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Hi guys I'm really confused by this electric field question. Basically for the given setup shown, draw a graph of how electric field varies with distance from the center, given that radius of smaller sphere is A and radius of bigger sphere is B.

  1. Does drawing the graph of the electric fields vs distance for both the smaller and bigger spheres individually and summing them give the answer?

  2. Is the field outside of both spheres 0? Does gauss' law apply in this case? In which cases does gauss' law apply cause I read that it only applies in situations with enough symmetry.

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  • $\begingroup$ You need to include more information about the setup. Are those insulators with uniform charge density? $\endgroup$ Commented Feb 17, 2014 at 3:56
  • $\begingroup$ the question i got does not state, but it says the setup is a capacitor. $\endgroup$ Commented Feb 17, 2014 at 4:00

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Does drawing the graph of the electric fields vs distance for both the smaller and bigger spheres individually and summing them give the answer?

Yes, that's the superposition principle. Note that if the spheres are conductors the charge of one will modify the charge on the other.

Is the field outside of both spheres 0? Does gauss' law apply in this case? In which cases does gauss' law apply cause I read that it only applies in situations with enough symmetry.

Gauss law always apply, but I there is no symmetry in will give really complicated integrals. So you tell me if the electric field is 0 (no net charge).

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You know about Gauss's law. Your outer shell is connected to earth, so we can assume it is conducting. Your inner shell is drawn in the same manner, but the + signs seem to be distributed in the interior. But since you also state that the setup "is a capacitor", a hollow conducting sphere is most likely.

For two concentric spheres around the origin there is rotational symmetry, so the field is radial and depends on |r| only. Gauss gives the field strength.

With the outer shell grounded, your capacitor will have a negative charge on the inside of the outer shell, equal to the positive charge of the inner sphere. So outside Gauss tells you E = 0.

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