Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/247

of two spherical segments meeting at a re-entering angle ${\tfrac {\pi }{n}}$, charged to potential unity and placed in free space.

For this purpose we invert the system with respect to $P$. The circle on which the images formerly lay now becomes a straight line through the centres of the spheres.

Fig. 11

If the figure (11) represents a section through the line of centres $AB$, and if $D,D'$ are the points where the circle of intersection cuts the plane of the paper, then, to find the successive images, draw $DA$ a radius of the first circle, and draw $DC,DB$, &c., making angles ${\frac {\pi }{n}},\ {\frac {2\pi }{n}}$, &c. with $DA$. The points $C,B$, &c. at which they cut the line of centres will he the positions of the positive images, and the charge of each will be represented by its distances from $D$. The last of these images will be at the centre of the second circle.

To find the negative images draw $DP,DQ$, &c., making angles ${\frac {\pi }{n}},\ {\frac {2\pi }{n}}$, &c. with the line of centres. The intersections of these lines with the line of centres will give the positions of the negative images, and the charge of each will be represented by its distance from $D$.

The surface-density at any point of either sphere is the sum of the surface-densities due to the system of images. For instance, the surface-density at any point $S$ of the sphere whose centre is $A$, is

$\sigma ={\frac {1}{4\pi DA}}\left\{1+\left(AD^{2}-AB^{2}\right){\frac {DB}{BS^{3}}}+\left(AD^{2}-AC^{2}\right){\frac {DC}{CS^{3}}}+etc.\right\},$

where $A,B,C$, &c. are the positive series of images.

When $S$ is on the circle of intersection the density is zero.

To find the total charge on each of the spherical segments, we may find the surface-integral of the induction through that segment due to each of the images.

The total charge on the segment whose centre is $A$ due to the image at $A$ whose charge is $DA$ is