Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/275

points from $O$, $e$ and $e'$ the total electrification of a body, $S$ and $S'$ superficial elements, $V$ and $V'$ solid elements, $\sigma$ and $\sigma '$ surface- densities, $\rho$ and $\rho '$ volume densities, $\phi$ and $\phi '$ corresponding potentials,

$\left.{\begin{array}{c}{\frac {r'}{r}}={\frac {S'}{S}}={\frac {a^{2}}{r^{2}}}={\frac {r'^{2}}{a^{2}}},\ {\frac {V'}{V}}={\frac {a^{4}}{r^{4}}}={\frac {r'^{4}}{a^{4}}},\\\\{\frac {e'}{e}}=1,\ {\frac {\sigma '}{\sigma }}={\frac {r^{2}}{a^{2}}}={\frac {a^{2}}{r'^{2}}},\ {\frac {\rho '}{\rho }}={\frac {r^{4}}{a^{4}}}={\frac {a^{4}}{r'^{4}}},\\\\{\frac {\phi '}{\phi }}=1.\end{array}}\right\}$

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EXAMPLE II. Electric Images in Two Dimensions.

189.] Let $A$ be the centre of a circle of radius $AQ=b$, and let $E$ be a charge at $A$, then the potential at any point $P$ is

$\phi =2E\log {\frac {b}{AP}}$

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Fig. 17.

and if the circle is a section of a hollow conducting cylinder, the surface-density at any point $Q$ is $-{\tfrac {E}{2\pi b}}$.

Invert the system with respect to a point $O$, making

$A=mb$ and $a^{2}=\left(m^{2}-1\right)b^{2}$;

then we have a charge at $A'$ equal to that at $A$, where $AA'={\tfrac {b}{m}}$.

The density at $Q'$ is

$-{\frac {E}{2\pi b}}{\frac {b^{2}-(AA')^{2}}{A'Q'^{2}}}$

and the potential at any point $P'$ within the circle is

${\begin{array}{ll}\phi '=\phi &=2E(\log b-\log AP),\\&=2E(\log OP'-\log A'P'-\log m).\end{array}}$

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This is equivalent to a combination of a charge $E$ at $A'$, and a charge $-E$ at $O$, which is the image of $A'$, with respect to the circle. The imaginary charge at $O$ is equal and opposite to that at $A'$.

If the point $P'$ is defined by its polar coordinates referred to the centre of the circle, and if we put

$\rho =\log r-\log b$ and $\rho _{0}=\log AA'-\log b$

then

$AP=be^{\rho },\ AA'=be^{\rho _{0}},\ AO=be^{-\rho _{0}}$;

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