Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/294

In the case of the grating with two conducting planes whose equations are $\eta =-\beta _{1}$ and $\eta =-\beta _{2}$, that of the plane of the grating being $\eta =0$, there will be two infinite series of images of the grating. The first series will consist of the grating itself together with an infinite series of images on both sides, equal and similarly electrified. The axes of these imaginary cylinders lie in planes whose equations are of the form

$\eta =\pm 2n\left(\beta _{1}+\beta _{2}\right)$

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$n$ being an integer.

The second series will consist of an infinite series of images for which the coefficients $A_{0},\ A_{1},\ A_{2}$, &c. are equal and opposite to the same quantities in the grating itself, while $A_{1},\ A_{3}$ &c. are equal and of the same sign. The axes of these images are in planes whose equations are of the form

$\eta =2\beta _{2}\pm 2m\left(\beta _{1}+\beta _{2}\right)$

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$m$ being an integer.

The potential due to any finite series of such images will depend on whether the number of images is odd or even. Hence the potential due to an infinite series is indeterminate, but if we add to it the function $B\eta +C$, the conditions of the problem will be sufficient to determine the electrical distribution.

We may first determine $V_{1}$ and $V_{2}$, the potentials of the two conducting planes, in terms of the coefficients $A_{0},\ A_{1}$, &c., and of $B$ and $C$. We must then determine $\sigma _{1}$ and $\sigma _{2}$, the surface-density at any point of these planes. The mean values of $\sigma _{1}$ and $\sigma _{2}$ are given by the equations

$4\pi \sigma _{1}=A_{0}-B,\ 4\pi \sigma _{2}=A_{0}+B,$

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We must then expand the potentials due to the grating itself and to all the images in terms of $\rho$ and cosines of multiples of $\theta$ adding to the result

$B\rho \cos \theta +C$

The terms independent of $\theta$ then give $V$ the potential of the grating, and the coefficient of the cosine of each multiple of $\theta$ equated to zero gives an equation between the indeterminate coefficients.

In this way as many equations may be found as are sufficient to eliminate all these coefficients and to leave two equations to determine $\sigma _{1}$ and $\sigma _{2}$ in terms of $V_{1},\ V_{2}$, and $V$.

These equations will be of the form

${\begin{array}{l}V_{1}-V=4\pi \sigma _{1}\left(b_{1}+\alpha -\gamma \right)+4\pi \sigma _{2}(\alpha +\gamma ),\\V_{2}-V=4\pi \sigma _{1}\left(\alpha +\gamma \right)+4\pi \sigma _{2}(b_{2}+\alpha -\gamma ),\end{array}}$

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