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If we place a charge +q at a distance a along the x-axis from a grounded plate, we can use the method of images to show the potential is $V(\textbf{r})=\frac{q}{4\pi \epsilon_0}(\frac{1}{r_1} -\frac{1}{r_2})$, where $\frac{1}{r_1}$ & $\frac{1}{r_2}$ are the distances from +q and the image charge -q respectively.

Following this reasoning, my notes(and Wikipedia) state that from $\textbf{E}=-\nabla V$ that

$$\textbf{E}_{x=0} = \frac{-qa}{2 \pi \epsilon_0 (a^2+y^2+z^2)^{3/2}}$$

So by $E=\frac{\sigma}{\epsilon_0}$ for a charged plate, the surface charge density is $$\sigma=\frac{-qa}{2 \pi (a^2+y^2+z^2)^{3/2}}$$

However, doesn’t this use the total field of the point charge and the plate, when we should be using only the field produced by the plate to calculate the surface charge density?

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You found a field distribution above the plane with the help of the method of images. That field distribution is unique and being an answer to both:

a) the original problem

b) the original charge with it's image.

However in a) you have a boundary (grounded plate) and the last thing to do in this problem is to find a boundary condition $E(+0,y,z) = \frac {\sigma(y,z)} {\varepsilon_0}$ that corresponds to the field configuration that is being an unique answer to that system. Field of the grounded plane without a charge above it is zero everywhere.

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No, you have to use both.

Intuitive, physical reason: the charges on the plate are attracted to the real charge, but they tend to exert a force on each other, causing them to repel a little bit.

Mathematical reason: For a given open set $U$ in which you want to determine the electric potential $V$, and given boundary condition on $\partial U$ (in our case $V$ on the plate and to infinity), the solution to the Poisson equation inside $U$: $$\nabla ^2 V \left(\vec{x}\right)= -\frac{1}{\epsilon_0}\rho \left(\vec{x}\right)$$

is unique.

Notice that $\rho$ is given in $U$, and isn't necessarily known on the boundary. After all, there must be charges somewhere outside $U$ to produce the given boundary condition.

(In the case of the plate problem, you are assuming that the charge is on the boundary itself since it can't be inside the conductor since $\vec{E} =0$ inside it).

Now, once you find a solution to the Poisson problem, you have the correct $V$, no matter how you found it. The method of images simply helps you find the solution to this problem in an intutive way. Once you have $V$, you can then use it to find the electric field $\vec{E}$ (take the gradient) and then relate the latter to the superficial density charge $\sigma$ on the plate itself by using: $$\vec{E} = \frac{\sigma}{\epsilon_0} \hat{n}$$.

Since the correct $V$ is the superposition of the potential of both the real and image charge, you have to use that.

The physics doesn't care if you used the method of images to find the potential or another trick up your sleeve. The only important thing is that the $V$ is the correct solution to the Poisson problem given the right boundary conditions (which is unique!).

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