Suppose a point charge $q$ is located at $(x=0,y=0,z=d)$, and that along the $x$-$y$ plane is a infinite plate of potential $V = 0$. Then the method of images solves Laplace's equation for the potential for $z>0$, $$V(x,y,z) = \frac{1}{4\pi\epsilon_0}\left(\frac{q}{\sqrt{x^2+y^2+\left(z-d\right)^2}} - \frac{q}{\sqrt{x^2 + y^2 + \left(z + d\right)^2}}\right).$$ This satisfies the boundary conditions $V(x,y,0) = 0$ and $V(r\rightarrow\infty) = 0$.

Suppose instead the plate has potential $V(x,y,0) = V_0 \ne0$. Although very simple, it seems to me that the method of images cannot be used in this case. Is this correct?

  • $\begingroup$ Is $V_0$ a constant? In that case there is no $V(r\to\infty)=0$ boundary condition. Instead it goes to $V_0 + \text{const.}\times z$ where the constant depends on the surface charge on the plate. $\endgroup$
    – Michael
    Oct 26 '13 at 2:28
  • $\begingroup$ Correct, $V_0$ is a constant, so the boundary condition at infinity must be adjusted accordingly (but it still should hold that $V(z\rightarrow\infty) = 0$). $\endgroup$
    – Doubt
    Oct 26 '13 at 2:38
  • 1
    $\begingroup$ Not if the plate is infinite. The solution of Laplace's equation for planar symmetry is linear in $z$. The field due to the point charge will go to zero but the field of the plate will remain the same at infinite distances. $\endgroup$
    – Michael
    Oct 26 '13 at 3:24
  • $\begingroup$ @MichaelBrown It's clear that if a solution to Laplace's equations has $x$ and $y$ translation symmetry, then the general solution in the upper half space satisfying the boundary condition $V(x,y,0) = V_0$ is $V(x,y,z) = V_0+Cz$. However, I can't convince myself that the solution needs to possess this symmetry given only that the boundary condition at $z=0$ has this symmetry. $\endgroup$ Oct 26 '13 at 5:51
  • $\begingroup$ @Doubt : If the plate has a surface charge density $\sigma$, the solution is $V'(x,y,z) = V(x,y,z) + V_0 - (\frac{\sigma}{2 \epsilon_0} z)$. $\endgroup$
    – Trimok
    Oct 26 '13 at 8:05

I got a bit carried away thinking about this; sorry for the unnecessarily long answer.

I'll comment on the utility of the method of images a little further down, but first note the following fact about Poisson's equation which will reduce this problem to that of solving Laplace's equation with the given boundary condition:

Lemma. Suppose that $V_1$ and $V_2$ are both solutions to Poisson's equation with charge density $\rho$ on the interior of a region $R$ (possibly satisfying different boundary conditions), then their exists some function $W$ satisfying Laplace's equation such that $V_2 = V_1+W$.

Proof. Simply note that their difference satisfies Laplace's equation; \begin{align} \nabla^2(V_2-V_1)=\nabla^2V_2 - \nabla^2V_1 = \rho -\rho = 0 \end{align} Now, let's apply this to the problem you posed. Let $V^{(0)}$ denote the solution for the boundary condition $V(x,y,0) = 0$. Let $V$ denote the solution for the boundary condition $V(x,y,0) = V_0$, then (using the above lemma) we note that there exists some solution $W$ to Laplace's equation in the region $z>0$ such that $V = V^{(0)} + W$, so the problem has been reduced to determining $W$. The boundary conditions on $V$ and $V^{(0)}$ imply that $W$ satisfies the same boundary condition as $V$; \begin{align} W(x,y,0) = V(x,y,0)-V^{(0)}(x,y,0) = V_0 - 0 = V_0 \end{align} Putting this together, we see that the problem has been reduced to solving Laplace's equation (for the function $W$) subject to the boundary condition given in the problem.

Notice, that the method of images was still useful here. In particular, it allowed us to reduce a Poisson equation problem to Laplace equation problem by "removing" the effect of the source on the interior of the upper-half space $z>0$. This was explicitly done by subtracting $V^{(0)}$ (which was obtained through the method of images) from $V$.

Solving for $W$.

Ok, so now let's solve for $W$. There is, of course, the simple solution $W(x,y,z) = V_0$, but it turns out that there are many more (infinitely many in fact) solutions satisfying the desired boundary condition. In order to choose one of these solutions, one needs more information.

Let's assume that $W$ is analytic in $z$ so that we can write \begin{align} W(x,y,z) = \sum_{k=0}^\infty w_k(x,y) z^k \end{align} Notice that we might have tried to include negative powers of $z$ in this expansion, but those would be prohibited by the boundary condition. Moreover, the boundary condition $W(x,y,0) = V_0$ implies that $w_0(x,y) = V_0$. Laplace's equation $\nabla^2W = 0$ then gives \begin{align} \sum_{k=0}^\infty \Big[(\partial_x^2 + \partial_y^2) w_k(x,y)+(k+2)(k+1)w_{k+2}(x,y)\Big]z^k =0 \end{align} Setting the coefficient of each power of $z$ in the series to $0$ gives us constraints on the functions $w_k$ for $k\geq 0$; \begin{align} (\partial_x^2 + \partial_y^2) w_k(x,y) = -(k+2)(k+1)w_{k+2}(x,y) \end{align} For $k=2$, we obtain \begin{align} 0=(\partial_x^2+\partial_y^2)w_0 = -2 w_2 \end{align} which gives $w_2=0$. Applying this repeatedly shows that $w_k = 0$ for all even $k$. For odd $k$, we obtain an infinite tower of equations that couples all $w_k$ for odd $k$ to $w_1$. A particularly simple solution to this tower is obtained by setting $w_k = 0$ for all $k\geq 3$ in which chase $w_1$ satisfies the two-dimensional Laplace equation. We could, for example, set $w_1 = C$ for some constant $C$. This would give the solution that exhibits translation invariance in $x$ and $y$ mentioned by Michael Brown in the comments.

  • $\begingroup$ Umm... I believe your Laplace equation is wrong. The $k(k-1)w_k$ term should have a power $z^{k-2}$, right? You can shift the $k$ in the sum for the second term and all the $w_{k>2}$ are coupled to $w_{0,1}$. $\endgroup$
    – Michael
    Oct 28 '13 at 6:09
  • $\begingroup$ Alternately I would suggest Fourier transforming $x,y$. You'll get exponentially growing & decaying modes in $z$. Also a linear mode for $k_x=k_y=0$. Physical b.c.'s rule out growing modes. The damped modes could be interesting... $\endgroup$
    – Michael
    Oct 28 '13 at 6:15
  • $\begingroup$ @MichaelBrown Thanks; I was sloppy in shifting the sum. The Fourier transform suggestion is really interesting. $\endgroup$ Oct 28 '13 at 7:25

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