# Electric Potential of Non-Uniformly Charged Infinite Plane

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?

• There’s a bit of a problem with your question: as there are charges at infinity where do you define the reference potential? – ZeroTheHero Oct 16 '18 at 4:49
• @ZeroTheHero It's a good question. I would assume that since the sheet has no net monopole-moment we may choose the potential at $z=\infty$ to be zero without any problems. – Geoffrey Oct 16 '18 at 5:28

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.

• You know, I actually think that this answer is on the right track. I just think that you played a little too fast and loose with the numbers and missed a few minor factors. I actually think that the answer is $V(x,z)=-\frac{\sigma_0}{2\epsilon_0\kappa}\sin(\kappa x)e^{-\kappa|z|}$. I may write up a break down of this solution later when I have clarified my thoughts. – Geoffrey Oct 16 '18 at 5:45
• Oh yeah right, I thought that the problem was a metallic "half space" and that you had to find the $V$ only on one side of the plate since in the other side $\vec{E}$ is $0$. I guess that this also solves the parallel $\vec{E}$ problem since we shouldn't impose any additional condition (in the plate case I think is necessary if we want to have $\nabla \times \vec{E} = 0$) – otillaf Oct 16 '18 at 6:17
• Right. Also, it's worthy pointing out that this problem actually can't take place on a conductor since the charge distribution varies dramatically between points with the same curvature. Some of the surface is even negatively charged while other parts are positive, which would be impossible on a conductor. – Geoffrey Oct 16 '18 at 6:28
• The only thing I don't agree with your solution is the sign. There are 2 signs when we calculate the field given $V$ in our case. One comes from taking the gradient along $z (negative exponent)$, and the other by the definition of $\vec{E} = -\nabla V$. In the end we should get a positive $E_z$ right above the plate if the charge is positive in a given point. of the plate – otillaf Oct 16 '18 at 7:06
• You're right. I forgot to do the negative gradient. – Geoffrey Oct 16 '18 at 16:01