Concentric conducting spherical shells

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Homework Statement

Consider three concentric conducting shells, with potentials $$0, ø_0, 0$$ and radius $$a, b ,c$$ where $$a < b < c$$.

(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 $$r → \infty |\vec {E}| → 0$$

Part (b)

Consider $$r → \infty , V → const.$$

$$\sum_{l=0}^{\infty} (A_l r^l)P_{l cosθ} → const.$$

This implies that $$l = 0$$ is the only solution.

Thus, general solution for ø:
$$ø = A + \frac {B}{r}$$

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

Middle shell
At r = b, $$ø_0 = A + \frac {B}{a}$$
Continuity: [tex]ø_{in} = ø_{out} <br /> so $$Ab = B$$<br /> Solving, we get $$ø_{in} = \frac {ø_0}{2}, ø_{out} = \frac{bø_0}{2r}$$<br /> <br /> <b>Outermost Shell</b><br /> Does nothing, as shown above.<br /> <br /> Using superposition principle, so far we have:<br /> $$ø = \frac {ø_0}{2} (0 < r <b)$$<br /> $$ø = \frac {ø_0b}{2r} (r > b)$$<br /> <br /> Using $$E = - \nabla ø$$<br /> <br /> $$E = 0 (0 < r < b)$$<br /> $$E = \frac {bø_0}{2r^2} (r > b)$$<br /> <br /> <b><u>Graphs</u></b><br /> <div class="bbImageWrapper js-lbImage" title="212cw15.png" data-src="https://www.physicsforums.com/attachments/212cw15-png.166574/" data-type="image" data-lb-sidebar-href="" data-lb-caption-extra-html="" data-single-image="1" > <img src="https://www.physicsforums.com/attachments/212cw15-png.166574/" data-url="" class="bbImage" data-zoom-target="1" style="" alt="212cw15.png" title="212cw15.png" width="712" height="321" loading="lazy" decoding="async" /> </div><br /> <br /> <u><b>Part(d)</b></u><br /> <br /> At $$r = b, \vec{E} = -\nabla V$$<br /> $$(\frac {\partial {V_{out}}}{\partial {r}} - \frac{\partial {V_{in}}}{\partial {r}}) = -\frac {σ_b}{ε_0}$$<br /> $$(\frac {-bø_0}{2b^2}) = -\frac {σ}{ε_0}$$<br /> $$σ_b = \frac {ø_0ε_0}{2b}$$<br /> <br /> For r = a,<br /> $$(\frac {\partial {V_{a,out}+V_{b,in}}}{\partial r} - \frac {\partial {V_{a,out}+V_{b,in}}}{\partial r}) = 0$$<br /> So $$σ_a = 0$$<br /> <br /> $$Energy = \frac{1}{2} ε_0 \int_{a}^{c} E^2 dV = \frac {1}{2} ε_0 \int_b^c \frac {bø_0}{2r^2} dr = \frac {1}{4}ε_0bø_0(\frac {c-b}{bc})$$[/tex]

 

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