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I have seen similar questions here but haven't understood this:

If we introduce a charge +Q inside a spherical conducting shell, this charge attracts an equal negative charge i.e. electrons towards the inside of the shell, so as the electrons are pulled away from the atoms of the metal constituting the shell a charge +Q is induced on the outside of the shell.

So there are 3 charges at work here, aren't there? A +Q charge at the centre of the spherical shell, a +Q on the outside surface of the shell and a -Q charge on the inside surface of the shell.

So, shouldn't the net charge enclosed by the shell be +Q (which is the charge at the centre), it hasn't disappeared somewhere has it?

If so, then why, when calculating the electric field inside the shell using Gauss' Law, do we take the net charge enclosed by the shell to be zero (and so, then proceed to calculate the electric field inside to be zero) ?

I would really like if someone could help me here.

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  • $\begingroup$ @MrReality...Look , you yourself have said that electric field intensir6inside is zero.That is , the induced positive charge outside has nothing to do with the interior intensity... $\endgroup$
    – Nehal Samee
    Feb 11, 2018 at 7:47
  • $\begingroup$ physics.stackexchange.com/questions/384868/… Look here... $\endgroup$
    – Nehal Samee
    Feb 11, 2018 at 7:49
  • $\begingroup$ @NehalSamee, what do you mean? What I asked here is why do we take the charge enclosed by a Gaussian surface inside a spherical conducting shell to be zero, when in fact we have introduced a charge at the centre of the shell ? I didn't say the induced charges at the inside and outside surfaces of the shell have anything to do in the electric field inside.. $\endgroup$
    – HeWhoMustBeNamed
    Feb 11, 2018 at 7:53
  • $\begingroup$ @NehalSamee, so (from the link you provided I gather) we don't take the charge enclosed by the sphere in the case to be zero? $\endgroup$
    – HeWhoMustBeNamed
    Feb 11, 2018 at 7:58
  • $\begingroup$ @MrReality...The electric field depends on the Gaussian surface you consider...If you take the Gaussian surface in the interior surface , you get the net charge zero inside...You can watch out the lecture of Walter Lewin in You Tube... $\endgroup$
    – Nehal Samee
    Feb 11, 2018 at 9:26

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