A point charge q is at the center of an uncharged spherical conducting shell, of inner radius a and outer radius b. Question: How much work would it take to move the charge out to infinity (through a tiny hole drilled in the shell)? [Answer: (q^2/8πε0)(1/a − 1/b).]

Question

A point charge q is at the center of an uncharged spherical conducting shell, of inner radius a and outer radius b. Question: How much work would it take to move the charge out to infinity (through a tiny hole drilled in the shell)? [Answer: [latex]\left.\left(q^{2} / 8 \pi \epsilon_{0}\right)(1 / a-1 / b) .\right][/latex]

Accepted Answer

The initial configuration consists of a point charge q at the center, “q induced on the inner surface, and +q on the outer surface. What is the energy of this configuration? Imagine assembling it piece-bypiece. First bring in q and place it at the origin—this takes no work. Now bring in “q and spread it over the surface at a—using the method in Prob. 2.35, this takes work $-q^{2} /\left(8 \pi \epsilon_{0} a\right)$ . Finally, bring in +q and spread it over the surface at b—this costs $q^{2} /\left(8 \pi \epsilon_{0} b\right)$ . Thus the energy of the initial configuration is

$W_{i}=-\frac{q^{2}}{8 \pi \epsilon_{0}}\left(\frac{1}{a}-\frac{1}{b}\right)$ .

The final configuration is a neutral shell and a distant point charge—the energy is zero. Thus the work necessary to go from the initial to the final state is

$W=W_{f}-W_{i}=\frac{q^{2}}{8 \pi \epsilon_{0}}\left(\frac{1}{a}-\frac{1}{b}\right)$ .

[latex]W_{i}=-\frac{q^{2}}{8 \pi \epsilon_{0}}\left(\frac{1}{a}-\frac{1}{b}\right)[/latex] .

The final configuration is a neutral shell and a distant point charge—the energy is zero. Thus the work necessary to go from the initial to the final state is

[latex]W=W_{f}-W_{i}=\frac{q^{2}}{8 \pi \epsilon_{0}}\left(\frac{1}{a}-\frac{1}{b}\right)[/latex].

Question 2.60: A point charge q is at the center of an uncharged spherical ...

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