The potential at the surface of a conducting sphere is KQ/R where 'R' is the radius of the conducting sphere and even the electric field on the surface of the conducting sphere is maximum. And while taking a test charge from the surface to the inside the potential neither increases nor decreases because electric field inside the conducting sphere is zero, so whatever potential is on the surface, it is the same inside. But my question is why don't we need to do work on the test charge kept on the surface against the electric field on the surface to move the test charge to the inside of the sphere where electric field is zero? And if we would need to do work against electric field on the surface to move charge inside then potential inside will also be different from the surface. Thank You! NOTE- i am assuming that the conducting sphere is positively charged.

  • $\begingroup$ That there is no electric field inside a conducting sphere that has a charge inside is trivially false. The charge itself generates an electric field inside the sphere. This charge causes a mirror charge which is attractive and that's why free charges are going to end up on the conductor rather than in the free volume. In general, if we want to separate a charge from a conducting surface then we have to expend energy. If we didn't, then charge carriers like electrons wouldn't stay bound to conductors but start floating in space all by themselves. $\endgroup$ Commented Jun 10, 2023 at 14:45
  • $\begingroup$ @FlatterMann....i have made an edit...so pls explain me according to that....pls explain. $\endgroup$ Commented Jun 10, 2023 at 15:07
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    $\begingroup$ The surface is an equipotential surface (that's the definition of "conduction"), so by definition of "potential" there can be no change in energy, hence no work can be done. What actually happens on and near the surface of an actual metal, however, is simply not covered by classical electrodynamics. I would warn you against taking these macroscopic statements too seriously. A real metal has a work function for electrons that depends on its microscopic structure and as soon as a charge leaves a metal surface there is an effective interaction even for the macroscopic case. $\endgroup$ Commented Jun 10, 2023 at 15:13

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You give the answer yourself: the electric field inside is zero, so there is no force F=qE=q0=0 on the charge. as experiment you probably have seen that in a faraday cave with very strong electric field outside you have no forces on charges inside. The field outside has no influence inside,

  • $\begingroup$ @trula....This what I want to understand that why there is no influence inside the charged conducting sphere by the field outside i.e. on it's surface....why we are not doing work against the field on the surface of the charged conducting surface to move inside. Pls explain it more deeply. $\endgroup$ Commented Jun 10, 2023 at 15:05
  • $\begingroup$ also tell me about Faraday cave...thank you $\endgroup$ Commented Jun 10, 2023 at 15:09
  • $\begingroup$ @SukritiSharma to your first comment, it's a symmetry argument, if you have equal influence on the charge from each part of the shell but it all sums up to zero. $\endgroup$
    – Triatticus
    Commented Jun 10, 2023 at 15:11
  • $\begingroup$ @Triatticus can you pls elaborate...it will be more clear to me then. $\endgroup$ Commented Jun 10, 2023 at 17:23
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    $\begingroup$ Newton Shell Theorem all the proof is right here, it's the same in the gravitational case as in the electrostatic case. $\endgroup$
    – Triatticus
    Commented Jun 10, 2023 at 18:56

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