It is well known that Gauss's Law & Coulomb's Law are inter-dependent. Any violation of Gauss's Law will indicate departure from inverse square law and vice versa.

Then how gaussian surface and continuous charge distribution overlap. Since when this happens distance between the 'point (where electric field has to be measured using Gauss's Law) and charge' is comparable to the 'size of the charge'. And in this condition Coulomb's Law is not applicable, so how Gauss's Law can be applied here.

Gauss's Law - The net electric flux through any hypothetical closed surface is equal to 1/ε times the net electric charge within that closed surface.

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    $\begingroup$ Just trying to clarify where your concern lies. You seem to be worried about the distance between $\mathbf{r}$ where we measure the field $\mathbf{E}(\mathbf{r})$, and $\mathbf{s}$, the location of the charge element $\rho(\mathbf{s}) d^3\mathbf{s}$ which we integrate over. You are concerned that this distance is small compared with the "size of the charge". But this should not be a concern: when we deal with a continuous charge distribution, the charge elements may be regarded as infinitesimally small. The resultant overall field is a superposition (integral) of infinitesimal contributions. $\endgroup$ – user197851 Jul 5 '18 at 10:08
  • $\begingroup$ @lonelyProf When gaussian surface completely overlaps charged distribution then distance between the point where electric field is to be measured and charge distribution is also infinitesimally small. $\endgroup$ – TontyTon Jul 6 '18 at 11:32
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    $\begingroup$ I think that it is generally accepted that the electric field due to a point charge exactly at the position of the charge is undefined, but that this does not affect the results of the integrations over all of space (such as the first integral in Sebastian Riese's answer). It's a vanishingly small contribution. Everywhere else, the distance is not infinitesimal, and Coulomb's law may be assumed to apply. Of course, there are mathematical subtleties in these derivations, e.g. the Dirac delta function, and the divergence theorem. But I'm guessing that these are not the source of your worries? $\endgroup$ – user197851 Jul 6 '18 at 13:46
  • $\begingroup$ @LonelyProf Thanks, "It's a vanishingly small contribution. Everywhere else, the distance is not infinitesimal, and Coulomb's law may be assumed to apply" This solved my query. Thanks. $\endgroup$ – TontyTon Jul 11 '18 at 5:28

The Coulomb law can be applied in the following manner by considering the continuous charge distribution to consists of infinitesimally small point charges and writing it as integral: $$ \vec E(\vec r) = \sum_i q_i \frac{\vec r_i - \vec r}{\left|\vec r_i - \vec r\right|^3} = \sum_i \rho(\vec r_i) \Delta V \frac{\vec r_i - \vec r}{\left|\vec r_i - \vec r\right|^3} = \int d^3r' \rho(\vec r') \frac{\vec r' - \vec r}{\left|\vec r' - \vec r\right|^3}. $$ Now, we can apply the Gauß integral theorem to prove: \begin{align*} \int_{\partial V} d\vec S \cdot \vec E &= \int_V d^3r\, \nabla \cdot \vec E(\vec r) = \int_V d^3r \nabla_{\vec r} \cdot \int d^3r' \rho(\vec r') \frac{\vec r' - \vec r}{\left|\vec r' - \vec r\right|^3} \\ &= \int_V d^3r \cdot \int d^3r' \rho(\vec r') \nabla_{\vec r} \cdot \frac{\vec r' - \vec r}{\left|\vec r' - \vec r\right|^3} = 4\pi \int_V d^3r\,\rho(\vec r) \end{align*} Here we use the formula $\nabla_{\vec r} \cdot \frac{\vec r' - \vec r}{\left|\vec r' - \vec r\right|^3} = 4\pi \delta(\vec r - \vec r')$. (You can easily prove that it is zero for $\vec r \ne \vec r'$ and then use the Gauß integral law to compute the strength of the $\delta$-peak by computing the flux.)

By replacing $\rho(\vec r) = q \delta(\vec r)$ we can consider a point charge and see that the Coulomb law implies the Gauß law.

The converse, however, is not true in general: The Gauß law holds in full electrodynamics, but there are additional electric fields induced by changing magnetic fields. You need to assume the electrostatic limit, where $\nabla \times \vec E = 0$ and that the field vanishes sufficiently fast at infinity to derive the Coulomb law from the Gauß law.

  • $\begingroup$ I asked about Gauss's Law,not Gauss's integral theorem. Refer en.m.wikipedia.org/wiki/Gauss%27s_law $\endgroup$ – TontyTon Jul 5 '18 at 9:44
  • $\begingroup$ I answered by deriving Gauß' law from Coulomb's law (and the assumption that electric fields superimpose linearly) by using Gauß' integral theorem. I just use Gauß' integral theorem in the process. I do, by no means, prove it, or explain it. $\endgroup$ – Sebastian Riese Jul 5 '18 at 11:03

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