enter image description here

In the text, it is said that the skewed distribution of positive charge on the inner wall cannot produce an eletric field in the shell to affect the distribution of charge on the outer wall. Why? Shouldn't the negative charges be more close to the positive charge concentration?

  • $\begingroup$ They also tend to be less concentrated because of the vicinity of the negative charge inside the sphere. That cancels the attraction made by the more concentrated positive charges and is in fact the point of $\nabla \cdot \stackrel{\to }{E}=\frac{\rho }{{\epsilon }_{0}}$. $\endgroup$ – Soba noodles Sep 1 '15 at 22:51

This can certainly seem confusing since one might reasonably expect that the positive charge which accumulates densely on the left side of the inner surface of the spherical shell will attract a denser accumulation of negative charge on the outer surface of the shell. However, that does not take into account the fact that the negative charge inside the shell will be repelling the accumulating negative charge on the outer surface. In fact, the attraction from the inner surface and the repulsion from the enclosed charge exactly cancel so that no net field from the internal structure of the shell is experienced by the outer surface.

As a result, the only electric fields that affect the charges on the outer surface will be from the other charges on the outer surface. For this reason, the charges will all spread out as much as possible on the outer surface due to their mutual repulsion. Obviously, the charge distribution for a spherically symmetric surface which minimizes these interactions is the uniform charge distribution depicted in the diagram.

  • 1
    $\begingroup$ Geoffrey, if I could give you more than one upvote, I would. That was an excellent answer, and I will try to use this explanation in the spring, when I am teaching AP Physics C students electricity and magnetism. $\endgroup$ – David White Aug 26 '17 at 2:07
  • $\begingroup$ @DavidWhite Thank you! That's very kind, David. I'm glad I could be of help. $\endgroup$ – Geoffrey Aug 26 '17 at 2:27

This is presumably because the shaded-blue region is a conductor and the electric field in there must be zero; all of the field lines of the inner charge must therefore terminate on positive charges at the surface of the conductor.

The field in a conductor must be 0 because if it weren't it would generate a large current (for a perfect conductor, an infinite current) which would tend to reduce that field. So very rapidly you get to a point where the field must be 0 because charge followed the E-field lines and built up in ways that oppose them.

Now the positive charge must create a negative charge due to charge conservation, but this negative charge shoots all the way to the outside of the conductor, essentially because the force of the positive charges has perfectly canceled the force due to the negative charge inside, so they're just trying to spread out away from each other. A uniform spherical shell of charge has no electric field inside, so this preserves the condition that the field in the conductor is 0.

  • $\begingroup$ I get it thanks. Just one more thing, suppose there is a hollow conductor sphere of radius R positively charged, if I put a positive charge in any point from a distance R/2 to the center, will it stay there? It will move, all charges will rearrange and them he eletric field will be zero again, correct? It thats correct, the eletric field really was never zero. I dont get it. $\endgroup$ – João Pedro Sep 1 '15 at 23:17
  • $\begingroup$ If the sphere is an insulator with symmetric positive charge, the particle will indeed stay wherever it is. However if it is a conductor, then even though the net positive charge has zero field inside the conductor, the charge placed inside will push charges around in the conductor, and these tend to attract it to whatever side of the conductor is closest. These are sometimes called "image charge effects" since if you place a charge above a neutral conducting plane, the fields are the same as they'd be if a mirror image of that charge had the opposite charge and attracted it. $\endgroup$ – CR Drost Sep 1 '15 at 23:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.