Electric potential everywhere around an infinite slab

Question

The problem:

Consider an infinite slab with uniform charge density $\rho$ between two planes $z=\frac{d}{2}$ and $z=-\frac{d}{2}$. Find the electric potential $V$ in the entire space and plot $V$ vs $z$.

I have been able to find the electric field using two Gaussian pillboxes:

Case 1: $z \geq \frac{d}{2}$ , $\vec{E}=\frac{\rho d}{2\varepsilon _{0}} \hat{k}$.

Case 2: $ \frac{d}{2} \geq z \geq -\frac{d}{2}$ , $\vec{E}=\frac{\rho z}{\varepsilon _{0}} \hat{k}$.

Case 3: $-\frac{d}{2} \geq z$ , $\vec{E}=\frac{-\rho d}{2\varepsilon _{0}} \hat{k}$.

But when I try to integrate those functions with respect to $z$ to find the electric potential, I end up getting two different values for $V(z=d/2)$ (case 1 and case 2) and two other different values for $V(z=-d/2)$ (case 2 and case 3).

I would appreciate a college-freshman-level explanation (preferable) and also a graduate-level one.

Answer 1

Since the charge distribution itself is infinite, we cannot use the usual formula for potential (which takes it to be zero at infinity): $$V(\mathbf{r}) = \frac{1}{4 \pi \epsilon_0} \int \frac{dq(\mathbf{r'})}{| \mathbf{r} - \mathbf{r'}|}$$ where $dq$ is a infinitesimal charge element, $\mathbf{r}$ is the position at which the field is being calculated, and $\mathbf{r'}$ is the location of the charge element. We must set the potential to be zero at some other point that is not at infinity.

Remember that $\mathbf{E} = -\nabla V$. By symmetry, we know that the field always points along the $z$-axis. In this case, $\mathbf{E} = -\nabla V$ becomes $\mathbf{E} = E \mathbf{\hat z} = -\frac{dV}{dz} \mathbf{\hat z}$, which becomes $E = -\frac{dV}{dz}$.

Inside the slab, the magnitude of the field is not uniform, so we must then integrate with respect to $z$ to obtain the function for $V$. Since $E$ inside the slab is a linear function of $z$ as you have obtained, we expect the potential to be a quadratic.

Outside the slab, since the magnitude of the field is uniform, this further reduces to $E = -\frac{\Delta V}{\Delta z}$ and we can conclude that $\Delta V = -E \Delta z$.

We can then proceed to set a location where the potential is zero, and calculate the respective values.