We know that the electric field for a point charge is $$ E = \frac{KQ}{R^2}. $$ If $R$, i.e. distance from the electric field producer to the point where we want to find the electric field becomes zero, then $E$ will tend to infinity. Now take a case where you have a uniformly charged sphere and you need to find the electric field on the charged sphere. The charged sphere is not a point charge, its rather a combination of point charges. What is electric field at any point, say $P$, on the charged sphere?

Now we know that when we use Gauss's Law, we select a Gaussian surface in such a way that all the points on the Gaussian surface should experience same electric field. Many sources say that if we use Gauss's Law then on any point on the charged sphere the electric field is going to be $$ E = \frac{1}{4\pi\epsilon_0}\frac{Q}{R^2} $$ where $R$ is the radius of the sphere and $\epsilon_0$ is the permittivity.

But now, don't consider Gauss's Law. As $P$ is at the surface of the charged sphere, then the electric field due to the small element of the charged sphere on which point $P$ lies is infinity, as the small element has charge $dq$ (let's also say we have assumed the small element of charge $dq$ on which point $P$ lies to be a point charge, and hence $R$ becomes zero, as the point $P$ lies on the small element). So electric field must be infinity due to that small element on which the point P lies. It doesn't really matter how much small the charge of the element is! When you do divide a large number and zero, you get infinity and similarly when you divide a very very small number and zero, you will again get infinity. Now for rest of the small elements which make up the whole sphere, the electric field won't be infinity; instead it would be some finite value. Say on the point $P$ for all the other small elements (except the element on which the point $P$ lies) net electric field is $X \rm\,N/C$. Still, $\text{infinite}+X = \text{infinite}$.

From here we conclude that on the point $P$ the electric field must be infinite. Then how did we get $$\frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}~?$$

My second question is: Similarly how can one say that we can find the potential on the surface of the charged conductor? I saw sources saying that the potential on the surface of the charged sphere is $V=KQ/R $ if $R$ is the radius of the sphere. But how is this possible? I mean on the surface again for the above reasons we would get it to be infinite!

Please explain these doubts to me. These might be silly doubts, but still. I don't know how far my thinking is correct.

  • 1
    $\begingroup$ Based in what I understood of your question you are confusing the concept of point and surface charges. Yes, if you disperse real point charges over the surface of a sphere at fixed locations, then the electric field will not be homogeneous at the surface and it will diverge at the point charges. That's not how surface charge works, though. $\endgroup$
    – CuriousOne
    May 21, 2015 at 18:45
  • $\begingroup$ @CuriousOne Please convert that comment to an answer. $\endgroup$
    – rob
    May 21, 2015 at 20:07
  • $\begingroup$ May be this helps: physics.stackexchange.com/q/210162 $\endgroup$
    – velut luna
    Apr 23, 2016 at 2:00
  • $\begingroup$ It seems this electrostatic question (v6) can be reformulated in terms of Newtonian gravity, and that it is essentially resolved by a version of the Newton's Shell Theorem. $\endgroup$
    – Qmechanic
    Apr 23, 2016 at 12:47

4 Answers 4


(Other) Rob's answer seems good to me, but let me offer another way of thinking.

As you approach the surface of the sphere very closely, the electric field should resemble more and more the electric field from an infinite plane of charge.

If you check Gauss's law (recalling that the field in the conductor is zero) you will see that if the surface charge density is $\sigma=Q/4\pi R^2$, then indeed the field at the surface is $\sigma/\epsilon_0$ as in the infinite charge of plane case.

Such a field is constant, the field lines are parallel and non-diverging, and the infinities associated with the field due to point charge do not arise.


You seem to be confused about the concept of the limit, which is normally covered in calculus courses. If you have, not a point charge, but a volume of charge with some density $\rho$, then the charge enclosed in a small sphere with radius $r$ is $$ dq = \rho\, dV = \rho \cdot \frac{4\pi}3 r^3 $$ and the electric field at the surface of the sphere will be proportional to $$ E \propto \frac{dq}{r^2} \propto \frac{\rho r^3}{r^2} = \rho r. $$ This field magnitude is well-behaved in the limit that $r$ is very small.

