Question:

Electric Potential of Non-Uniformly Charged Infinite Plane

Chloe: 3 days ago

A little background: I was tutoring an undergrad upperclassman when we came to a problem that he had been assigned which I couldn't make heads or tails of - at least in terms of what was being expected of him.

The problem asks to find the electric potential above an infinite sheet lying in the $xy$-plane and carrying a surface charge density of $\sigma=\sigma_0 \sin(\kappa \ x)$. The answer must be in terms of $\sigma_0$ and $\kappa$.

From the statement of the problem and the context of the class, it is clear that the solution is expected to be analytical, which immediately rules out a numerical or series solution.

The current topic of the class is solving Laplace's Equation using separation of variables, but the associated Poisson's Equation for this problem (viz. $\frac{\partial^2 V}{\partial x^2}+\frac{\partial^2 V}{\partial z^2}=-\frac{\sigma_0}{\epsilon_0}\sin(\kappa \ x)\delta(z)$) is clearly not separable.

On the other hand, a more straightforward approach such as integrating for the potential over the entire sheet leads to an intractable integral. Like this: $$V(x',z')=\frac{\sigma_0}{4\pi\epsilon_0}\int_{-\infty}^{\infty}dy\int_{-\infty}^{\infty}dx \ \Big[ \frac{\sin(\kappa \ x)}{\sqrt{(x'-x)^2+(y)^2+(z')^2}} \Big] $$ I also tried cutting the sheet into infinitesimally wide, infinitely long strips along the $y$-direction and then integrating over the potential of an infinite wire, but of course, this results in the same sorts of weird integrals involving the natural log.

Is there an analytical method for solving this problem? Am I forgetting a technique, or is there perhaps a trick to evaluating one of these weird integrals?

Answer:
Lucas: 3 days ago

The final solution should be something like $$V\left (x, y, z \right) = \frac{\sigma_0}{2\kappa \epsilon_0}\sin\left(\kappa x \right)e^{-\kappa |z|}$$.

The main idea is to solve the Poisson equation outside the plane, $$\nabla ^2 V = 0$$. and to apply boundary conditions only after having found the general form for the potential.

Since the solution is unique, we can guess a form and if it solves the equation we have solved the problem.

To do this, we will study the symmetries of the problem and write down the consequences of these:

  • Translation invariance along $y$

This means that the potential is function of $x$,$z$ only.

$$V\left (x,y,z\right) = V\left(x,z\right)$$

  • Translation invariance along $x$ by $\frac{2\pi}{\kappa}$

$$V\left(x, z \right) = V\left(x+\frac{2\pi}{\kappa}, z\right)$$

  • Symmetry under spatial reflection with respect to the $x-y$ plane.

$$V\left(x, z \right) = V \left(x, -z \right)$$

Notice that the latter condition can be used to study the problem in the half space $z>0$. Indeed once we find $V$ in the superior half space, we get it also in the $z<0$ half space by making the substitution $z \rightarrow |z|$ in $V\left( x,z\right)$.

Thanks to this consideration, we will focus on the $z>0$ half space from now on we.

We will look for a separate variable solution (we are in the vacuum outside the plate!) $$V\left( x,z\right) = A\left(x \right)B\left( z\right)$$. Plugging in Poisson equation and dividing by $V$ we get $$\frac{A''(x)}{A(x)} + \frac{B''(z)}{B(z)} = 0 $$

A possible solution which respect the symmetry we want the solution to have is given by $$V\left(x,z\right)=V_0\sin(\kappa x) e ^{-\kappa z}$$

We only need to check if the electric field (calculated by taking the derivative along the normal direction on the plate) is equal to $\frac{\sigma}{2\epsilon_0}$ (the electric field near a charged sheet) This can be done by choosing $V_0 = \frac{\sigma_0}{2\kappa\epsilon_0}$. This is the potential in the upper half space ($z>0$).

In order to find $V$ in the whole space, we can do the substitution $z\rightarrow |z|$ as we discussed earlier.