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,

{{MathForm2|(7)|$$\left.\begin{array}{c} \frac{r'}{r}=\frac{S'}{S}=\frac{a^{2}}{r^{2}}=\frac{r'^{2}}{a^{2}},\ \frac{V'}{V}=\frac{a^{4}}{r^{4}}=\frac{r'^{4}}{a^{4}},\\ \\ \frac{e'}{e}=1,\ \frac{\sigma'}{\sigma}=\frac{r^{2}}{a^{2}}=\frac{a^{2}}{r'^{2}},\ \frac{\rho'}{\rho}=\frac{r^{4}}{a^{4}}=\frac{a^{4}}{r'^{4}},\\ \\ \frac{\phi'}{\phi}=1. \end{array}\right\} $$}}

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

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}} $|undefined

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

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