Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/278

EXAMPLE IV. – Distribution of Electricity near an Edge of a Conductor formed by Two Plane Faces.

191.] In the case of an infinite plane face of a conductor charged with electricity to the surface-density $\sigma _{0}$, we find for the potential at a distance $y$ from the plane

$V=C-4\pi \sigma _{0}y$

where $C$ is the value of the potential of the conductor itself.

Assume a straight line in the plane as a polar axis, and transform into polar coordinates, and we find for the potential

$V=C-4\pi \sigma _{0}ae^{\rho }\sin \theta$

and for the quantity of electricity on a parallelogram of breadth unity, and length $ae^{\rho }$ measured from the axis

$E=\sigma _{0}ae^{\rho }$

Now let us make $\rho =n\rho '$ and $\theta =n\theta '$, then, since $\rho '$ and $\theta '$ are conjugate to $\rho$ and $\theta$, the equations

$V=C-4\pi \sigma _{0}ae^{n\rho '}\sin n\theta '$

and

$E=\sigma _{0}ae^{n\rho '}$

express a possible distribution of electricity and of potential.

If we write $ae^{\rho '}$, $r$ will be the distance from the axis, and $\theta$ the angle, and we shall have

${\begin{array}{l}V=C-4\pi \sigma _{0}{\frac {r^{n}}{a^{n-1}}}\sin n\theta \\\\E=\sigma _{0}{\frac {r^{n}}{a^{n-1}}}\end{array}}$

$V$ will be equal to $C$ whenever $n\theta =\pi$ or a multiple of $\pi$.

Let the edge be a salient angle of the conductor, the inclination of the faces being $\alpha$, then the angle of the dielectric is $2\pi -\alpha$, so that when $\theta =2\pi -\alpha$ the point is in the other face of the conductor. We must therefore make

$n(2\pi -\alpha )=\pi$ or $n={\frac {\pi }{2\pi -\alpha }}$.

Then

${\begin{array}{l}V=C-4\pi \sigma _{0}a\left({\frac {r}{a}}\right)^{\frac {\pi }{2\pi -\alpha }}\sin {\frac {\pi \theta }{2\pi -a}}\\\\E=\sigma _{0}a\left({\frac {r}{a}}\right)^{\frac {\pi }{2\pi -\alpha }}\end{array}}$

The surface-density $\sigma$ at any distance $r$ from the edge is

$\sigma ={\frac {dE}{dr}}={\frac {\pi }{2\pi -\alpha }}\sigma _{0}\left({\frac {r}{a}}\right)^{\frac {\alpha -\pi }{2\pi -\alpha }}$