# Can Gauss Law for any arbitary closed surface be derived from Coulomb's Law? [duplicate]

As far as I know, Gauss' Law is useful for calculating the electric field where Coulomb's Law doesn't work. For example, it is used for calculating the field produced by a thin sheet of charge and also used for infinitely long thin line of charges.

But in both of these cases, the field can also be obtained using Coulomb's inverse square law. This method just needs some extra integration in contrast with Gauss' Law.

So my question is that is this true in general? Can we calculate the field produced by any arbitary charge distribution using Coulomb's Law?

Also in the both cases mentioned above, one can use Coulomb's Law for field and then use this to get to Gauss' Law. So is it possible to derive Gauss' Law for any arbitrary closed serface using Coulomb's Law?

Coulomb's Law works in all cases and Gauss' Law is derived from it. Indeed, by Gauss' Theorem,

$$\oint_{\partial\Omega}\mathbf E\cdot\hat{\mathbf n}\ \text dS = \int_\Omega\nabla\cdot\mathbf E\ \text d\Omega,$$

and by the expression of $$\mathbf E$$ given by Coulomb's Law, viz.

$$\mathbf E(\mathbf r) \propto \int \rho(\mathbf r')\frac{\mathbf r-\mathbf r'}{|\mathbf r - \mathbf r'|^3}\text d\Omega$$

you then derive

$$\oint_{\partial\Omega}\mathbf E\cdot\hat{\mathbf n}\ \text dS = \int_\Omega\rho(\mathbf r')\ \text d\Omega$$

by computing the divergence.

• I can't understand how you got to the last line. I'll appreciate if you elaborate it. May 18 '20 at 11:17