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Concentric conducting spherical shells

  1. Jan 8, 2014 #1
    1. The problem statement, all variables and given/known data

    Consider three concentric conducting shells, with potentials [tex] 0, ø_0, 0 [/tex] and radius [tex] a, b ,c [/tex] where [tex]a < b < c[/tex].

    (a)State conditions for Laplace to work and boundary conditions for E

    (b)Show ø is of the form:

    (c) Find ø and E everywhere.

    (d) Find the charge density and electrostatic energy.

    29bey38.png


    2. Relevant equations



    3. The attempt at a solution

    Part (a)

    Condition for laplace: Potential at surface of shells must be constant
    Boundary Conditions for E: As [tex] r → \infty |\vec {E}| → 0 [/tex]

    Part (b)
    29vxs8n.png

    Consider [tex] r → \infty , V → const.[/tex]

    [tex] \sum_{l=0}^{\infty} (A_l r^l)P_{l cosθ} → const. [/tex]

    This implies that [tex] l = 0 [/tex] is the only solution.

    Thus, general solution for ø:
    [tex]ø = A + \frac {B}{r} [/tex]

    Part (c)
    1. ø must be continuous at r= a, so [tex]ø_{in} = ø_{out} [/tex] at r=a.
    2. Potential at r = a is 0.
    We obtain two simultaneous equations:
    [tex] A = \frac {B}{a} [/tex]
    [tex] 0 = A + \frac {B}{a} [/tex]
    So this implies [tex] A = B = 0 [/tex] for the first shell.

    Middle shell
    At r = b, [tex] ø_0 = A + \frac {B}{a}[/tex]
    Continuity: [tex]ø_{in} = ø_{out}
    so [tex]Ab = B [/tex]
    Solving, we get [tex]ø_{in} = \frac {ø_0}{2}, ø_{out} = \frac{bø_0}{2r} [/tex]

    Outermost Shell
    Does nothing, as shown above.

    Using superposition principle, so far we have:
    [tex] ø = \frac {ø_0}{2} (0 < r <b) [/tex]
    [tex] ø = \frac {ø_0b}{2r} (r > b) [/tex]

    Using [tex] E = - \nabla ø [/tex]

    [tex] E = 0 (0 < r < b) [/tex]
    [tex] E = \frac {bø_0}{2r^2} (r > b)[/tex]

    Graphs
    212cw15.png

    Part(d)

    At [tex] r = b, \vec{E} = -\nabla V [/tex]
    [tex] (\frac {\partial {V_{out}}}{\partial {r}} - \frac{\partial {V_{in}}}{\partial {r}}) = -\frac {σ_b}{ε_0} [/tex]
    [tex] (\frac {-bø_0}{2b^2}) = -\frac {σ}{ε_0} [/tex]
    [tex] σ_b = \frac {ø_0ε_0}{2b} [/tex]

    For r = a,
    [tex] (\frac {\partial {V_{a,out}+V_{b,in}}}{\partial r} - \frac {\partial {V_{a,out}+V_{b,in}}}{\partial r}) = 0 [/tex]
    So [tex] σ_a = 0 [/tex]

    [tex] Energy = \frac{1}{2} ε_0 \int_{a}^{c} E^2 dV = \frac {1}{2} ε_0 \int_b^c \frac {bø_0}{2r^2} dr = \frac {1}{4}ε_0bø_0(\frac {c-b}{bc}) [/tex]
     
    Last edited: Jan 8, 2014
  2. jcsd
  3. Jan 9, 2014 #2
  4. Jan 11, 2014 #3
    bumppp
     
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