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: 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 &amp;lt; r &amp;lt;b)<br /> ø = \frac {ø_0b}{2r} (r &amp;gt; b)<br /> <br /> Using E = - \nabla ø<br /> <br /> E = 0 (0 &amp;lt; r &amp;lt; b)<br /> E = \frac {bø_0}{2r^2} (r &amp;gt; 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-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})

 

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