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Potential of a grounded spherical shell inside of a metal one

  1. Feb 4, 2012 #1
    Picture:
    http://img263.imageshack.us/img263/7361/ongelma.jpg [Broken]


    1. The problem statement, all variables and given/known data
    Solve the potential energy of a charged sphere-shaped metal shell. What happens when you place a smaller, grounded sphere-shaped metal shell inside of that shell? What is the potential energy then? How about the electric potential?

    R is the radius of the bigger metal shell
    Q is the total charge of the bigger metal shell
    W is potential energy
    c=kQ/R

    2. Relevant equations
    (1) V(r)=k*∫(dQ)/|r-r'|
    Where V(r) is electric potential, k is a constant and dQ is a very small charge

    (2) W=∫F*dr
    Where W is potential energy and F is force

    (3) W=q*∫E*dr
    Where q is a charge and E is the electric field

    (4) W=q*V(r)
    -

    (5) E=-(nabla)*V(r)

    (6) W=1/2*V(r)*∫dQ

    3. The attempt at a solution

    Okay so the first one is easy. From the equation (1) it can be calculated that the electric potential inside the metal shell is constant, V=kQ/R.
    Let's say c=kQ/R

    The potential energy can be calculated from equation (4)

    This is equation (6)
    W=1/2*V(r)*∫dq

    The half is there because we are calculating the potential of the spherical shell from its own electric potential, which means charge a's potential will be calculated in comparison to charge b and charge b's potential will be calculated in comparison to charge a, so everything gets calculated twice. We don't want that.

    Now we place the grounded sphere-shaped shell with no charge inside of the metal ball.

    So we can calculate V(r) again:

    V=c, still! Because the grounded ball is neutral.

    So W=1/2*V(r)*∫dq, still!

    But here's the catch:

    If we have a sphere-shaped shell with no charge inside its electric potential is:
    V=0, this is obvious because it has no charge.


    So if we place it inside, we can calculate that when it is inside the charge is in fact not 0, but c=kQ/R.

    So what happens is that the electric potential inside changes.

    It should follow that some sort of electric field must have been created by the grounded metal shell, as stated by equation (5). So E=-(nabla)*V(r). If V(r) changes, (nabla)*V(r) should not be zero, at some point anyway.

    So now let's look at equation (3):
    W=q*∫E*dr

    Because we have electric field when the grounded metal shell moves, it MUST have done work, let's call this work W'

    So now we have contradicting results.

    W should be: (6)
    W=1/2*V(r)*∫dq = 1/2*c*Q

    But at the same time we have:

    W= 1/2*c*Q + W'

    Where W' is not zero

    So where did I go wrong?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
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