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Consider a point charge between two grounded parallel conducting planes $V=0$. Let the first plane be at $x=0$, the second one at $x=D$ and the charge be at $x=x_0$.

Like in a double mirror, the total electric potential is obtained by method of image charges and adding an infinite series of sign alternating charges that are at a distance $nD\pm x_0$, from the plates where $n$ is a non-negative integer.

What are the right conditions to find this result? I was considering putting one charge at a time and setting $V=0$ on one of the plates at a time. But that only fixes the charge or the distance but not both.

Is there a method I could use to fix all the variables and find the solutions with a minimal number of ansatz? (I would not mind something based on Green functions but I guess it is going to be equivalent to add infinite terms to cover the requirements of the boundary conditions).

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What are the right conditions to find this result? I was considering putting one charge at a time and setting V=0 on one of the plates at a time. But that only fixes the charge or the distance but not both.

Maybe you have some misconceptions about the image solution for the charge in front of a conducting plate.

Let's ignore the plate at $x=D$ first. The fact that the plate at $x=0$ is conducting means that an electric field has to stand perpendicular on it. Equivalently you can say that the potential is constant on the plate, for example $V=0$, but any other constant value will do. You achieve this by choosing the image charge as the exact mirror of the existing charge, i.e. same absolute value of the charge, different sign, and the same distance to the plate. Both fixed!

Now to your double plate problem: We'll be setting pairs of charges at a time. To achieve constant potential in the plate at $x=0$ we start off by mirroring the charge $q$ at $x=x_0$ with a charge $-q$ at $x=-x_0$. Now we look at the plate at $x=D$ and realize that we have to mirror both charges we now have (the original and the image) to achieve constant potential. Thus we put a charge $-q$ at $2D-x_0$ and a charge $q$ at $2D+x_0$. All's well now at this second plate, but we've messed up for the first one. To restore constant potential on the plate at $x=0$ we put a charge $q$ at $-2D+x_0$ and $-q$ at $-2D-x_0$. Great, but we've messed up the constant potential on the plate at $x=D$... You get the idea... With increasing numbers of charges and thus increasing distance from the plates the changes to the potential on the plates caused by the addition of another pair of charges gets smaller and smaller.

So charges are $q$ at $$ 2nD+x_0 $$ and charges $-q$ at $$ 2nD-x_0\,. $$

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  • $\begingroup$ So you have two add a single image charge and then add more by pairs? Is there any other method to calculate this without using image methods? $\endgroup$
    – Mauricio
    Commented Mar 21, 2022 at 17:51

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