1
$\begingroup$

We know that divergence of Electrostatic field is

$$ \overrightarrow{\nabla}\cdot\overrightarrow{E}=\dfrac{\rho(\overrightarrow{r})}{\epsilon_0} $$

in case of a point charge, the divergence would be the Dirac-delta function.

What, similarly could be said about the curl at the location of point charge, we know that curl is zero for an electrostatic field , but that's at the points other than the location of the point charge.

If one tries to use Stoke's theorem to say that curl is zero everywhere, but Stoke's theorem requires that curl be define everywhere, how can we presume curl to be defined everywhere.

$\endgroup$
3
  • $\begingroup$ Very related: physics.stackexchange.com/questions/168905/… $\endgroup$
    – user258881
    May 31 '20 at 8:21
  • $\begingroup$ i had read that before asking the question, it still doesn't answer me. What i am asking is , as there is discontinuity in Electric Field around the source point charge, it's divergence at the location of point charge comes out to be a delta function, as we would expect the derivatives to increase beyond all bounds. Similarly, what about the curl at the location of the point charge. $\endgroup$ May 31 '20 at 8:24
  • $\begingroup$ Try this discussion thread: physics.stackexchange.com/questions/126366/… $\endgroup$ May 31 '20 at 8:41
0
$\begingroup$

Use Faraday's Law:

$$\nabla\times\mathbf{E}=\frac{\partial\mathbf{B}}{\partial t}$$

For a point charge at rest, we know that $\mathbf{B}=0$ everywhere, including at the location of the point charge. Thus $\frac{\partial\mathbf{B}}{\partial t}=\nabla\times\mathbf{E}=0$ everywhere, including at the location of the point charge.

$\endgroup$
3
  • $\begingroup$ That's good , but can you, please, give me some resource to get the proof of Faraday's Law. $\endgroup$ May 31 '20 at 8:37
  • $\begingroup$ @uditnarayanpandey A proof starting from which axioms? $\endgroup$ May 31 '20 at 8:40
  • $\begingroup$ okay , i got you, thanks $\endgroup$ May 31 '20 at 8:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.