0
$\begingroup$

In its integral form, the Gauss' law in the vacuum says that the total flux entering a closed surface is proportional to the net charge.

In its differential form, if the charge distribution is zero within the volume, does it mean that there are no charges there or can there be positive and negative charges with a total net charge equal to zero?

$\endgroup$
  • $\begingroup$ What matters - both in the integral and differential form - is the NET charge, not the individual amounts of positive or negative charges. $\endgroup$ – ZeroTheHero Mar 21 '17 at 20:56
  • $\begingroup$ Comment to the post (v2): A fixed finite volume does not appear in the differential version, so the last sentence seems meaningless. $\endgroup$ – Qmechanic Mar 21 '17 at 21:56
  • $\begingroup$ @ZeroTheHero The differential form is a punctual relationship (no volume). In the equation the charge density appears instead of the total charge. How can be the NET charge what matters? $\endgroup$ – baister Mar 21 '17 at 22:10
  • $\begingroup$ @Qmechanic OK, but if I take a volume and, in that volume, $\rho(\vec{r}) = 0$, does it mean that there are no sources/sinks in that volume? $\endgroup$ – baister Mar 21 '17 at 22:11
  • 1
    $\begingroup$ The differential form applies to arbitrarily small but non-zero volumes. After all, you are taking derivatives, which means you are comparing values at arbitrarily close points, but not at a single point, i.e. you need an infinitesimal volume to make sense of the differential form. (Of course not so small as to reach quantum granularity.) The book by Purcell on E&M is an excellent read on this very topic. $\endgroup$ – ZeroTheHero Mar 21 '17 at 22:17
1
$\begingroup$

Net charge is what's important. Flux has to do entirely with net charge. Individual electrons are not important if they are canceled by protons.

$\endgroup$

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.