This is a self study question based on two videos from Khan's academy here:




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?

  • $\begingroup$ An infinite size charged plate is physically impossible. Everybody who teaches electrostatics with potentials of infinite objects does you a disservice. That is not the correct way of thinking about physics when you are trying to simplify something. The correct way is to say I have a finite plate with a finite charge and I am so close that the fringe fields that are being cause by the geometry of the border do not matter. Let's say you have a 1m size plate, then the formula you got there will work well enough for objects that are up to 1cm from the plate and there are no infinities. $\endgroup$
    – CuriousOne
    Sep 22, 2015 at 18:44
  • $\begingroup$ Which formula? $KQ/r$? What about my math and the rest? Does the logic follow for an actual infinite sized plates? I'm approaching this from more of a pure math point of view so I am curious if I get the derivations right? $\endgroup$ Sep 22, 2015 at 18:57
  • $\begingroup$ There are no infinite sized plates. :-) Again, your problem is that you start with a non-existent setup, you preform mathematical operations on it and you end up with nonsense. Your math teacher would have given you an F for doing that, just as I am giving physics teachers who make students believe in infinite size plates an F. The correct approximation is that the force on a charge over a finite size plate is only constant when the charge is very close to the center of the plate. At a large distance that force will be smaller and it will go down with $1/r^2$, which makes the integral finite. $\endgroup$
    – CuriousOne
    Sep 22, 2015 at 19:01
  • 2
    $\begingroup$ @CuriousOne You can definitely imagine an infinite plate though. There is nothing logically inconsistent about it. I am pretty sure he is just confused about what $Q$ and $r$ are supposed to be and how to apply $V=kQ/r$ in this situation. I will try to write an answer. $\endgroup$ Sep 22, 2015 at 19:05
  • $\begingroup$ @NowIGetToLearnWhatAHeadIs: No, actually I can't imagine that. Your assignment for tomorrow is to get me an infinite size plate so I see what it looks like. :-) $\endgroup$
    – CuriousOne
    Sep 22, 2015 at 19:06

3 Answers 3


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: $$E=-\dfrac{dV}{dz} = -2 \pi k \sigma \left( \dfrac{z}{\infty + z^2} - 1\right) \hat{z} = 2 \pi k \sigma \hat{z}.$$

  • $\begingroup$ Excellent answer. I wonder what happens when you start with a finite total charge Q that is spread evenly over the infinite area. Do you still get infinite potential, finite but constant potential, or a potential gradient like a normal electric field? $\endgroup$
    – fishinear
    Jan 17 at 12:33

you should purchase or download "introduction to electromagnetism by David j. Griffiths". He talks about this problem in the 2nd chapter, basically his answer is that in this problem our convention of taking infinity as "zero potential" breaks down.. from his textbook , chapter 2 section 2.3.1(comments on potential):-

Evidently potential as such carries no real physical significance, for at any given point we can adjust its value at will by a suitable relocation of O. In this sense it is rather like altitude: If I ask you how high Denver is, you will probably tell me its height above sea level, because that is a convenient and traditional reference point. But we could as well agree to measure altitude above Washington D.C., or Greenwich, or wherever. That would add (or, rather, subtract) a fixed amount from all our sea-level readings, but it wouldn't change anything about the real world. The only quantity of intrinsic interest is the difference in altitude between two points, and that is the same whatever your reference level. Having said this, however, there is a "natural" spot to use for 0 in electrostaticsanalogous to sea level for altitude-and that is a point infinitely far from the charge. Ordinarily, then, we "set the zero of potential at infinity." (Since V (0) = 0, choosing a reference point is equivalent to selecting a place where V is to be zero.) But I must warn you that there is one special circumstance in which this convention fails: when the charge distribution itself extends to infinity. The symptom of trouble, in such cases, is that the potential blows up. The remedy is simply to choose some other reference point (in this problem you might use the origin). Notice that the difficulty occurs only in textbook problems; in "real life" there is no such thing as a charge distribution that goes on forever, and we can always use infinity as our reference point.


Electric fields due to infinite sheet of charge : $$E_{sheet}=2πkσ$$ $$dV=-E.dr$$ Integrating $$V=-2πkσr$$

  • $\begingroup$ I also want to ask what will be the limits while integrating. $\endgroup$ Aug 6, 2016 at 7:39
  • $\begingroup$ Limits could be 0 to r in my point of view to get the required answer. Also how can we think of it pratically without using maths. $\endgroup$ Aug 6, 2016 at 7:45
  • $\begingroup$ Latex is a simple markup language, and considering your physics skill, you probably wouldn't find it hard to learn. $\endgroup$
    – peterh
    Aug 6, 2016 at 11:01

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