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?
