# Electrostatic Energy of a thin spherical shell

hey,

i have a question from an exam paper which isn't worded too nicely (most of the questions on the exam are worded in similar ways :grumpy:)

Determine the electrostatic energy U of a thin spherical shell of radius R which carries chare Q uniformly distributed over it's surface [Hint: imagine to assemble the sperical shell by superimposing sheels of radius R and infintesimal charge dq. ].

The way i've done it is to first put in my first shell of infintesimal charge and then treat it as a point charge. Then i treat the next shell as just a point, and place it at a distance R from the point (electrostatic energy of this;
$$U = \frac{n dq^{2}}{4 \pi \epsilon _{0} R}$$
where n is the number of shells already added.

so the total Electrostatic energy of the whole thing (with all the shells assembled) is going to be:

$$U=\Sigma^{n}_{i=1} \frac{i dq^{2}}{4 \pi \epsilon_{0} R}$$ Where $$n dq = Q$$

I can turn that into an integral (Replacing one of the dqs with $$\frac{q}{dq}$$) ;

$$U = \int ^{Q}_{0} \frac{q dq}{4 \pi \epsilon_{0} R} dq$$

This goes to $$U=\frac{Q^{2}}{8 \pi \epsilon_{0} R}$$

now i make it so it's shells rather than points;

$$U=\int^{2\pi}_{0}\int^{\pi}_{0}\frac{Q^{2}}{8 \pi \epsilon_{0} R} d\theta d\varphi$$

$$U=\frac{\pi Q^{2}}{4 \epsilon_0 R}$$

Could someone confirm whether or not this is correct please?

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## Answers and Replies

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Why don't you distribute your 'infinitesimal charge' equally over the spherical shell. Then, the electrinc field generated by it outside the shell is the same as when it is concentrated in the center.

The answer is U=Q^2/(8 pi epsilon R), as you obtained in the middle of your calculation. The reasoning is the same as yours. Though I can't understand what you mean by replacing a 'dq' with 'q/dq'. I think it should be replacing 'idq' with 'q'.

Just discard the last part - integrating over angles.

Defennder
Homework Helper
It should be possible to do this question by integrating the square of the magnitude of the E-field over all space, isn't it? The hint given doesn't cover this possibility. It seems easier to do this since Gauss law allows you to exploit symmetry to get E-field.

The answer is U=Q^2/(8 pi epsilon R), as you obtained in the middle of your calculation. The reasoning is the same as yours. Though I can't understand what you mean by replacing a 'dq' with 'q/dq'. I think it should be replacing 'idq' with 'q'.

Just discard the last part - integrating over angles.

Yeah, i meant replace the i with a q/dq - which is what you said. never used latex before, so it's kinda hard to skim through it and check everything's correct.

and it probably would be easier just to have integrated the field twice. :grumpy:

thanks.

So I know I'm bringing up an old one but I actually have this exact same problem, the only difference is its surrounded by a vaccume. What changes will that cause to the work/ answer?