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As shown in the picture below, there is a thin infinite conducting grounded sheet in the plane $z=0$, a charge $q$ in $(0,0,d)$ and another charge $-q$ in $(0,0,-2d)$. I need to calculate the potential at the point $P$.

enter image description here

The charge in $(0,0,-2d)$ does not apply an electric field in the upper part, which I think means that I should only consider the charge in the upper part, and solve it by adding one image charge $-q$ in $(0,0,-d)$.

But again, since the plane is grounded, putting a negative charge and positive charge next to it, makes it get positive and negative charges from the earth, and it will then have a certain charge density $\sigma$. So I thought again of putting two image charges, $q$ in $(0,0,2d)$, and $-q$ in $(0,0,-d)$, if we calculate the potential at $z=0$, the boundary conditions are satisfied. Therefore the potential on the point $P$ will just be the sum of the potentials of the four charges.

I am really confused on whether I should consider the charge in $(0,0,-2d)$ or not. Any help would be very much appreciated.

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2 Answers 2

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The method of images consists on getting a potential which satisfies your boundary conditions and the charge distribution for a region.

If $P$ is above the plane ($z_P > 0$), your boundary conditions are $0$ at $z=0$ (grounded) and $0$ at infinity and you only have a charge $q$ at $(0,0,d)$. What charge distribution satisfies both conditions? Well, mirroring your charge $q$ at $(0,0,d)$ to $-q$ at $(0,0,-d)$ (and forgetting about what's under the plane).

For $z_p < 0$, you should ask yourself the same question. It's analogous.

It may feels somehow wrong but if you think of the conducting plate with thickness $\epsilon$, it's the classic shell effect of a grounded conductor. So, I would say that even with no thickness that's what would happen.

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  • $\begingroup$ This makes the most sense to me, Thank you very much for your time! I apologize I do not have enough reputation to upvote your response $\endgroup$
    – Chihiro
    Commented Jan 29, 2021 at 8:29
  • $\begingroup$ @Chihiro, no problem at all. I'm way too far from being an expert thoguht. I've been thaught lightly about this so, if you end realising there might be a problem, consult with your professor. $\endgroup$
    – Gilgamesh
    Commented Jan 29, 2021 at 9:06
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I don't really get why you are considering whether or not to take the charge at -2D into consideration? I mean if you just ignore it and solve the problem like in the first case you mentioned above, it is not like ignoring the charge (the one at -2D) will make its field disappear. It will still be applying some force and some potential on the plate which will give you a non 0 potential on a grounded system.

So I do think that you should consider the charge (at -2D) while solving the problem like you did in the 2nd case.

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  • $\begingroup$ Thank you very much for your answer! I was thinking that the electric field of -q does not get to the point P, and the positive charge density on the conducting plane due -q is on the opposite side so it won't have any effect on the potential on P. that is why I thought maybe I should ignore the -q charge $\endgroup$
    – Chihiro
    Commented Jan 28, 2021 at 10:18
  • $\begingroup$ Actually the presence of the sheet does not affect the field produced by the charge -q(at -2D). The potential at p is affected because the charge -q(at -2D) induces an extra charge distribution on the sheet which produces its own electric field. The final result is the superposition of all the fields produced by resulting charge distributions. If you were to replace the conducting sheet by a plastic sheet, the resulting field at p will simply be the superposition of the two charges at D and -2D $\endgroup$
    – Aseem Mittal
    Commented Jan 28, 2021 at 10:38
  • $\begingroup$ Ahh I see! So the positive extra charge distribution due to -q, does not affect only one side (the one opposing the point P), but it affects the potential of every point, including P. Did I understand right? $\endgroup$
    – Chihiro
    Commented Jan 28, 2021 at 10:56
  • $\begingroup$ Yes, but the math will be complicated if we consider every point on the sheet so we represent the distribution with an equivalent q (at +2D) which gives the same net effect on the potential as will the distribution of charge on the conducting sheet. That is why we are able to get the result by neglecting the charge distribution of the sheet after replacing it with the +q charge (at +2d). It can be seen as a trick to ease the pre-requiste mathematical knowledge for solving the problem $\endgroup$
    – Aseem Mittal
    Commented Jan 28, 2021 at 12:39

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