# Pre-Quantum Electrodynamics
Author: Jean-Sébastien Caux

##### 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. Now if, however, we change the value of the charge at \(z = -d\) from \(q\) to \(q'\): what do we get ? The potential is \[ V(x,y,z) = \frac{1}{4\pi \varepsilon_0} \left[ \frac{q}{[x^2 + y^2 + (z-d)^2]^{1/2}}

- \frac{q'}{[x^2 + y^2 + (z+d)^2]^{1/2}} \right]

\] This vanishes when

\begin{align} \left(\frac{q}{q'}\right)^2 \frac{ x^2 + y^2 + z^2 + d^2 + 2dz}{x^2 + y^2 + z^2 + d^2 - 2dz} = 1, \rightarrow x^2 + y^2 + z^2 + d^2 + 2d \frac{\left(\frac{q}{q'}\right)^2 + 1}{\left(\frac{q}{q'}\right)^2 - 1} z = 0 \nonumber \\ \rightarrow x^2 + y^2 + (z + d\alpha)^2 = d^2 (\alpha^2 - 1), \hspace{1cm} \alpha \equiv \frac{\left(\frac{q}{q'}\right)^2 + 1}{\left(\frac{q}{q'}\right)^2 - 1}. \end{align}

This only makes sense if \(\alpha > 1\), {\it i.e.} \(q' < q\), otherwise the radius is not positive definite. Therefore, the equipotential \(V = 0\) is a sphere of radius \(R = d\sqrt{\alpha^2 - 1}\) centered at \(z = -d\alpha\). The problem is therefore equivalent to a grounded metal sphere of that radius centered at this position, with a single point charge \(q\) at \(d\hat{z}\) ! (Griffiths pulls that out of a hat, but you now see where it comes from). Correspondence with Griffiths Ex 3.2: his \(a-b\) is my \(2d\), but his \(b\) is (3.16) \(R^2/a\), so we get \(a - R^2/a = 2d\), whose solution is \(a = d[1 + \alpha]\).

Author: Jean-Sébastien Caux

Created: 2022-02-08 Tue 06:55

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