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This is a conductive sphere with two holes in it, they put charge $q_{a}$ in one of the holes and $q_{b}$ in the other hole (The holes are surronded by conductive material and not open to the outside).

Now after I found the electric field they tell me the sphere has been grounded in relation to infinity so the potential in infinity which is $0$ is now the same on the sphere. enter image description here

while I found for $r>R$

$E=\frac{K*(q_{a}+q_{b})}{r^{2}}$

The integral limits is from infinity to $R$

$potential = -\int_{\infty}^{R} E*dr = \frac{K*(q_{a}+q_{b})}{R} = 0 \Rightarrow q_{a} = -q_{b}$

So the weird thing for me is that from grounding the sphere, mathematically I got that $q_{a} = -q_{b}$ but it doesn't make sense that grounding the sphere decides the charges that have been placed are with opposite signs or are 0, grounding could not change the charges that have been already placed, they can't go anywhere.

I know that all the surface charge is being drained away, but still how come I get $q_{a} = -q_{b}$ which is a false statement if the two charges are positive. So my question is what am I missing here ?

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You are missing the fact that when the sphere is grounded, charge can flow towards it (or away from it) so that the NET charge of sphere plus the two internal charges is zero. Add to that the fact that the charge on the sphere-with-holes will not be uniformly distributed (so there is no electric field inside) and you have your answer.

Make sense?

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  • $\begingroup$ Thanks for the answer ! I understood the charge can flow, but about not being uniformly distributed, you mean in general ? Because if we look at every surface individually the outer surface is $0$ after grounding and the charge on the cavities surface is uniformly distributed to cancel out the charges is that right ? But my biggest misunderstanding is still that I can't understand how the math gives me a false statement which is $q_{a} = -q_{b}$ is it incorrect because this equation is only correct if charge were unable to move ? (more in next comment) $\endgroup$ – user3575645 Jul 7 '17 at 18:05
  • $\begingroup$ I should just count on the logic that before grounding, there is $q_{a} + q_{b}$ lying on the surface, and i get $q_{a} = -q_{b}$ not because it is actually true but because there is now $0$ on the surface and this is the only way for the math to tell me that from that equation ? $\endgroup$ – user3575645 Jul 7 '17 at 18:06
  • $\begingroup$ If you don't ground the conducting sphere, then there will be polarization of the charges in the sphere so that the electric field is normal to the surfaces of the conductor - with the net charge still +2q. When you ground the sphere, the net charge will become zero. Your integral doesn't account for the charge on the sphere, which is why you are getting the wrong answer. $\endgroup$ – Floris Jul 7 '17 at 18:11
  • $\begingroup$ I see so after the grounding I should change my equation because the net charge isn't necessarily the same, and add another unknown charge ( that after the integral calculation i'll know it which will be $q_{uknown} = -(q_{a}+q_{b})$, that will actually explain that charge has moved which cancels out the charge on the surface ? Is that logic right ? Thanks alot for your time ! $\endgroup$ – user3575645 Jul 7 '17 at 18:17
  • $\begingroup$ Yes, that's the correct way to think about this. $\endgroup$ – Floris Jul 7 '17 at 18:56

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