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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'.

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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.

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  • $\begingroup$ Thank you I understand the logic but I still have some ambiguity regarding the ‘thin’ conductor. Is it really a 2d ‘surface’, or it is just a good approximation of a 3d volume? If the conductor was not grounded, by your explanation, the positive charges would reside on the other side of the plate, right? But I couldn’t imagine the charge configuration if it were literally thin ( exact 2d), so I was confused about its dimension. $\endgroup$ Commented Dec 28, 2023 at 5:00
  • $\begingroup$ 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. $\endgroup$
    – JohnA.
    Commented Dec 28, 2023 at 5:10

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