My question may be very basic, but I can't think of a reasonable explanation for this.

Consider a solid charged sphere. Now, we have an electric field inside the solid sphere, but at any particular point there are charges infinitesimally close to the point. As I see it, these charges should contribute to an infinite value of the electric field.

One might possibly argue that at any given point, infinitely-close charges are distributed uniformly everywhere and hence their effects cancel out. But at the surface of the sphere the uniformity of infinitely-close charges is not present and hence here we must have infinite field, but that is not the case as well! Where are we going wrong?

Also, if we think of electric field at that point in continuous charge distribution, the contribution of electric field might exist finitely from nearby charges with both charge and distance between them tending to zero, but for the very small charge present exactly there, the electric field contribution from itself must be infinite with charge being finitely small and distance being exactly 0. What's wrong here?

I have strong suspicions, is it sensible and correct to define electric fields at such a close vicinity of other continuous charges?

  • $\begingroup$ If you apply Gauss's Law to a tiny shrinking sphere of uniform charge density, the volume of charge decreases faster than the area of flux. As r gets smaller, E approaches zero. $\endgroup$
    – R.W. Bird
    Oct 25, 2019 at 18:38
  • $\begingroup$ @Yashkalp Sharma For further discussion please see physics.stackexchange.com/questions/510427/… $\endgroup$ Oct 27, 2019 at 13:51

3 Answers 3


Your question is interesting but I think there is a mistake in your reasoning. Yes, if you look closely at a point inside a continuous charge distribution there are going to be "charges" arbitrarily close to the point, but the value of this electric charge is also arbitrarily small. This is because continuous charge distributions are given by densities, not point charges. The total charge is given by a volume integral:

$$Q =\int_V\rho(\mathbf x)~\mathrm{d}v$$

If you take a very small volume $V$, the charge $Q$ is equally small. A more useful way to look at it is with Gauss's law.


Where $S$ is the surface bounding $V$, and $Q_V$ is the total charge inside of $V$. As long as the distribution $\rho(\mathbf{x})$ is finite (i.e., not infinite) you can use this to show that $\mathbf{E}$ is finite as well (you use the divergence theorem and the definition of $Q$, above).

There are some cases of relevant non-finite charge distributions (e.g., dipole layers), but the case you propose of a uniformly charged sphere has a finite $\rho$ and hence $\mathbf E$ is as well.

  • $\begingroup$ @S V I have edited my answer considering yours. $\endgroup$ Oct 23, 2019 at 22:44

One might possibly argue that at any given point, infinitely-close charges are surrounded uniformly everywhere and hence cancel out their effects, but at surface of the sphere the uniformity of infinitely-close charges is not present and hence here we must have infinite field, but that is not the case as well!

Why do you think that the uniformity of charge distribution breaks on the surface of the sphere? When we say a solid sphere is uniformly charged, we mean that every infinitesimally thin shield of this sphere has the same charge density as that for the entire sphere. However, the mutual effects of charge fields which likely tend to cancel out each other are no longer symmetric on the outer shield (sphere's surface).

@S V 's answer is correct for when we have no point charge but rather uniform densities. However, remember that charge is quantized, and no such a continuous charge distribution does really exist. Continuity is a good approximation for modeling these quantized charges located very close to each other for an easy calculation. Your confusion is more valid if we consider real quantized cases. That is, if a neutral solid sphere is charged by, say, $10^{23}$ number of electrons, according to your suspicion, the electric field very close to each electron located (at least) on the surface of this sphere is very great based on Coulomb's force. It is said that the radius of electron is smaller than $10^{-16}\space m$, and hence, for a point with a distance of the order of $r\approx 10^{-15}\space m$ outside the electron, the Coulomb's field implies:

$$E=\frac{1}{4\pi \epsilon_0} \frac{e}{r^2} \approx 10^{21} \space V/m \space ,$$

which is a great number. I think even if we consider the electron as a tiny solid sphere with a uniform charge density, I am really doubtful that it can tangibly affect the great order of $E=10^{21} \space V/m$. This probably means that near the quantized charges in a so-called uniformly charged solid sphere our approximations are not good enough.

Your confusion can also arise for uniform mass distribution. In this case, in contrast to the electric field, Newton's law of gravitation does not imply gravitations with great values very close to an electron of, say, a solid sphere. The gravitation at a distance of the order of $r\approx 10^{-15}\space m$ outside the electron is:

$$g=G\frac{m_e}{r^2}\approx 10^{-11}\space m/s^2 \space,$$

which is a very small number, and can be neglected compared to the total gravitation inside the sphere. Nevertheless, as shown previously, this number is not that small for the electric fields!

I think taking account of the non-localization of some fundamental particles cannot resolve this problem either. A single proton or, generally, nuclei are massive enough to be easily localized. However, if a proton is enclosed by a spherical Gaussian surface with a radius very close to the proton's radius ($\approx 10^{-15}\space m$), the electric field is extremely great:

$$E=\frac{\sigma}{\epsilon_0}=\frac{e^+}{4\pi r^2 \epsilon_0}\approx 10^{22} \space V/m$$

This question is very good and, to me at least, very challenging!

  • $\begingroup$ Correct me if I am wrong, but I think that even if you look close enough to appreciate the fact that electric charge is quantized, the electric field is continuous. This because you can't directly apply Coulomb's law since quantum particles are not localized. (I vaguely remember something about electric fields at quantum scale in the introductory chapter of Jackson's book) $\endgroup$
    – user137661
    Oct 23, 2019 at 23:27
  • 1
    $\begingroup$ @S V I've edited my answer considering your viewpoint. $\endgroup$ Oct 25, 2019 at 14:54

For a charged Solid Sphere: we actually have two options to find the continuous charge distributions ,i.e., a disc and a hollow sphere. If density ($\rho$) varies, we use hollow sphere as it is convenient to use it. Let density be a function of $r$ and $\rho=\rho_0 r$ where $r$ is distance from centre and $\rho_0$=positive quantity then

$$\mathrm dq=\rho_0 r(4\pi r^2 \mathrm dr)$$

integrating both sides with limits extending from $0$ to $R$:

$$Q= \pi \rho_0 R^4$$


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