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Charge inside of Conducting Shell

  • #1

Homework Statement


A positive point charge q is located inside a neutral hollow spherical conducting shell. The shell has inner radius a and outer radius b; b − a is not negligible. The shell is centered on the origin. Assume that the point charge q is located at the origin in the very center of the shell.

a. iii. Determine the electric potential at x = a.

http://www.aapt.org/physicsteam/2015/upload/E3-2-2-solutions.pdf

Homework Equations


V = Kq/r, gauss's law


The Attempt at a Solution


Ok so the solution is in the link above. However, I don't understand why the potential at a is Kq/b. I know that the potential at A and B must be the same. If you look at it from outside the shell, the potential is Kq/r for r>b and this predicts Va = Vb = Kq/b. However, if you look at it from inside the shell, the potential is Kq/r for r<a and this predicts Va = Kq/a which is not correct. What is wrong with the second approach?
 

Answers and Replies

  • #2
Doc Al
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However, if you look at it from inside the shell, the potential is Kq/r for r<a and this predicts Va = Kq/a which is not correct.
That would be true if the field from the charge extended unbroken to infinity. But it does not. The field has a gap from r = a to r = b.
 
  • #3
SammyS
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Ok so the solution is in the link above. However, I don't understand why the potential at a is Kq/b. I know that the potential at A and B must be the same. If you look at it from outside the shell, the potential is Kq/r for r>b and this predicts Va = Vb = Kq/b. However, if you look at it from inside the shell, the potential is Kq/r for r<a and this predicts Va = Kq/a which is not correct. What is wrong with the second approach?
The usual convention is for the potential to be zero as r → ∞ .

If you take Va = kq/a, then the potential on the outer surface, r = b, is also kq/a rather than being kq/b. This gives a potential as r → ∞ of kq(1/a - 1/b) .
 

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