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F=\frac{\mu ep}{4\pi}\int\int\int\frac{1}{PQ}\frac{d^{2}}{dx^{2}}\frac{1}{\rho}dx\ dy\ dz;

Now \frac{d^{2}}{dx^{2}}\frac{1}{\rho}=\frac{Y_{2}}{\rho^{3}}, where Y2 is a surface harmonic of the second order. And when ρ > R,


and when ρ < R,


where Q1, Q2, &c. are zonal harmonics of the first and second orders respectively referred to OP as axis.

Let Y'2 denote the value of Y2 along OP. Then, since \int Y_{n}Q_{m}ds, integrated over a sphere of unit radius, is zero when n and m are different, and \frac{4\pi}{2n+1}Y_{n}^{'} when n=m, Y'n being the value of Yn at the pole of Qn, and since there is no electric displacement within the sphere,

F=\frac{\mu ep}{4\pi}\times\frac{2\pi Y_{2}^{'}}{5}\left\{ \int_{R}^{\infty}\frac{R^{2}}{\rho^{4}}d\rho+\int_{a}^{R}\frac{\rho\ d\rho}{R^{3}}\right\}

=\frac{\mu ep}{5}Y_{2}^{'}\left(\frac{5}{6R}-\frac{a^{2}}{2R^{3}}\right),

or, as it is more convenient to write it,

=\frac{\mu ep}{5}\left(\frac{5R^{2}}{6}-\frac{a^{2}}{2}\right)\frac{d^{2}}{dx^{2}}\frac{1}{R}.

By symmetry, the corresponding values of G and H are

G=\frac{\mu ep}{5}\left(\frac{5R^{2}}{6}-\frac{a^{2}}{2}\right)\frac{d^{2}}{dx\ dy}\frac{1}{R},

H=\frac{\mu ep}{5}\left(\frac{5R^{2}}{6}-\frac{a^{2}}{2}\right)\frac{d^{2}}{dx\ dz}\frac{1}{R}.

These values, however, do not satisfy the condition


If, however, we add to F the term \frac{2\mu ep}{3R}, this condition will be satisfied; while, since the term satisfies Laplace's equation, the other conditions will not be affected: thus we have finally