- #1
wbrigg
- 15
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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 )
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?
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 )
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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