The zero-size point charge, zero-thickness surface charge, etc. are useful approximations when the size of the charge distribution is much smaller than the size of the field distribution that you care about. Physical point charges, surface charges, and line charges are all actually distributed over a finite volume. Even the charge of a single electron is distributed over a finite volume, thanks to quantum-mechanical uncertainty in the electron's position.

  • 1
    $\begingroup$ Sir,we know that when the distance between the electric field producer and the point at which we want to calculate the electric field becomes zero, electric field tends to infinity. In the case of electron if I take the point at which I want to find the electric field on the surface of the electron then the EF rands to infinity. Similarly if you imagine the sphere of radius 6 cm as the larger version of the electron, similarly if you take a point on the surface of the sphere and find the EF,as the R in the formula KQ/R^2 becomes zero, EF must tend to infinty. The formula itself says it! $\endgroup$ May 23, 2015 at 3:24
  • $\begingroup$ And also sir in the above small derivation you did, E∝dq/r2∝ρr3/r2=ρr, you assumed the point where you want to find the electric field equal to the radius of the sphere. Actually it should be: E= dq/x^2 => E =(ρ⋅4πr^3/3)/ x^2 . If x tends to zero then E would tend to infinity. Sir, then where did I did wrong? Sir I am a student and these are completely new to me. And as these are new things i always get confused. So sir please help me out! My doubts might be weird but still..... $\endgroup$ May 23, 2015 at 3:40
  • $\begingroup$ Have you studied limits? For instance you might expect that the function $\mathrm{sinc}\,x = \frac{\sin x}{x}$ should misbehave at zero, but plotting the function for values very near zero reveals no misbehavior at all, and it's possible to rewrite $\mathrm{sinc}\,x$ as a series that's even well-behaved exactly at $x=0$. What I'm arguing is that if the charges are smeared out in space (which is true for all physical charges) then the electric field near any particular blob of charge is finite. You might also read about the "shell theorem" which lets me ignore nearby-but-symmetric charges. $\endgroup$
    – rob
    May 23, 2015 at 4:36

For a charge distribution $\rho({\bf r'})$, the electric field at ${\bf r}$ is $${\bf E}({\bf r})=k\iiint\frac{\rho({\bf r'})}{|{\bf r}-{\bf r'}|^2}\hat{\bf R}d^3{\bf r'}$$ where ${\bf R}={\bf r}-{\bf r'}$.

I think the OP's claim is at positions where $\rho \ne 0$, the above integral is infinite because the integrand blows up at ${\bf R}={\bf 0}$.

I think the OP should first ask yourself whether you are asking a mathematics question or a physics question.

If you are asking a mathematics question, I think your question is similar to whether $$\int_0^1\frac{1}{\sqrt{x}}dx$$ is undefined (or infinite according to your argument), or equals $2$.

In my opinion, strictly speaking, the above integral is undefined, because the integrand is undefined at $x=0$.

However, since $$\lim_{\epsilon\rightarrow0^+}\int_\epsilon^1\frac{1}{\sqrt{x}}dx=2$$ we usually just say $$\int_0^1\frac{1}{\sqrt{x}}dx=2$$

So in my opinion, when we write $${\bf E}({\bf r})=k\iiint\frac{\rho({\bf r'})}{|{\bf r}-{\bf r'}|^2}\hat{\bf R}d^3{\bf r'}$$ we are actually saying $${\bf E}({\bf r})=\lim_{\epsilon\rightarrow0^+}k\iiint_{D_{\epsilon}}\frac{\rho({\bf r'})}{|{\bf r}-{\bf r'}|^2}\hat{\bf R}d^3{\bf r'}$$ where $D_\epsilon$ is the space minus a ball with radius $\epsilon$ centered at ${\bf r}$.

