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I have a spherical conductor with a cavity. A point charge +q is placed in the cavity. Now, on the outside of the sphere, I bring a charge of -q close to the sphere. I understand that the amount of charge on the inner surface will remain the same as by Gauss' Law, considering a gaussian surface enclosing the inner surface and cavity, the total enclosed charged should be zero as the electric field on the interior of a conductor is zero. The question now is, does the electric field at every point on the interior of the cavity remain the same before and after the external charge -q was brought into the neighbourhood? How can I use Gauss' law to show that the field remains the same, as gauss' law only talks about total flux through a surface?

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  • $\begingroup$ Isn't it electrostatic shielding? $\endgroup$ – Anubhav Goel Feb 12 '16 at 3:21
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Yes the electric field remains the same in the presence of the negative charge. Apply gauss' Law inside of the cavity. The enclosed charge is the same therefore the field is the same. In a conductor, charge redistributes itself to kill off any field inside. If you bring a $-q$ charge close to the surface there will be redistribution of charge on the outer surface of the sphere and inner surface of the cavity. However remember that the field at a point inside determined only by the total flux through the appropriate gaussian surface which is determined only by the charge distribution inside. The presence of the conductor completely eliminates the effects of outside charges.

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  • $\begingroup$ Your last line is exactly the statement I have trouble understanding. How do we deduce that a conductor has the property of electrical shielding? $\endgroup$ – Zhanfeng Lim Feb 12 '16 at 9:13
  • $\begingroup$ There is no electric field in the conductor. Thus, if there is no charge in the cavity, there can be no field. Otherwise a closed loop integral that goes through the boundary of a conductor would be nonzero which is contractidtory to the fact that the curl of the E field is zero (in electrostatics). $\endgroup$ – bremsstrahlung Feb 17 '16 at 23:59

protected by ACuriousMind Apr 9 '17 at 16:41

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