Charge near an ungrounded infinite conducting plane

[jsphy]

Sorry if this has been answered already, i searched for a while. I know how to solve the problem of the potential of a point charge near a grounded infinite conducting plane, and a line charge near an infinite conducting plane. If the plane isn't necessarily grounded, say its at some potential V_0, the how would you solve the problem Then. I was tempted to just say that its a constant term added to the original problem, since that wouldn't affect the electric field, or the surface charge on the plane, because those are derivatives of the potential and a constant term won't matter, but I am not sure if that is a valid argument, or if its correct. Thanks in advance

 

[andrewkirk]

Specifying a potential for the plane conveys no information, because only potential differences have physical effect. If we add any constant to all potentials in a problem, no physical change occurs.

Perhaps the scenario you want to consider is, what if the plane has an average excess positive charge density of $\rho$ Coulombs per square metre. The question is then, what is it that you want to calculate, eg the electric field near the charge, or the distribution of charge on the plane near the charge, or some other measurable quantity?

 

[vanhees71]

You can first solve the problem with the method of images for the homogeneous boundary condition, i.e., for a grounded plane $V(\vec{r})=0$ for $z=0$ (making the $xy$ plane of the Cartesian coordinate system the conducting plane. On the side, where your point charge is located, you have the corresponding potential (assuming the charge is located at $(0,0,z_Q)$ with $z_Q>0$
$$V_0(\vec{x})=\frac{Q}{4 \pi} \left (\frac{1}{|\vec{x}-z_Q \vec{e}_z|}+\frac{1}{|\vec{x}+z_Q \vec{e}_z|} \right ) \quad \text{for} \quad z>0.$$
On the other side of the plane, i.e., for $z<0$ you have $V_0=0$.

Using Gauss's Law to the half-space $z<0^+$, you immediately get the total influence charge as $-Q$. You can also verify this by directly integrating over the surface charge, i.e., $\sigma(x,y)=E_z(x,y,0^+)$.

Now you need to add for $z>0$ a field which doesn't spoil the boundary conditions to add an additional homogeneous surface charge (of course that's rather unphysical, because the total charge on the plane is infinite then). This obviously is achieved by adding a homogeneous field at $z>0$:
$$V(\vec{x})=V_0(\vec{x})-\frac{\sigma_{\text{add}}}{2} z \quad \text{for} \quad z>0$$
and
$$V(\vec{x})=+\frac{\sigma_{\text{add}}}{2} z \quad \text{for} \quad z>0.$$
The latter solution can be found by first integrating a homogeneous surface-charge distribution along a finite square of length $L$ and then taking looking at the limit $L \rightarrow 0$. You can subtract a constant diverging contribution from the potential to finally get the above solution.

Of course, the potential is still $0$ along the conducting plane, but as was stressed already above, an additive constant of the potential is physically irrelevant anyway.

The above solution has to be read as the approximate solution for a large conducting finite plane with the additional charge not too far away and looking at the field also not too close and beyond the edges. Then you just have a finite additional charge on the plane compared to the "grounded-plane solution".