It can be shown that if $\rho({\bf r})$ is finite or does not go to infinity $\textit{too fast}$, then the limit exists.

If you accept that, then the physics question one should ask is whether this gives the correct answer in physics.

In classical EM, the $\textit{macroscopic}$ field ${\bf E}$ is in fact the average of the $\textit{microscopic}$ field ${\bf e}$. (Landau and Lifshitz, $\textit{Electrodynamics of Continuous Media}$, p.1.)

So you can consider ${\bf E}({\bf r})$ as the average of ${\bf e}$ over a ball with radius $\epsilon$ centered at ${\bf r}$.

Now, it can be argue that $${\bf E}({\bf r})=k\iiint_{D_{\epsilon}}\frac{\rho({\bf r'})}{|{\bf r}-{\bf r'}|^2}\hat{\bf R}d^3{\bf r'}$$ as follows.

It can be shown that (1) the average field over the ball due to charges outside the ball is the same as the total field due to all charges outside the ball at the center of the ball. , and (2) the average field over the ball due to charges inside the ball is $$-k\frac{{\bf p}}{\epsilon^3}$$ where ${\bf p}$ is the total dipole moment inside the ball. (Griffith, $\textit{Introduction to Electrodynamics}$, p. 156-157, problem 3.41).

So when $\epsilon$ is small macroscopically, ${\bf p}\rightarrow{\bf 0}$.

  • $\begingroup$ Because $1/0$ is undefined in math. $\endgroup$
    – velut luna
    Apr 23, 2016 at 12:59
  • $\begingroup$ It's not $\infty$. It's undefined. $\endgroup$
    – velut luna
    Apr 23, 2016 at 13:03
  • $\begingroup$ @AnubhavGoel Explain what? $\endgroup$
    – velut luna
    Apr 23, 2016 at 13:11
  • $\begingroup$ When x→0, function is well defined and E should become ∞. $\endgroup$ Apr 23, 2016 at 13:16
  • $\begingroup$ Sorry? You mean the electric field AT the position of a point charge is well-defined and is infinity? $\endgroup$
    – velut luna
    Apr 23, 2016 at 13:18

Your thinking is correct. At least, if the sphere is made up of small point charges, then the field will be infinite as you approach them.
There is a point you are missing when you say that this contradicts Gauss' law: Gauss' law only gives you the flux of the field. To get the field of the sphere from it in the textbook-way, you have to use symmetry. You argue then, that since a sphere is symmetric, the flux is simply field times area. But this is not true with your assumptions. A sphere consisting of discrete points is no sphere, it has at least no spherical symmetry. The contradiction is gone.

If we say, that the field equals a certain value, we mean the mean value. In your reasoning you are aiming exactly at a point charge. Then the field is infinite. But if you have points, you must also have gaps between them, and in those gaps the field is less than that which you get with the usual formula. If you say, that the points are dense (in the sense in which rational numbers are: between any two there is still one more), then they are infinitely many and have the charge of zero each.
You argue, that a value divided by zero is infinite ... but zero divided by zero is not! It can be, but it doesn't have to. And in this case you have the same zero-ness, if you divide them you get something finite.

In reality the electrons have a non-zero charge of course. But there are also finitely many of them. So there are real finite gaps.

But you are right - point charges are a strange thing. They are not consistent with classical electromagnetism. One could argue, that it's no problem that the potential gets infinite - just don't go there, than it's finite, one could say. But in the classical theory there is the concept of energy density of the field. And the energy of a point charge is infinite, if you do the calculation.
Well, this, too, would be no problem - as long as the number of charges is conserved, at least. But it isn't ... But the only thing that it proves is, that the world is not classical.

  • $\begingroup$ I accept your answer. I had same feelings but couldn't explain it to myself. $\endgroup$ Apr 22, 2016 at 15:20
  • $\begingroup$ thank you :) ah, you asked me to edit something (I read it on mobile), so it worked like this already $\endgroup$
    – Ilja
    Apr 23, 2016 at 17:38

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