Find the energy of a uniformly charged spherical shell of total charge q and radius R.

Question

Accepted Answer

1

Use Eq. 2.43, in the version appropriate to surface charges:

[latex]W=\frac{1}{2}\int{ ho Vd au }.[/latex] (2.43)

[latex]W=\frac{1}{2}\int{\sigma Vd a}.[/latex]

Now, the potential at the surface of this sphere is [latex](1/4\pi \epsilon _{0})q/R[/latex] (a constant—Ex. 2.7), so

[latex]W=\frac{1}{8\pi \epsilon _{0}} \frac{q}{R}\int{\sigma da} = \frac{1}{8\pi \epsilon _{0}} \frac{q^{2}}{R}[/latex]

2

Use Eq. 2.45. Inside the sphere, E = 0; outside,

[latex]W=\frac{\epsilon _{0}}{2}\int{E^{2}d au } [/latex] (all space). (2.45)

[latex]E=\frac{1}{4\pi \epsilon _{0}}\frac{q}{r^{2}}\hat{r}, [/latex] so [latex]E^{2}=\frac{q^{2}}{\left(4\pi \epsilon _{0} ight)^{2}r^{4} }, [/latex]

Therefore,

[latex]W_{tot}=\frac{\epsilon _{0}}{2\left(4\pi \epsilon _{0} ight)^{2} }\int\limits_{outside}^{}{} {\left(\frac{q^{2}}{r^{4}} ight)\left(r^{2}\sin heta drd heta d\phi ight) } [/latex]

[latex]=\frac{1}{32\pi ^{2}\epsilon _{0}}q^{2}4\pi \int_{R}^{\infty }{\frac{1}{r^{2}}dr }=\frac{1}{8\pi \epsilon _{0}}\frac{q^{2}}{R} [/latex]

Question 2.9: Find the energy of a uniformly charged spherical shell of to...

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