enter image description here

I have a spherical conductor with a charge $+q$ place inside the cavity, now the charges redistribute as shown, If I apply gauss law where my guassian surface is such the $q$ inside is non zero now , $\oint \vec{E}.\vec{da}= \dfrac{q}{\epsilon} $ we can say since $q≠0$ , $\vec{E}≠0$. Now this contradicts the fact we already know that electric field inside a conductor is zero , please tell where I went,is it something wrong with my guassian surface (why?)any help would be appreciated, Thanks.

$\textbf{EDIT}$: To avoid any confusion my guassian surface is only about the boundary of conductor it neither goes inside the cavity nor outside the conductor but it includes the charges present at the periphery of conductor that is some positive charge and a lesser negative charge making overall net charge non zero.

  • $\begingroup$ You are not wrong; if you want, you might have more luck attracting answers if you rephrase your question in terms of the physics of how electric charge is distributed on conducting interfaces. $\endgroup$
    – TLDR
    May 16, 2021 at 21:32
  • 1
    $\begingroup$ I think your diagram misses a (+q) charge in the middle of cavity. Then only will you get such a charge distribution in a spherical conductor. $\endgroup$
    – Satwik
    May 16, 2021 at 21:55
  • 1
    $\begingroup$ I don't understand your gaussian surface. $\endgroup$
    – Bhavay
    May 17, 2021 at 4:40
  • $\begingroup$ An electric field may be present inside a wire with current $I$ if the wire has a resistance. $\endgroup$ May 17, 2021 at 7:23
  • $\begingroup$ The edit is not clear. The surface charge is infinitely thin normal to the surface. If your Gaussian surface includes the charges it lies outside the conductor, where it picks up the electric field. $\endgroup$
    – my2cts
    May 17, 2021 at 16:20

4 Answers 4


The charge density on the conductor surface is singular, so gauss' law is not well defined if the surface you draw goes through the conductor surface. You can put the surface slightly within the conductor, and the surface charge will not be picked up, giving no field inside the conductor, or you can put the surface just outside the conductor and the charge will be picked up, reflecting that the surface charge creates a field outside the conductor.

  • $\begingroup$ I also think the same that we cannot draw such a guassian surface that would only pass through the surface of conductor $\endgroup$ May 17, 2021 at 12:07
  • 2
    $\begingroup$ I disagree with this. The OP specified that "it includes the charges present at the periphery of conductor". So it is not really singular, it does include the surface charge. $\endgroup$
    – Dale
    May 17, 2021 at 14:11
  • $\begingroup$ In that case, it means part of the Gauss surface lies outside the conductor, and it makes sense for the flux to be non-zero, since you pick up electric field outside the conductor. The point is that that you cannot have part of the Gauss surface coincide with the conductor boundary in a manner that is not transverse. $\endgroup$ May 17, 2021 at 16:29

To include the surface charge densities, the Gaussian surface must be just outside of the surfaces. The flux in will be proportional to the included negative charge on that surface, and the flux out is proportional to the included positive charge on the other surface (with no field or flux in the conductor).


Gauss' law tells the total charge inside a surface. In special cases you can also draw conclusions on the value of $\bf E$ at the surface. In your example it is not possible to draw the conclusion that $\bf E\neq 0$ inside the conductor.

Note that the statement that $\bf E = 0$ inside a conductor is only true at a scale where the underlying ionic system can be treated as a continuum. At this scale the surface charge has zero thickness. At atomic scale $\bf E = 0$ only on average over large distances.

  • $\begingroup$ I draw that conclusion because if E=0 then wouldnt be charge inside the guassian surface =0 , which is not here. $\endgroup$ May 17, 2021 at 12:05
  • $\begingroup$ @AkshajBansal You should explain your assumptions on the location of the gaussian surface and the surface charge with respect to the conductor surface. $\endgroup$
    – my2cts
    May 17, 2021 at 12:13
  • $\begingroup$ I take my gussian surface only covering the surface of conductor so it includes the charges at the boundary but not go beyond the conductor $\endgroup$ May 17, 2021 at 12:17
  • $\begingroup$ @AkshajBansal There the field is non-zero. $\endgroup$
    – my2cts
    May 17, 2021 at 12:27
  • $\begingroup$ But guass law tells us the net field by all charges which should have been zero since its a conductor ,had there been any electric field charges would start flowing $\endgroup$ May 17, 2021 at 12:29

Gauss law only can only be used to evaluate electric field of charges contained inside gaussian surface. The contribution due to charges outside always dies.

You can prove that any external field outside the gaussian surfaces dies on taking the integral. Therefore, you can't speak of net field using gauss law.

OP in comments:

You are working the maths wrong here in the counter example , firstly you are taking the E out of the integral considering it constant which is not true , the fact is neither is the magnitude of E nor the direction is same at all points odlf the guassian surface so you cant perform the maths so simply.

Consider Electric field defined on the gaussian surface decomposed the following way: Enet=E+E′ E is due to charge inside the gaussian surface and E′ is due to the charge outside. Consider the case of removing the charge inside and evaluating the flux across the boundary due to the external charge, in this case we find that the flux due to external charge is zero by gauss law. Now, reinsert the old charge, we find that ∫Enet⋅dS=∫EdS and we can evaluate it's field as needed by pulling out the E.

Ultimately the point is that gauss law is useless for making any claims for fields caused by charges outside the gaussian surface because they will always cause zero flux independent of the distribution

  • $\begingroup$ Guass law acounts for the net field by all the charges inside and outside . $\endgroup$ May 17, 2021 at 4:41
  • $\begingroup$ Pl see this post. Counter example of the claim: Consider two point charges q placed a distance 2R apart, at the midpoint of the point charge, take any one of the charge and consider a gaussian surface of radius R through them and evaluate field at the distance R. You will find it's only $\frac{kq}{r^2 }$, this is not the true field as because the other charge's field is not accounted for. The true field would be zero at the midpoint of two pointcharges. @AkshajBansal $\endgroup$ May 17, 2021 at 7:05
  • $\begingroup$ you are working the maths wrong here in the counter example , firstly you are taking the E out of the integral considering it constant which is not true , the fact is neither is the magnitude of E nor the direction is same at all points odlf the guassian surface so you cant perform the maths so simply. $\endgroup$ May 17, 2021 at 12:12
  • $\begingroup$ I've added the required in the answer @AkshajBansal $\endgroup$ May 17, 2021 at 12:32
  • $\begingroup$ In the link you provided the first answer only tells that Electric field by all the charges inside or outside the guassian surface. do you disagree with that answer too? $\endgroup$ May 17, 2021 at 12:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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