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Imagine you have hollow concentric spheres A and B with radius a and b (b>a), respectively. If A is a conductor and B has a certain density charge, I´ve been taught that B will induce some net charge over A (appart from what it could have yet). But, if, according to Gauss law, the contribution to electric field inside sphere B from its surface charge distribution is zero, How can this interaccion ocurr? How can be any charge displacement inside B?

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  • $\begingroup$ Just my own thought: Your sphere is grounded which means it connects to something large far away. The E field of B push the charge from the ground to A? $\endgroup$ – velut luna Apr 25 '16 at 15:53
  • $\begingroup$ This question come to me in an exercise where, as you say, A sphere was grounded, but I don´t know if it has much to do with my question. Anyway, any answer is welcome. $\endgroup$ – amejmar Apr 25 '16 at 15:59
  • $\begingroup$ If it's not grounded but isolated, there won't be induced charge. $\endgroup$ – velut luna Apr 25 '16 at 16:01
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I assume that A is neutral to begin with. Then inside of A cannot be any charge by Gauss law. This means that the inside of A must neutral.

Then you can take the volume between B and A. The boundary of that are the metal spheres A and B. On the whole boundary surface, there cannot be any electric field as that surface lies inside the metal. If there was any electric field, the electrons would move to counteract it. Therefore there is no total charge between the two spheres. There could be charge on the inside of B and outside of A. The two spheres form a spherical capacitor.

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  • $\begingroup$ There´s a point I don´t understand in your argument. Your penultime and final sentence seem to me to be in contradiction, because firstly you say there´s no total charge between the two spheres, but thereupon you say that there could be charge inside B but outside A, which is the same region than the other one, Is it what you actually wanted to say or is it a confusion? $\endgroup$ – amejmar Apr 25 '16 at 17:23
  • $\begingroup$ I think I haven´t explained accurately my question before. My doubt is how is electrostatic induction possible if, given the fact that A is neutral and has no surface charge density at first, the electric field inside B is zero (according to Gauss law) $\endgroup$ – amejmar Apr 25 '16 at 17:28
  • $\begingroup$ With the last two sentences I mean that the total charge must be zero. But you can have $Q$ on B and $-Q$ and A and that requirement would still be fulfilled. You do have a point with your second comment. Adding charge to B does not influence A in any way. If you however but charge $Q$ on A, then charge $-Q$ must be on the inside of B to shield the field; there must not be any field within B. Does that make sense now? $\endgroup$ – Martin Ueding Apr 25 '16 at 17:52
  • $\begingroup$ Ok, I´m catching up with the idea. So, if there´s Q charge on A, in order to fulfill Gauss law, -Q charge have to appear on the inside of B, just because A is a conductor and can´t suffer any electric field, otherwise it will react so as to cancel off its effects). But there´s something that still eludes me. At first, there´s only the conductor (A) with no charge and the greater sphere (B), in such a way that A is inside B. According to Gauss law, B creates no electric field inside itself, therefore it wouldn´t be necessary for A to make any charge displacement, Why induction ocurrs? $\endgroup$ – amejmar Apr 25 '16 at 19:09
  • $\begingroup$ If there is no charge on A, no induction will occur. Putting charge on B does not alter anything on the inside. $\endgroup$ – Martin Ueding Apr 25 '16 at 20:04

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