The Coulomb's law is an expertimental law which calcuates the electrostatic force between distinct two electrically charged point particles (point charge) at rest.

And Point particle doesn't take up space.

However, there is following sentence in Content 'Limitation' in an the article in wikipedia:

There are three conditions to be fulfilled for the validity of Coulomb's inverse square law:

  1. The charges must have a spherically symmetric distribution (e.g. be point charges, or a charged metal sphere).

In here, i can't understand this sentence. Why charges must have a spherically symmetric distribution?


2 Answers 2


Why charges must have a spherically symmetric distribution?

You can see in the Wikipedia article there is a section "Deriving Coulomb's Law from Gauss's Law". Gauss's law is a fairly fundamental result in multivariable calculus, not limited at all to the field of physics, so this is the direction we should be going when deriving one law from the other.

Notice that this derivation depends on assuming the source is spherically symmetrical. Without this assumption (or pre-condition on the physical situation to be analyzed), Gauss's law doesn't give the result that we call Coulomb's law and Coulomb's law is not valid.

That said, we can in fact use Coulomb's law to analyze non-symmetric systems. We just have to use the section of the Wiki article that says,

A system ''N'' of charges $q_i$ stationed at $\mathbf{r}_i$ produces an electric field whose magnitude and direction is, by superposition $$\mathbf{E}(\mathbf{r}) = {1\over4\pi\varepsilon_0} \sum_{i=1}^N q_i \frac{\mathbf{r}-\mathbf{r}_i}{|\mathbf{r}-\mathbf{r}_i|^3}$$

For systems with continuously distributed charge rather than a finite number of point charges, we can extend this by taking infinitesimal elements of charge and turning the summation into an integral.

$$\mathbf{E}(\mathbf{r}) = {1\over4\pi\varepsilon_0} \int_{R} q'(\mathbf{r'}) \frac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3}d\mathbf{r'}$$

Where $R$ is the region containing charge and $q'(\mathbf{r})$ is the charge density at point $\mathbf{r}$.

  • $\begingroup$ Thank you for good answer $\endgroup$
    – KHJ
    Nov 5, 2023 at 6:41

Gravity has a law very much like Coulomb's law. The force of gravity is usually calculated as if the Earth was a point particle located at the center. Earth is not. It is a sphere with a dense spherical core, lighter rocks in spherical shell shaped mantle and crust. But gravity is the same for both of those situations, even for us right on the surface.

Suppose the Earth had a non-spherical shape, say a disk. Gravity would not be the same as from a sphere or point.

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