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Homework Help: Electric potential of concentric spheres

  1. May 12, 2012 #1
    1. The problem statement, all variables and given/known data
    attached image
    http://sphotos.xx.fbcdn.net/hphotos-snc7/148934_353406858048194_100001366491698_897020_688558166_n.jpg [Broken]

    2. Relevant equations
    V = integral of E * dr

    3. The attempt at a solution
    I do not completely understand the solution to part B. I was able to solve it with the use of hints.

    My guess to the explanation is: the electric field is 0 in the hollow conducting sphere/shell and thus the potential should at the inner surface and the outer surface should be equal.

    However, I do not know why the potential is kq/c and not kq/b.

    Also, I do not know the solution to part C (the hint says it involves both variables a and b).

    My attempt at part C was kq/a
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 12, 2012 #2
    Since the electric field for b<r<c is zero, charge is uniformly distributed on the outer part of the sphere. So the potential for points B and C(and anything in between) should be....?

    Edit: I thought this would be a better way to solve. Do you know Shell Theorem? What does that imply on the potential inside the sphere?
    Last edited: May 12, 2012
  4. May 12, 2012 #3
    Since E=-dV/dr, and E inside the sphere is zero, therefore -dV/dr is zero and therefore V=constant as the rate of change of potential is zero. Hence, the potential inside the sphere is equal to the potential at the surface of sphere.
  5. May 12, 2012 #4


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    Homework Helper

    You missed a minus sign, and do not forget that the integration involves an additional constant.
    If E=kq/r2 the potential is


    It is very important to note that the potential is continuous. It does not jump at an interface. You need two find the constant C for each domain.

    The potential is zero at infinity. That means C=0 for r≥c, V=kq/r for r≥c,so it is V(c)=kq/c at the outer surface of the hollow sphere.

    The electric field is zero inside a conductor so the potential is constant.
    The potential is continuous, it is the same at the inner side of the outer surface as outside: V=kq/c, and stays the same in the whole shell, even at radius b: V(b)=kq/c.

    The potential in the empty space between r=a and r=b is of the form V(r)=kq/r+C' again, with a different constant as in the domain r>c. And the continuity requires that V(b)=kq/b+C'=kq/c at r=b. Find C', and then calculate V(a).

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