Im having trouble understanding the electric field within a sphere of charge. Here's the problem. Suppose we have a hollow conduction sphere of negative charge, say its inner radius is 5cm. Inside this hollow sphere we have another solid charged spherical conductor whose radius is 2cm. Now i want to find the electric field 3cm away from the solid inner conducting sphere (the r=2cm sphere). Also lets say the charge is opposite in sign to the outer shell.

My understanding, which seems to be incorrect is that the inner solid sphere polarizes the outer one, thus there is also an electric field due to the shells inner surface where there is negative charge residing. Why is this assumption incorrect and why is there only an electric field due to the solid inner sphere at this distance.

  • $\begingroup$ You want to find the field between the two charged surfaces or on the outer surface? $\endgroup$ – Sanya Oct 1 '16 at 16:43
  • $\begingroup$ between the solid sphere and the inner shell $\endgroup$ – Teyash Arjun Oct 1 '16 at 16:53
  • $\begingroup$ and the solid outer sphere has a finite thickness, is charged and not grounded? $\endgroup$ – Sanya Oct 1 '16 at 16:58
  • $\begingroup$ correct just charged, not grounded $\endgroup$ – Teyash Arjun Oct 1 '16 at 17:20

According to Gauss Law and the symmetry of the problem the electric field at any radius between the inner sphere and the inner surface of the outer conducting sphere is only given by the total charge on the inner sphere. Of course, a positive inner sphere charge will induce a negative surface charge on the inner surface of the outer conducting sphere and a corresponding positive charge on the outer surface of the outer spherical shell. There is, however, no influence of these induced charges on the electric field in the inner space between the spheres. If you calculate the electric field (e.g. by using Coulomb's Law) produced by any spherical symmetric charge distribution outside the inner space between the spheres you will find that the field produced by the sum of all charge elements is exactly zero in the inner space.


The inner shell can't polarise the outer one because of the symmetry of the problem. A conductor's surface needs to be an equipotential one and if there is a charge at its centre/ a spherically symmetric charge distribution inside it, the only way for it to be an equipotential surface is to have a uniform surface charge.
There is indeed of course also a field from those uniform charges inside the outer sphere - but they cancel out to give a zero net field in total.

Edit: The last statement can be formally shown by using Gauß' law: the charge included in any volume inside a hollow sphere of uniform surface charge is zero and the field can, due to symmetry, only have a component in the radial direction, thus, the field is zero inside. To actually see this "intuitively" is not that easy in my opinion. There are three questions here on Stackexchange that are worth reading on this (keeping in mind that Newton's gravitational law and the Coulomb law are very similar):
Gravitational field intensity inside a hollow sphere
What is the electric field inside a hollow sphere?
Why is the field inside a hollow sphere zero?

  • $\begingroup$ I understand what you've said. It makes a bit more sense now, but I still feel that the only point that has an electric field of zero due to the inner part of the shell is at the center of the shell, where the radius is 0. That would make sense as because of the symmetry all the electric fields would cancel. I'm still having difficulty understanding why at a point between these two surfaces there would not be a field. $\endgroup$ – Teyash Arjun Oct 2 '16 at 16:31
  • $\begingroup$ @TeyashArjun There is a field from the inner sphere, but not from the outer sphere. I will update my answer as soon as I find the time to elaborate on this point. $\endgroup$ – Sanya Oct 2 '16 at 19:54
  • $\begingroup$ @TeyashArjun I hope I have made that a bit clearer now $\endgroup$ – Sanya Oct 3 '16 at 14:03

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