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Homework Help: Potential Problems

  1. Jan 4, 2010 #1
    1. The problem statement, all variables and given/known data

    2. The potential at the center of a circle of radius 90 cm with three charges of +1 nC, -2 nC, and +2 nC placed 120 degrees apart on the circumference is:
    a) 8.0 V
    b) 12
    c) 16
    d) 9
    Answer: d
    6. The electrostatic energy stored in the electric field around a conducting sphere of radius R carrying a net charge of Q is:

    a)[tex] kQ^{2} / 2R [/tex]
    b)[tex] kQ^{2} / 4R [/tex]
    c)[tex] kQ^{2} / 2R^{2} [/tex]
    d) [tex]kQ^{2} / 4R^{3} [/tex]
    Answer: a

    2. Relevant equations

    3. The attempt at a solution

    2) [tex] V = \frac{k}{0.9}*10^{-9}*(1-2+2) = 10 V[/tex]
    That obviously doesn't agree with the answer.

    6) I was thinking of doing U = qV, and then using kq/r for the potential of a sphere, except then I don't have a point charge to multiply by. If i multiply by q, I just get kq^2 / r, which isn't the right answer.
  2. jcsd
  3. Jan 4, 2010 #2


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    Homework Helper

    For the first one, I'm not sure what's going on. My calculations seem to agree with yours. Maybe it's one of those "pick the closest answer" things? (Or maybe whoever made up the question is really bad at rounding)

    For the second one, [itex]U = qV[/itex] applies to the case where you're moving a single point charge (q) through an existing potential difference. That's not really what you're dealing with here. However, you could consider assembling this sphere of charge one bit at a time. Roughly speaking, the first bit q takes no energy (because there's no existing potential), the second bit takes [itex]U = qV = q\cdot kq/r[/itex], the third bit takes [itex]U = qV = q\cdot 2kq/r[/itex], etc. If you add up all these energy contributions and extrapolate to the total amount required to build up a charge Q, you'll find how much energy is involved in creating the sphere, which is the same amount of energy stored in its electric field (since there's nowhere else for that energy to go).
  4. Jan 4, 2010 #3
    So I'm guessing this will involve an integral of sorts. Would I integrate the little bits of potential energy? So like the integral of V * q?
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