Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/297

$x$ and to $y$ at the positive surface, and $\alpha ^{\prime }$, $\beta ^{\prime }$ those on the negative surface

Within the sheet the components vary continuously from $\alpha$ and $\beta$ to $\alpha ^{\prime }$ and $\beta ^{\prime }$.

The equations

${\frac {dH}{dy}}-{\frac {dG}{dz}}=-{\frac {d\Omega }{dx}}$, ${\frac {dF}{dz}}-{\frac {dH}{dx}}=-{\frac {d\Omega }{dy}}$, ${\frac {dG}{dx}}-{\frac {dF}{dy}}=-{\frac {d\Omega }{dx}}$,

(8)

which connect the components $F$, $G$, $H$ of the vector-potential due to the current-sheet with the scalar potential $\Omega$, are satisfied if we make

We may also obtain these values by direct integration, thus for $F$,

$F$	${}=\iint {\frac {u}{r}}\,dx^{\prime }\,dy^{\prime }=\iint {\frac {1}{r}}{\frac {d\phi }{dy}}\,dx^{\prime }\,dy^{\prime }$,
	${}=\int {\frac {\phi }{r}}\,dx^{\prime }-\iint \phi {\frac {d}{dy^{\prime }}}{\frac {1}{r}}\,dx^{\prime }\,dy^{\prime }$.

Since the integration is to be estimated over the infinite plane sheet, and since the first term vanishes at infinity, the expression is reduced to the second term; and by substituting

${\frac {d}{dy}}{\frac {1}{r}}$ for $-{\frac {d}{dy^{\prime }}}{\frac {1}{r}}$,

and remembering that $\phi$ depends on $x^{\prime }$ and $y^{\prime }$ and not on $x$, $y$, $z$, we obtain

$F$	${}={\frac {d}{dy}}\iint {\frac {\phi }{r}}\,dx^{\prime }\,dy^{\prime }$,
	${}={\frac {dP}{dy}}$, by (1).

If $\Omega ^{\prime }$ is the magnetic potential due to any magnetic or electric system external to the sheet, we may write

and we shall then have

for the components of the vector-potential due to this system.