Electric Potential of a Spherical Shell

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Homework Statement


A conducting spherical shell has inner radius a, outer radius b, and has a +Q point charge at the center. A charge of -Q is put on the conductor.

a) What is the charge on the inner and outer surface of the shell?

b) What is the electric field everywhere?

c) What is the electric potential everywhere?


Homework Equations


Gauss's Law
V = -∫E[itex]\bullet[/itex]dl

The Attempt at a Solution


a) From conservation of charge and the fact that it's a conductor, the charge on the inner surface of the shell is -Q and the charge on the outer surface is 0. In addition, there is no charge inside the shell.

b) For 0 < r < a, Gauss's Law gives E = kQ/(r^2)

For a < r < b, E = 0

For r > b, E = 0

c) This is where I'm stuck. E = 0 for r > b, so the potential difference is 0. But the question asks for the actual potential function. Normally, I would use V = kQ/r. But Q = 0 in this case, so V = 0. So that would mean V = 0 for a < r < b Lastly, for 0 < r < a,
V(r) = V(a) - kQ((1/a)-(1/r)) = 0 - kQ((1/a)-(1/r)) = kQ((1/r)-(1/a)).

I believe this is what the answer should be. Please, can anyone tell me if they agree?
 
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Answers and Replies

  • #2
Simon Bridge
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a) From conservation of charge and the fact that it's a conductor, the inside of the shell is -Q and the outside is 0. In addition, there is no charge inside the shell.
First you say there is a charge of -Q inside the shell, then you say there is no charge inside the shell ... which is it?

The question is a tad vaguely worded - it is unclear what is meant by "inside" and "outside" in each case. Taken literally, "inside" would commonly be considered r<a and "outside" would be r>b ... so that would exclude the surface charges. a≤r≤b would be "within the shell itself". It is possible to answer all the sections using that definition ... it's an exercise in Gauss' Law.

You'll have to use the context provided by your lessons to correctly interpret the question.
Whatever you choose, you need to be precise in your answer.

b) For 0 ≤ r ≤ a, Gauss's Law gives E = kQ/(r^2)
... that is only for r<a, at r=a, isn't there an additional surface charge?

Being clear about this should help you with the last part.
Note: what is the relationship between electric potential and electric feild?
 
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Yes,sorry about that. I rephrased the question.
 
  • #4
Simon Bridge
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Gauss's Law gives E = kQ/(r^2)
No it doesn't. Check. Hint: E is a vector.

Also:
What you have written means that at r=a, E=kQ/a^2 (1st relation) and E=0 (2nd relation), at the same time: how can this be?
 
  • #5
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Right. E is indeed a vector that points radially outward in this case. I'm focused on its magnitude here. In vector form, there is an r roof vector included.

Ok I see that you are saying that E(r) is discontinuous at r = a. I'm guessing that E must be 0 at r = a since the enclosed net charge is 0, so the first relation should be E(0 < r < a) = kQ((1/r^2) - (1/a^2)). That would make it continuous. However, that contradicts Gauss's Law. That doesn't make sense. Also, why does E have to be continuous anyway? There are several examples in my textbook in which it is not. Are the charges correct?

Lastly, do you agree with my reasoning in the part on finding the potential? Suppose my electric fields are correct, then did I get the correct potentials?
 
  • #6
Simon Bridge
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You have the idea, it is just you are not being careful about the boundaries.
The trick is to be careful about dividing the volume up.

For a gaussian surface that is a sphere centered on the origin of radius ##0<r<a##,
What is the total charge enclosed?
What is the electric field due to that charge?

Again for ##a\leq r < b## ?

Again for ##r\geq b## ?

It may help to think of the surface charge as being infinitesimally inside the surface.

For the potentials, you seem to have chosen your reference potential at r=a ... I'm not really sure I've understood your reasoning: was there a specific reason for this?

Note: there is a relationship between the electric field and the potential.
Why not use it?
 
  • #7
ehild
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For r > b, E = 0

c) This is where I'm stuck. E = 0 for r > b, so the potential difference is 0.
You should state with respect to what is the potential difference zero. But you are right, if the electric field is zero which is the negative gradient of the potential, the potential does not change between infinity and r=b.
But the question asks for the actual potential function. Normally, I would use V = kQ/r. But Q = 0 in this case, so V = 0. So that would mean V = 0 for a < r < b
The potential of a point charge Q is kQ/r if the zero of the potential is at infinity.
Yes, the potential is also zero inside the conductor, as the electric field is zero, the potential is constant and the potential is a continuous function. If it is zero just outside the conductive shell, it is zero also inside. For the same reason, the potential is also zero inside the void at r=a.

Lastly, for 0 < r < a,
V(r) = V(a) - kQ((1/a)-(1/r)) = 0 - kQ((1/a)-(1/r)) = kQ((1/r)-(1/a)).

I believe this is what the answer should be. Please, can anyone tell me if they agree?
Yes, your result is correct.

ehild
 

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