# 'Thin' conducting plate ambiguity

The situation at hand:

• We have an infinite, thin conducting, grounded ($$V=0$$) plate at $$z=0$$.
• Point charge (with charge +$$Q$$), at $$z = a$$.

How exactly are the charges distributed? I used the method of images and found the surface charge density is negative everywhere at $$(x,y,0)$$. Where did all the positive charges go? I can't imagine their route since the condition was a 'thin' plate - I thought they could not move to the other side since it is 'thin'.

The positive charges went to ground, which is possible since the plate is grounded. The connection to ground will carry away or bring in as much as needed to keep $$V=0$$ in the plate.
The potential due to a charge at $$z=a$$ is given by $$V\left(\rho,\varphi,z\right) = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q}{\sqrt{\rho^2 + \left(z-a \right)^2}} + \frac{-q}{\sqrt{\rho^2 + \left(z+a \right)^2}} \right) \,$$ which can be obtained by method of images noting we can add an additional charge $$-q$$ at $$z = -a$$. And thus charge surface density is given by $$\sigma = -\varepsilon_0 \left.\frac{\partial V}{\partial z} \right|_{z=0} = \frac{-q a}{2 \pi \left(\rho^2 + a^2\right)^{3/2} }$$ where $$\rho^2 = x^2 +y^2$$
If you find the total charge induced on the plate, it turns out to be $$-q$$ and so the residual $$+q$$ from the plate charge was taken away by the connection to ground, assuming the plate started off neutral.
• Mathematically speaking this is a $2D$ surface and should be treated as such in integrals etc... You can also think about it as a slab with thickness $\epsilon$ and take $\epsilon \rightarrow 0$. If the plate were ungrounded, the situation is different enough that it warrants its own separate analysis (and it is non-trivial) so I suggest you open a question about that. If the question has been solved please mark it so. Commented Dec 28, 2023 at 5:10