Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/298

658.] Let us now determine the electromotive force at any point of the sheet, supposing the sheet fixed.

Let $X$ and $Y$ be the components of the electromotive force parallel to $x$ and to $y$ respectively, then, by Art. 598, we have

If the electric resistance of the sheet is uniform and equal to $\sigma$,

where $u$ and $v$ are the components of the current, and if $\phi$ is the current-function,

But, by equation (3),

$2\pi \phi =-{\frac {dP}{dz}}$

at the positive surface of the current-sheet. Hence, equations (12) and (13) may be written

$-{}$	${\frac {\sigma }{2\pi }}{\frac {d^{2}P}{dy\,dz}}=-{\frac {d^{2}}{dy\,dt}}(P+P^{\prime })-{\frac {d\psi }{dx}}$,
	${\frac {\sigma }{2\pi }}{\frac {d^{2}P}{dx\,dz}}={\frac {d^{2}}{dx\,dt}}(P+P^{\prime })-{\frac {d\psi }{dy}}$,

(16)

(17)

where the values of the expressions are those corresponding to the positive surface of the sheet.

If we differentiate the first of these equations with respect to $x$, and the second with respect to $y$, and add the results, we obtain

The only value of $\psi$ which satisfies this equation, and is finite and continuous at every point of the plane, and vanishes at an infinite distance, is

Hence the induction of electric currents in an infinite plane sheet of uniform conductivity is not accompanied with differences of electric potential in different parts of the sheet.

Substituting this value of $\psi$, and integrating equations (16), (17), we obtain

Since the values of the currents in the sheet are found by