Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/269

When there is no free electricity, this is reduced to the equation of Laplace,

${\frac {d^{2}V}{dx^{2}}}+{\frac {d^{2}V}{dy^{2}}}=0$

The general problem of electric equilibrium may be stated as follows:–

A continuous space of two dimensions, bounded by closed curves $C_{1},\ C_{2}$, &c. being given, to find the form of a function, $V$, such that at these boundaries its value may be $V_{1},\ V_{2}$, &c. respectively, being constant for each boundary, and that within this space $V$ may be everywhere finite, continuous, and single valued, and may satisfy Laplace's equation.

I am not aware that any perfectly general solution of even this question has been given, but the method of transformation given in Art. 190 is applicable to this case, and is much more powerful than any known method applicable to three dimensions.

The method depends on the properties of conjugate functions of two variables.

Definition of Conjugate Functions.

183.] Two quantities $\alpha$ and $\beta$ are said to be conjugate functions of $x$ and $y$, if $\alpha +{\sqrt {-1}}\beta$ is a function of $x+{\sqrt {-1}}y$.

It follows from this definition that

${\frac {d\alpha }{dx}}={\frac {d\beta }{dy}},\ \mathrm {and} \ {\frac {d\alpha }{dy}}+{\frac {d\beta }{dx}}=0$;

(1)

${\frac {d^{2}\alpha }{dx^{2}}}+{\frac {d^{2}\alpha }{dy^{2}}}=0,\ {\frac {d^{2}\beta }{dx^{2}}}+{\frac {d^{2}\beta }{dy^{2}}}=0$

(2)

Hence both functions satisfy Laplace's equation. Also

${\frac {d\alpha }{dx}}{\frac {d\beta }{dy}}-{\frac {d\alpha }{dy}}{\frac {d\beta }{dx}}=\left({\frac {d\alpha }{dx}}\right)^{2}+\left({\frac {d\alpha }{dy}}\right)^{2}=\left({\frac {d\beta }{dx}}\right)^{2}+\left({\frac {d\beta }{dy}}\right)^{2}=R^{2}$.

(3)

If $x$ and $y$ are rectangular coordinates, and if $ds_{1}$ is the intercept of the curve ($\beta$ = constant) between the curves $\alpha$ and $\alpha +d\alpha$, and $ds_{2}$ the intercept of a between the curves $\beta$ and $\beta +d\beta$, then

${\frac {ds_{1}}{d\alpha }}={\frac {ds_{2}}{d\beta }}={\frac {1}{R}}$

(4)

and the curves intersect at right angles.

If we suppose the potential $V=V_{0}+k\alpha$, where $k$ is some constant, then $V$ will satisfy Laplace's equation, and the curves ($\alpha$) will be equipotential curves. The curves ($\beta$) will be lines of force, and