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This is a charge q (unknown sign) fixed inside a conductor so it cannot move, the goal is to find the electric field at A and to find the electric field at B

My attempt at solution:

I drew a gaussian surface passing by the point A and having q in the center of the gaussian surface and just applied gauss's law, having a conductor, the electric field is perpendicular to the surface at all points of the surface, so the electric field and a small segment of the area are always parallel and so

E da = Qin/epsilon0

continuing with the same steps with point B, but I'm hesitant because I've never applied gauss's law to a fixed charge inside a conductor, and the surface being a conductor makes me think that what I'm doing is wrong and there's another way to approach it, can you enlighten me please (the conductor was uncharged before placing the charge q)

  • $\begingroup$ Does the big ellipse represent the conductor? What is its shape in the third dimension? $\endgroup$ – hmakholm left over Monica Jul 17 '18 at 17:43
  • $\begingroup$ yes this is the conductor, it's a sphere with radius R $\endgroup$ – khaled014z Jul 17 '18 at 17:46

Well your method seems correct to me as gauss law deals with flux and so we can imagine a sphere at whose centre the charge is placed and of radius $r/2 + 3r/4$ and so field is equal to $16kq/25r^2$
There can be a different approach which is to imagine a negative chargeq on the inner surface of the conductor and as charge is conserved there will be a positive charge on the outer surface of the conductor and also the charge density will be diff at diff points on sphere but net charge on sphere is equal to 0 and so field by conductor is 0 so you can simply write $16kq/25r^2$

  • $\begingroup$ but what would be the difference if we let this sphere be a non conductor with a charge placed at the same location $\endgroup$ – khaled014z Jul 17 '18 at 17:34
  • $\begingroup$ No difference in this case $\endgroup$ – Rinki Dwivedi Jul 17 '18 at 18:23

This seems to be an electrostatics problem where everything is in equilibrium. Then it is impossible for there to be an electric field inside the conductor, so some of the charge carriers in the conductor will cluster around the $q$ and neutralise it completely. A matching amount of opposite charges will distribute themselves on the surface of the conductor, as if we had charged the conductor itself with charge $+q$.

By symmetry (and Gauss's law) the field outside the conductor is then the same as the field of a point charge of size $q$ located at the geometric center of the spherical conductor -- no matter where in the conductor your original trapped charge is located, the conduction will hide its location from the outside world.

  • $\begingroup$ i forgot to mention that this conductor was uncharged before placing the charge q $\endgroup$ – khaled014z Jul 17 '18 at 17:56
  • $\begingroup$ @khaled014z: Good thing that was what I assumed too. $\endgroup$ – hmakholm left over Monica Jul 17 '18 at 17:56
  • $\begingroup$ so is your statement correct $\endgroup$ – khaled014z Jul 17 '18 at 18:23
  • $\begingroup$ @khaled014z: Of course I think my statement is correct; otherwise I wouldn't have made it. If you want someone else's opinion on the correctness of my answer, I'm not the one you should be asking for it. $\endgroup$ – hmakholm left over Monica Jul 17 '18 at 18:24

The field is the same as with and without the conducting sphere.

  • $\begingroup$ Please make your negative vote explicit :-) $\endgroup$ – my2cts Jul 18 '18 at 10:26

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