Concentric conducting spherical shells

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Homework Statement

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.

Homework Equations

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)

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

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]

 

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