A capacitor consisting of two concentric spheres of radius R1 and R2 = 2.50·R1 has a capacitance of C = 6.00 picoFarads and is charged to a potential difference of 74.0 Volts.

  1. Calculate the energy stored in the capacitor.

  2. Calculate the charge on this capacitor, when the electrical energy stored is 19.9·10-8 Joules.

  3. If the radius of the outer sphere is increased by a factor of 3.00 while keeping the charge on the plates constant, by what factor does the stored energy change?

I've already answered #1 and #2 Questions, but I don't know how to calculate the third one. Help please!

  • 1
    \$\begingroup\$ What are the formulas you found for #1? This is part of the solution for #3. \$\endgroup\$ – jippie Feb 26 '13 at 8:02
  • 2
    \$\begingroup\$ This sounds homeworkesque \$\endgroup\$ – Grady Player Feb 26 '13 at 8:09
  • \$\begingroup\$ If this is homework, you better mention this in your question. \$\endgroup\$ – Johan.A Feb 26 '13 at 8:48
  • \$\begingroup\$ There is nothing wrong with asking a homework question, generally users are expected to show some work on their homework, but this user seems to have calculated 1 and 2. \$\endgroup\$ – Kortuk Feb 26 '13 at 16:28

What you missing for part C is the formulae for the capacitance of two concentric spheres: $$ C= \frac{4\pi\epsilon}{1/a - 1/b}$$ with $$ b>a $$ From that you should be able to determine the ratio of C wrt the change in R


Energy stored in a capacitor is: \$U = \frac{1}{2}\cdot\frac{Q^2}{C}\$

You know the amount of charge.

What effect would tripling the radius of the outer "plate" of the capacitor have on the structure's capacitance?

  • \$\begingroup\$ @Kortuk - If he has calculated #1 he can calculate #3, it's a bit lazy... and the fact it's obviously "please do my homework for me" makes it doubly so. \$\endgroup\$ – John U Feb 26 '13 at 13:27
  • \$\begingroup\$ My point, which I am worrying I am not making clear, is the user knows how to solve the first 2 parts but not the third, and this just reads as restating it. However, I remember when I have had to restate a question before to have a student understand it, so I will remove the -1. \$\endgroup\$ – Kortuk Feb 27 '13 at 1:25

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