Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/292

line can be transformed by continuous motion from one form to the other without passing through an electrode. For the two forms of the line will enclose an area within which there is no electrode, and therefore the same quantity of electricity which enters the area across one of the lines must issue across the other.

If $s$ denote the length of the line $AP$, the current across $ds$ from left to right will be ${\frac {d\phi }{ds}}\,ds$.

If $\phi$ is constant for any curve, there is no current across it. Such a curve is called a Current-line or a Stream-line.

649.] Let $\psi$ be the electric potential at any point of the sheet, then the electromotive force along any element $ds$ of a curve will be

$-{\frac {d\psi }{ds}}ds$,

provided no electromotive force exists except that which arises from differences of potential.

If $\psi$ is constant for any curve, the curve is called an Equipotential Line.

650.] We may now suppose that the position of a point on the sheet is defined by the values of $\phi$ and $\psi$ at that point. Let $ds_{1}$ be the length of the element of the equipotential line $\psi$ intercepted between the two current lines $\phi$ and $\phi +d\phi$, and let $ds_{2}$ be the length of the element of the current line $\phi$ intercepted between the two equipotential lines $\psi$ and $\psi +d\psi$. We may consider $ds_{1}$ and $ds_{2}$ as the sides of the element d<p d^\r of the sheet. The electromotive force $-d\psi$ in the direction of $ds_{2}$ produces the current $d\phi$ across $ds_{1}$.

Let the resistance of a portion of the sheet whose length is $ds_{2}$, and whose breadth is $ds_{1}$, be

$\sigma {\frac {ds_{2}}{ds_{1}}}$,

where $\sigma$ is the specific resistance of the sheet referred to unit of area, then

$d\psi =\sigma {\frac {ds_{2}}{ds_{1}}}d\phi$,

whence

${\frac {ds_{1}}{d\phi }}=\sigma {\frac {ds_{2}}{d\psi }}$.

651.] If the sheet is of a substance which conducts equally well in all directions, $ds_{1}$ is perpendicular to $ds_{2}$. In the case of a sheet of uniform resistance $\sigma$ is constant, and if we make $\psi ^{\prime }=\sigma \psi$, we shall have

${\frac {ds_{1}}{ds_{2}}}={\frac {d\phi }{d\psi }}$,

and the stream-lines and equipotential lines will cut the surface into little squares.