Let's say we have a grounded conducting plane at $z=0$ and a charge moving above it with some position $\vec{r}_q(t)$ and velocity $\vec{v}_q(t)$. I know that in the particular case when the velocity is just zero the potential is

\begin{equation} \phi(\vec{r})=\frac{q}{\sqrt{(x-x_q)^2+(y-y_q)^2+(z-z_q)^2}}-\frac{q}{\sqrt{(x-x_q)^2+(y-y_q)^2+(z+z_q)^2}} \end{equation}

where we can think of the second term as being sourced by an image charge $-q$ at $\vec{r}_{im}=(x_q,y_q,-z_q)$.

The question is: Can we also use an image charge if the source is moving? I know that the solution in free space is given by the Lienard-Wiechert potential

\begin{equation} \phi_{LW}(\vec{r},t)=\frac{q}{|\vec{r}-\vec{r}_q(t')|\big(1-\frac{\vec{v}_q(t')}{c}\cdot\hat{n}\big)}|_{t_{ret}} \end{equation} where we are evaluating at $t'=t_{ret}$. Is the general solution just the Lienard-Wiechert potential plus a reflection potential sourced by a source with the same velocity but below the $z=0$ plane?

  • $\begingroup$ Guess: probably you can get an idealized solution as long as the image in the mirror is properly extrapolated backwards in time and you assume that the charge flow in the conductor is instantaneous, but the latter assumption seems highly suspect for very fast-moving charges. $\endgroup$
    – CR Drost
    Commented Oct 15, 2020 at 21:15
  • $\begingroup$ Thanks for the comment! I'm thinking about this in a very theoretical way, I'm not really interested in whether actual conductors are able to do it or not. $\endgroup$ Commented Oct 16, 2020 at 2:04

1 Answer 1


The method of images works if the charges and currents are mirror images (with opposite sign) about the x-y plane. One can use the way fields transform under reflection ($z\mapsto -z$ ) to verify that at $z=0$, the parallel E-field is $0$ and the perpendicular B field is $0$. These are the boundary conditions for a conducting plane.


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