In a lecture, I heard that even if a charge ( suppose a positive ) is kept inside a cavity of a hollow sphere, the net charge inside would be zero through induction. Then , by Gauss' law , the electric flux will be zero.

But when I see that all the field lines emerging from the charge inside end into the sphrere's inner wall , then I can accertain that all field lines pass through a surface inside cavity. So , flux should have been maximum...

Where am I mistaken?

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    $\begingroup$ You said that if there is a charge inside a cavity, then the charge inside would be zero. Perhaps you want to rephrase this? $\endgroup$ – garyp Feb 7 '18 at 16:50
  • $\begingroup$ @garyp...I told the "net charge" , not "charge"... $\endgroup$ – user182687 Feb 8 '18 at 9:41
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    $\begingroup$ I still don't understand. If you put a charge in the cavity, there is a charge in the cavity. The net charge won't be zero unless there is another charge in the cavity. Is there another charge that you didn't mention? $\endgroup$ – garyp Feb 8 '18 at 14:54
  • $\begingroup$ @garyp ... The net charge might be zero through "induction"...However,the electric field still remains zero,whereas all field lines pass through..You can see the lectures of Walter Lewin (802)...I've heard it there... $\endgroup$ – user182687 Feb 8 '18 at 18:20
  • $\begingroup$ I think I see what you are saying. $\endgroup$ – garyp Feb 8 '18 at 20:08

If your Gaussian surface is inside the inner surface of the shell and contains the charge, then there will be a net flux through the surface.

If the Gaussian surface is in the shell, between the inner and outer surfaces, there will be no flux because the static electric field in the interior of a conductor is zero. From that we can conclude that the total charge inside the surface is zero, and that there is an induced charge on the inner surface equal in magnitude to the enclosed charge, but opposite sign.

If the Gaussian surface is outside of the outer surface of the shell, it will be enclosing the charge that was placed in the shell. The shell itself is uncharged. So there will be a net flux through that surface. We can also conclude that there is an induced charge on the outer surface of the conductor equal in magnitude and sign to the charge that was placed inside the shell.

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  • $\begingroup$ @garyp...Once the flux is maximum and once it's zero...Which one is actually true...? $\endgroup$ – Nehal Samee Feb 9 '18 at 17:39

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