,