Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/245

If we now superpose on this system a charge $e$ at $A$, where

$e=-{\frac {e'}{a'}}R,$

(23)

the potential on the spherical surface, and at all points on the same side as $B$, will be reduced to zero. At all points on the same side as $A$ the potential will be that due to a charge $e$ at $A$, and a charge $e'{\frac {R}{f'}}$ at $B$.

But

$e'{\frac {R}{f'}}=-e{\frac {a'}{f'}}=-e{\frac {a}{f}},$

(24)

as we found before for the charge of the image at $B$.

To find the density at any point of the first sphere we have

$\sigma =\sigma '{\frac {R^{3}}{AP^{3}}}.$

(25)

Substituting for the value of $\sigma '$ in terms of the quantities be longing to the first sphere, we find the same value as in Art. 158,

$\sigma =-{\frac {e\left(f^{2}-a^{2}\right)}{4\pi aAP^{3}}}.$

(26)

On Finite Systems of Successive Images.

165.] If two conducting planes intersect at an angle which is a submultiple of two right angles, there will be a finite system of images which will completely determine the electrification.

Fig. 10

For let $AOB$ be a section of the two conducting planes per pendicular to their line of inter section, and let the angle of intersection $AOB={\frac {\pi }{n}},$, let $P$ be an electrified point, and let $PO=r$, and $POB=\theta$. Then, if we draw a circle with centre and radius $OP$, and find points which are the successive images of $P$ in the two planes beginning with $OB$, we shall find $Q_{1}$ for the image of $P$ in $OB$, $P_{2}$ for the image of $Q_{1}$ in $OA$, $Q_{3}$ for that of $P_{2}$ in $OB$, $P_{3}$ for that of $Q_{3}$ in $OA$, and $Q_{2}$ for that of $P_{3}$ in $OB$.

If we had begun with the image of $P$ in $AO$ we should have found the same points in the reverse order $Q_{2},P_{3},Q_{3},P_{2},Q_{1}$, provided $AOB$ is a submultiple of two right angles.