This is a self study question based on two videos from Khan's academy here:
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First the set up. Let's assume I have an infinite charged plate with some constant charge density over the plate, say $\sigma$. That means that I have $\sigma \frac{C}{m^2}$ over the plate where C is Coulomb and m is meters.
I did the math and found that the electric field at any point is $2 \pi K \sigma$ where $K$ is Coulomb's constant. That means the Force at any given point doesn't depend upon the distance from the plate and we get $F_e = 2 \pi K \sigma q$ for some other particle with charge $q$.
Now, I learned that Electric Potential is equal to $\frac{KQ}{r}$. I tried to derive this and I think it comes from taking the Force formula $F = \frac{KQq}{d^2}$ dividing by $q$ to get a "per unit charge" and then integrating out from $\infty$ to $r$. Basically I integrate out the work per charge to move the particle from an infinite distance to r away from the particle with charge Q.
However, I don't think that the formula works universally. If I have that same infinite plate, then $F = 2 \pi K \sigma q$. Doing the a related calculation for work on some charge coming in from infinity to r is:
$ W = - \int _{\infty}^r F ds = - \int _{\infty}^r 2 \pi K \sigma q ds =- 2 \pi K \sigma q \int _{\infty}^r ds = \infty$.
Since the work is $\infty$ that means the electric potential is infinite? That seems to me to be intuitively right. I need to add up a finite fixed amount infinitely many times as I move the charged particle in. As long as the finite fixed amount is some $\epsilon >0$, that would be infinity.
But that means the electric potential is infinite which is a direct contradiction to the formula $\frac{KQ}{r} < \infty$. So does that formula no longer hold for a plate? It's only for a point charge somewhere in space?
Answer
You know that if you have a point charge with charge $Q$, then the potential difference $V$ between spatial infinity and any point a distance $r$ from the charge is given by $$V_\textrm{point}=\frac{kQ}{r}.$$ You also know that the electric field from an infinite sheet of charge with charge density $\sigma$ is given by $$E_\textrm{sheet}=2 \pi k \sigma. $$
Because the electric field is uniform, you correctly concluded that there must be an infinite potential difference between any point and spatial infinity. You are surprised because this seems at odds with the first formula for $V_\textrm{point}$.
However, there is a good explanation. If $\dfrac{kQ}{r}$ is originally for a point charge, what values of $Q$ and $r$ should we plug in for the case of a sheet? Well, notice that the sheet has an infinite amount of charge, so that perhaps $Q$ should be infinite. This explains why we might get an infinite potential difference. However, there is a competing effect occuring with $r$. As you go farther out on the infinite sheet, you get farther and farther away from the point where you are trying to compute the potential, so it seems like maybe $r$ should be very big, maybe infinitely big as well. Let's see how to do the problem correctly.
To do the problem correctly, you need to realize that each point on the infinite sheet acts like a little point charge, so each point gives its own $\dfrac{kQ}{r}$ contribution. The total potential, by superposition, is the sum of these contributions. We can sum up the contributions by integration. Let's first pick a coordinate system where the plate is on the $x$-$y$ plane, and the point where we want to know the potential is on the $z$ axis. We can switch to cylindrical coordinates where $\rho = \sqrt{x^2+y^2}$. Then the distance $r$ between the point with coordinate $z$ on the $z$ axis and a point with coordinate $\rho$ is given by $r = \sqrt{z^2 + \rho^2}$, and so, applying the $kQ/r$ formula, the contribution $dV$ to the potential from a bit of charge $dQ$ a distance $\rho$ from the origin is given by $$dV = \frac{kdQ}{\sqrt{z^2+\rho^2}}.$$ Integrating this over all $\rho$ we find
$\begin{equation} \begin{aligned} V&=\int^\infty_0 \frac{2 \pi k \sigma \rho d \rho}{\sqrt{z^2+\rho^2}} \\ &= \pi k \sigma \int^\infty_0 \frac{du}{\sqrt{z^2+u}}\\ &=2 \pi k \sigma \left( \sqrt{\infty + z^2} - |z| \right). \end{aligned} \end{equation}$
Because of the infinity in the square root, the potential above is in fact infinite, even though were started with a finite $kQ/r$ law. This infinity was possible because we had infinitely much $Q$. Notice the electric field still works out because the infinite part does not have a spatial gradient: $$E=-\dfrac{dV}{dz} = -2 \pi k \sigma \left( \dfrac{z}{\infty + z^2} - 1\right) \hat{z} = 2 \pi k \sigma \hat{z}.$$
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