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Suppose one wanted to find the electric field around a point charge located above an infinite ground plane as in the classic method of images example, but without using the method of images itself. (I.e. do not use the analogy of boundary conditions and the uniqueness of a solution to obtain the answer by solving the image charge arrangement.)

Problem.

For a charge above a ground plane, with the charge $q$ located at $\mathbf{r}'$, and the $z=0$ plane as the ground plane, find the electric field around the charge in the electrostatic case without using the method of images.


Note: Everything below this point is my own attempt at a solution, and is not part of the question per se.

Attempt at a Solution.

This post previously contained my attempt at a solution, but it was quite far from the mark, so I've removed it.

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2 Answers 2

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Divide the potential into two parts, one arising from the point charge and one arising from the surface charges on the grounded surface $z = 0$: $$ V = V_p + V_s $$ where we have $$ V_p = \frac{q}{4 \pi \epsilon_0 \sqrt{x^2 + y^2 + (z - z_0)^2}} $$ with $(0, 0, z_0)$ being the location of the charge; and $$ \nabla^2 V_s = 0 $$ for $z > 0$. Since we know that $V_s + V_p = 0$ on the plane $z = 0$, we conclude that $V_s$ must satisfy $$ V_s(x, y, 0) = -\frac{q}{4 \pi \epsilon_0 \sqrt{x^2 + y^2 + z_0^2}} $$ on the $z = 0$ plane.

From here, the technique in the example on p. 5-2 of these lecture notes can be adapted; I will just sketch it out from this point. We apply a Fourier transform to $V_s$ in $x$ and $y$ only. The transformed function $\bar{V}_s(\vec{k}, z)$ (with $\vec{k} \equiv (k_x, k_y)$) then will be of the form $$ \bar{V}_s(\vec{k}, z) = C(\vec{k}) e^{-|\vec{k}| z} $$ with boundary condition $$ \bar{V}_s(\vec{k},0) = - \frac{q}{4 \pi \epsilon_0} \frac{e^{-|\vec{k}|z_0}}{|\vec{k}|} $$ (this latter function being the Fourier transform of $V_s(x,y,0)$ above.) Thus, $$ \bar{V}_s(\vec{k}, z) = - \frac{q}{4 \pi \epsilon_0} \frac{ e^{-|\vec{k}| (z + z_0)}}{|\vec{k}|} $$ and applying the inverse Fourier transform to this function yields $V_s(x,y,z)$. Unsurprisingly, it works out to be the potential produced by a negative point charge $-q$ at $(0,0, -z_0)$.

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I already accepted the answer above, but I also wanted to provide an answer using the Green's function for these boundary conditions without explicitly using the method of images. It's a nice basic exercise in Green's functions.

The general solution of Poisson's equation, derived from Green's second identity, for some function $G(\mathbf{x},\mathbf{x}')$ such that $\nabla'^2 G(\mathbf{x},\mathbf{x}')=-4\pi\delta(\mathbf{x}-\mathbf{x}')$ (assuming $\mathbf{x}$ to be the observation point, and $\mathbf{x}'$ to be the location of a source), is: $$ \Phi(\mathbf{x}) = \int_V G\frac{\rho(\mathbf{x}')}{4\pi\epsilon_0}d^3x' + \frac{1}{4\pi}\oint_S \left(\Phi(\mathbf{x}')\frac{\partial G}{\partial n'} - G\frac{\partial \Phi}{\partial n'}\right)dA' $$ We take the volume $V$ to be the $z\ge0$ region, and impose the conditions that $\Phi(z=0)=0$ and $\Phi\rightarrow 0$ as $\mathbf{x}\rightarrow\infty$. Therefore the integral relation reduces to $$ \Phi(\mathbf{x}) = \int_V G\frac{\rho(\mathbf{x}')}{4\pi\epsilon_0}d^3x' $$ And $G(z=0)=0$. The general form of $G(\mathbf{x},\mathbf{x}')$ is $$ G(\mathbf{x},\mathbf{x}') = \frac{1}{|\mathbf{x}-\mathbf{x}'|}+F(\mathbf{x},\mathbf{x}') $$ Such that $\nabla'^2 F(\mathbf{x},\mathbf{x}')=0$, i.e. $F$ is a solution to Laplace's equation.

In the general case, the boundary condition on $G$ requires $$ \frac{1}{\sqrt{(x-x')^2+(y-y')^2+z^2}}+F(\mathbf{x},\mathbf{x}')=0 $$ And including the condition that $F$ satisfy Laplace's equation, we are left with two possible solutions: $$ F(\mathbf{x},\mathbf{x}')=\frac{-1}{\sqrt{(x-x')^2+(y-y'^2)+(z\pm z')^2}} $$ Using $(z-z')^2$ forces $G=0$ and hence $\Phi=0$ everywhere, which we know is invalid. Therefore the solution must be the one with $(z+z')^2$, giving $$ G=\frac{1}{|\mathbf{x}-\mathbf{x}'|}-\frac{1}{|\mathbf{x}-\mathbf{\bar x}'|} $$ Where $\mathbf{\bar x} = x'\hat x + y' \hat y - z' \hat z$ (notation in analogy with complex conjugates).

Using the solution to Poisson's equation given above, and taking $\rho(\mathbf{x}) = q\delta(z \hat z-z_0 \hat z)$, we obtain the solution for $\Phi$, $$ \Phi(\mathbf{x}) = \frac{q}{4\pi\epsilon_0}\left(\frac{1}{|\mathbf{x}-z_0\hat z|}-\frac{1}{|\mathbf{x}+z_0\hat z|}\right) $$ Which is equivalent to the potential due to a charge above the ground plane and its mirror charge, as expected.

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