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Why does uniqueness of Poisson equation guarantee that method of imaging works?

If say, we find a distribution that matches the boundary potential (thus matches the boundary condition) we still need to make sure that our potential that arises from this distribution still satisfies Poisson equation to make sure that we found the solution to the boundary value problem, however for us to be able to do this we need to know the charge density, which we don't actually know do we? So then how can we check that our "trick" actually gave us the sought for potential, since we cant actually verify it satisfies Poisson equation.

I.e we would like to check: $\nabla^2V=-\frac{\rho}{\epsilon_0}$ but how do we know $\rho$ if we don't know the charge distribution?

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Usually, the method of images is used when the charge distribution is known. You're handed some configuration of charges in some region with a conducting boundary, and your job is to find the potential. The method of images gives you a potential with the same charge distribution in the region of interest, but with the conducting boundaries replaced by some extended region containing additional "image" charges. Since the charge distribution is the same as the original problem in the region of interest, and since the solution to Poisson's equation is unique, you have then successfully found the potential.

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  • $\begingroup$ say in the classical example of a plane and a charge a distance above the plane. Once we used the method of images and got the potential, what is $\rho$ that we have so that we can check that our V satisfies $\nabla^2 V = -\frac{\rho}{\epsilon_0}$ $\endgroup$
    – Luca Ion
    Feb 12, 2021 at 13:03
  • $\begingroup$ As stated, you have to assume from the start that $\rho$ is known. In electrostatics, the problem is almost always stated as, "Here is some configuration of charge $\rho(\vec{r})$ and some boundary conditions. What is the electric potential in the region surrounding the charge?" For example, if you happen to be working out of Griffith's book, the section on the method of images on page 124 starts: "Suppose a point charge $q$ is held a distance $d$ above an infinite grounded conducting plane. Question: what is the potential in the region above the plane?" In other words, $\rho$ is given to you. $\endgroup$
    – Zack
    Feb 13, 2021 at 5:26
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    $\begingroup$ The reason you don't need $\rho$ at the boundary is because you are already given the boundary conditions in the form of the potential. In PDEs, specifying the potential is known as a Dirichlet boundary condition, whereas specifying the charge (which thereby specifies $dV/dn$) is known as a Robin boundary condition. If you specified both, then your system would be overconstrained, and there would be no solution. $\endgroup$
    – Zack
    Feb 16, 2021 at 20:06
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    $\begingroup$ Perhaps if I specified the problem more carefully, your concern would be alleviated: given $\rho(\vec{r})$ in the region $z > 0$ and the boundary condition $V(z=0) = 0$, find $V(\vec{r})$ in the region $z > 0$. In other words, you are indeed given $\rho$ everywhere, since the region of interest is $z > 0$. The charge at the boundary is found by the additional physical assumption that the region $z \leq 0$ is a conductor (or mathematically, that $V(z \leq 0) = 0$), but you don't need any knowledge of this charge if your goal is to find the potential in the region $z > 0$. $\endgroup$
    – Zack
    Feb 16, 2021 at 20:12
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    $\begingroup$ Kind of. More accurately, $\rho(\vec{r}) = q\, \delta(\vec{r} - \vec{r}_0)$, where $q$ is the charge of the point charge, $\vec{r}_0$ is its location, and $\delta(\vec{r} - \vec{r}_0)$ is the Dirac delta function. In case you are not familiar, you can roughly think of it as a function which is zero everywhere except $\vec{r}_0$, where it is infinite, and integrating over the function gives 1. $\endgroup$
    – Zack
    Feb 17, 2021 at 21:59

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