##### Other Image Problems
ems.ca.mi.o

We can also play around differently. Elaborating on the grounded conducting plane above: if we just change \(d\), the problem stays of the same nature. If we however change the value of the charge at \(z = -d/2\) from \(q\) to \(q'\), what do we get?

The potential is

\begin{equation*} \phi(x,y,z) = \frac{1}{4\pi \varepsilon_0} \left[ \frac{q}{[x^2 + y^2 + (z-d/2)^2]^{1/2}} - \frac{q'}{[x^2 + y^2 + (z+d/2)^2]^{1/2}} \right] \end{equation*}

This vanishes when

\begin{equation*} \left(\frac{q}{q'}\right)^2 \frac{ x^2 + y^2 + z^2 + (d/2)^2 + dz}{x^2 + y^2 + z^2 + (d/2)^2 - dz} = 1, \end{equation*}

or in other words

\begin{equation*} x^2 + y^2 + z^2 + (d/2)^2 + d \frac{\left(\frac{q}{q'}\right)^2 + 1}{\left(\frac{q}{q'}\right)^2 - 1} z = 0 \end{equation*}

Choosing (without loss of generality) \(q' < q\), we can write

\begin{equation*} \rightarrow x^2 + y^2 + (z + d\alpha/2)^2 = \frac{d^2}{4} (\alpha^2 - 1), \hspace{1cm} \alpha \equiv \frac{\left(\frac{q}{q'}\right)^2 + 1}{\left(\frac{q}{q'}\right)^2 - 1}. \end{equation*}

Therefore, the equipotential \(\phi = 0\) is a sphere of radius \(R = \frac{d}{2}\sqrt{\alpha^2 - 1}\) centered at \(z = -d\alpha/2\). The problem is therefore equivalent to a grounded metal sphere of that radius centered at this position, with a single point charge \(q\) at \(\frac{d}{2}~\hat{z}\).

N.B. Correspondence with Griffiths Ex 3.2: his \(a-b\) is my \(d\), but his \(b\) is (3.16) \(R^2/a\), so we get \(a - R^2/a = d\), whose solution is \(a = \frac{d}{2}[1 + \alpha]\).

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42