Page:Thomson1881.djvu/2

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§ 2. The first case we shall consider is that of a charged sphere moving through an unlimited space filled with a medium of specific inductive capacity K.

The charged sphere will produce an electric displacement throughout the field; and as the sphere moves the magnitude of this displacement at any point will vary. Now, according to Maxwell's theory, a variation in the electric displacement produces the same effect as an electric current; and a field in which electric currents exist is a seat of energy; hence the motion of the charged sphere has developed energy, and consequently the charged sphere must experience a resistance as it moves through the dielectric. But as the theory of the variation of the electric displacement does not take into account any thing corresponding to resistance in conductors, there can he no dissipation of energy through the medium; hence the resistance cannot be analogous to an ordinary frictional" resistance, but must correspond to the resistance theoretically experienced by a solid in moving through a perfect fluid. In other words, it must be equivalent to an increase in the mass of the charged moving sphere, which we now proceed to calculate.

Let a be the radius of the moving sphere, e the charge on the sphere, and let us suppose that the sphere is moving parallel to the axis of x with the velocity p; let ξ, η, ζ be the coordinates of the centre of the sphere; let f, g, h be the components of the electric displacement along the axes of x, y, z respectively at a point whose distance from the centre of the sphere is ρ, ρ being greater than a. Then, neglecting the self-induction of the system (since the electromotive forces it produces are small compared with those due to the direct action of the charged sphere), we have

f=-\frac{e}{4\pi}\frac{d}{dx}\frac{1}{\rho},

g=-\frac{e}{4\pi}\frac{d}{dy}\frac{1}{\rho},

h=-\frac{e}{4\pi}\frac{d}{dz}\frac{1}{\rho};

therefore

\frac{df}{dt}=-\frac{ep}{4\pi}\frac{d^{2}}{dx\ d\xi}\frac{1}{\rho},

\frac{dg}{dt}=-\frac{ep}{4\pi}\frac{d^{2}}{d\xi\ dy}\frac{1}{\rho},

\frac{dh}{dt}=-\frac{e}{4\pi}\frac{d^{2}}{d\xi\ dz}\frac{1}{\rho};