Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/252

 Distribution of Electricity on Three Spherical Surfaces which Intersect at Right Angles.

169.] Let the radii of the spheres be $$\alpha, \beta, \gamma$$, then

$BC=\sqrt{\beta^{2}+\gamma^{2}},\ CA=\sqrt{\gamma^{2}+\alpha^{2}},\ AB=\sqrt{\alpha^{2}+\beta^{2}}.$|undefined

Let $$PQR$$, Fig. 13, be the feet of the perpendiculars from $$ABC$$ on the opposite sides of the triangle, and let $$O$$ be the intersection of perpendiculars.

Then $$P$$ is the image of $$B$$ in the sphere $$\gamma$$, and also the image of $$C$$ in the sphere $$\beta$$. Also $$O$$ is the image of $$P$$ in the sphere $$\alpha$$.

Let charges $$\alpha, \beta$$, and $$\gamma$$ be placed at $$A, B,$$ and $$C$$.

Then the charge to be placed at $$P$$ is

$-\frac{\beta\gamma}{\sqrt{\beta^{2}+\gamma^{2}}}=-\frac{1}{\sqrt{\frac{1}{\beta^{2}}+\frac{1}{\gamma^{2}}}}.$|undefined

Also $$AP=\tfrac{\sqrt{\beta^{2}\gamma^{2}+\gamma^{2}\alpha^{2}+\alpha^{2}\beta^{2}}}{\sqrt{\beta^{2}+\gamma^{2}}}$$, so that the charge at $$O$$, considered as the image of $$P$$, is

$\frac{\alpha\beta\gamma}{\sqrt{\beta^{2}\gamma^{2}+\gamma^{2}\alpha^{2}+\alpha^{2}\beta^{2}}}=\frac{1}{\sqrt{\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}+\frac{1}{\gamma^{2}}}}.$|undefined

In the same way we may find the system of images which are electrically equivalent to four spherical surfaces at potential unity intersecting at right angles.

If the radius of the fourth sphere is $$\delta$$, and if we make the charge at the centre of this sphere =$$\delta$$, then the charge at the intersection of the line of centres of any two spheres, say $$\alpha$$ and $$\beta$$, with their plane of intersection, is

$-\frac{1}{\sqrt{\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}}}$|undefined

The charge at the intersection of the plane of any three centres $$ABC$$ with the perpendicular from $$D$$ is

$+\frac{1}{\sqrt{\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}+\frac{1}{\gamma^{2}}}},$|undefined