Two infinite parallel grounded conducting planes are held a distance$\mathbf{a}$part. A point charge$\mathbf{q}$is placed in the region between them, a distance $\mathbf{x}$fromone plate. Find the force on ${\mathbf{q}}^{\mathbf{20}}$Check that your answer is correct for the special

cases $\mathbf{a}\mathbf{\to }\mathbf{\infty }$and $\mathbf{x}\mathbf{=}\frac{\mathbf{a}}{\mathbf{2}}$.

Here, using the concept of images charges to obtain the value of force.

The infinite parallel grounded conducting plates are separately by a distance a and the charge is placed at a distance $\mathrm{x}$from the left side of the plate.

Here, the dotted line is introduced so as to separate the induced charge.

From the reciprocation theorem the total induced charge on one of the plane is equal to $-\mathrm{q}$times the fractional perpendicular distance of the point charge from the other plane.

If the original charge is at a distance $(\mathrm{a}-\mathrm{x})$from the left side of the plate, the image charge$-\mathrm{q}$will be at a distance of $-(\mathrm{a}-\mathrm{x})$on the other side of the plate.

Write the expression for the force.

$\mathbf{F}\mathbf{=}\frac{{\mathbf{kq}}^{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}$ …… (1)

Here,$\mathrm{q}$ is the charge,$\mathrm{k}$ is the proportionality constant and $\mathrm{r}$is the distance.

Here, the positive image charge forces cancel in pairs, thus the net force of the negative image charges,

$\mathrm{F}=\frac{{\mathrm{q}}^2}{4{\mathrm{\pi \epsilon }}_0}\left\{\begin{array}{l}\frac{1}{\left[2\right(\mathrm{a}-\mathrm{x}){]}^2}+\frac{1}{[2\mathrm{a}+2(\mathrm{a}-\mathrm{x}){]}^2}+\frac{1}{[4\mathrm{a}+2(\mathrm{a}-\mathrm{x}){]}^2}+\dots \dots \\ -\frac{1}{(2\mathrm{x}{)}^2}-\frac{1}{(2\mathrm{a}+2\mathrm{x}{)}^2}-\frac{1}{(4\mathrm{a}+2\mathrm{x}{)}^2}-\dots ..\end{array}\right\}$

$=\frac{{\mathrm{q}}^2}{4{\mathrm{\pi \epsilon }}_0}\left\{\begin{array}{l}\frac{1}{4\left[\right(\mathrm{a}-\mathrm{x}){]}^2}+\frac{1}{4[\mathrm{a}+(\mathrm{a}-\mathrm{x}){]}^2}+\frac{1}{4[\mathrm{a}+2(\mathrm{a}-\mathrm{x}){]}^2}+\dots \dots \\ -\frac{1}{(4\mathrm{x}{)}^2}-\frac{1}{4(\mathrm{a}+\mathrm{x}{)}^2}-\frac{1}{4(2\mathrm{a}+\mathrm{x}{)}^2}-\dots \dots \end{array}\right\}$

$=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}^2}{4}\left\{\begin{array}{l}\frac{1}{\left[\right(\mathrm{a}-\mathrm{x}){]}^2}+\frac{1}{[\mathrm{a}+(\mathrm{a}-\mathrm{x}){]}^2}+\frac{1}{[\mathrm{a}+2(\mathrm{a}-\mathrm{x}){]}^2}+\dots \dots \\ -\frac{1}{(\mathrm{x}{)}^2}-\frac{1}{(\mathrm{a}+\mathrm{x}{)}^2}-\frac{1}{(2\mathrm{a}+\mathrm{x}{)}^2}-\dots \dots \end{array}\right\}$

Therefore, the equation of force is determined.

Rewrite the equation of force

$\mathrm{F}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}^2}{4}\left\{\begin{array}{l}\frac{1}{\left[\right(\mathrm{a}-\mathrm{x}){]}^2}+\frac{1}{[\mathrm{a}+(\mathrm{a}-\mathrm{x}){]}^2}+\frac{1}{[\mathrm{a}+2(\mathrm{a}-\mathrm{x}){]}^2}+\dots \dots \\ -\left(\frac{1}{(\mathrm{x}{)}^2}+\frac{1}{(\mathrm{a}+\mathrm{x}{)}^2}+\frac{1}{(2\mathrm{a}+\mathrm{x}{)}^2}+\dots \dots \right)\end{array}\right\}$…… (2)

Apply the condition $\mathrm{a}>>\mathrm{x}$and solve.

$\mathrm{F}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}^2}{4}\left\{\frac{1}{{\mathrm{a}}^2}+\frac{1}{[2\mathrm{a}{]}^2}+\frac{1}{[3\mathrm{a}{]}^2}+\dots ..-\left(\frac{1}{(\mathrm{x}{)}^2}+\frac{1}{(\mathrm{a}{)}^2}+\frac{1}{(2\mathrm{a}{)}^2}+\dots \dots \right)\right\}$

Now, take the value of $\mathrm{a}$to be infinity and determine the value of force.

$\mathrm{F}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}^2}{4}\left\{0+\dots \dots -\left(\frac{1}{(\mathrm{x}{)}^2}+0+\dots \dots \right)\right\}$

$=-\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}^2}{(2\mathrm{x}{)}^2}$

After that, substitute $\mathrm{x}=\frac{\mathrm{a}}{2}$in equation (2)

$\mathrm{F}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}^2}{4}\left\{\begin{array}{l}\frac{1}{{\left[\left(\mathrm{a}-\frac{\mathrm{a}}{2}\right)\right]}^2}+\frac{1}{{\left[\mathrm{a}+\left(\mathrm{a}-\frac{\mathrm{a}}{2}\right)\right]}^2}+\frac{1}{{\left[\mathrm{a}+2\left(\mathrm{a}-\frac{\mathrm{a}}{2}\right)\right]}^2}+\dots \dots \\ -\left(\frac{1}{{\left(\frac{\mathrm{a}}{2}\right)}^2}+\frac{1}{{\left(\mathrm{a}+\frac{\mathrm{a}}{2}\right)}^2}+\frac{1}{{\left(2\mathrm{a}+\frac{\mathrm{a}}{2}\right)}^2}+\dots \dots \right.\end{array}\right\}$

$\left.=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}^2}{4}\left\{\begin{array}{l}\frac{1}{{\left[\frac{\mathrm{a}}{2}\right]}^2}+\frac{1}{{\left[\frac{3\mathrm{a}}{2}\right]}^2}+\frac{1}{{\left[\frac{5\mathrm{a}}{2}\right]}^2}+\dots \dots \\ -\left(\frac{1}{{\left(\frac{\mathrm{a}}{2}\right)}^2}+\frac{1}{{\left(\frac{3\mathrm{a}}{2}\right)}^2}+\frac{1}{{\left(\frac{5\mathrm{a}}{2}\right)}^2}+\right.\end{array}\right\}\right\}$

$=0$

Thus, the force in between the infinite parallel grounded conductor is zero.