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Homework Help: Classic Hemispherical Analyzer

  1. Dec 30, 2012 #1
    1. The problem statement, all variables and given/known data


    I remember learning about this in Chemistry but boy I did not expect to actually prove it works haha. I've actually solved the problem but I feel a little uncomfortable with certain aspects.

    2. Relevant equations

    E = [itex]\Delta[/itex]Vq

    For uniform circular motion:
    a = v^2/r

    For two concentric cylinders with equal but opposite charge, the potential difference can be calculated out via Gaussian surface to be:

    2kQ/L * ln(b/a) where L is the length of the cylinder.

    3. The attempt at a solution

    First, I checked the potential difference and found it to be 2V * ln(b/a) which immediately tells me a lot about the system. Therefore, kQ/L = V (units seem to work out).

    Conservation of Energy:

    Vq = 0.5mv^2 --> v^2 = 2Vq/m

    Using Gaussian surfaces, E(r) = 2kQ/Lr = 2V/r.
    F(r) = E(r)q = 2Vq/r
    a(r) = 2Vq/mr = v^2/r which is the equation of motion for uniform circular motion of radius r.

    What's bothering me is that the reason I was able to solve this problem is because I'm *very* familiar with the coaxial cable conductor and quickly recognized it.

    Provided the inner and the outer conductor did not have equal but opposite charges, would the potential still be the same or would it somehow be different? I think I solved this problem operating under the assumption they do but is it necessary to make that assumption?
    Last edited: Dec 30, 2012
  2. jcsd
  3. Dec 31, 2012 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    This E is energy but later you used E for electric field ... ? Anyway, energy is not relevant here.
    That is correct.
    What the ...? What is your V? It's not the potential difference between the two shells and it's not V0 either ... I am confused ...
    This V must be V0.
    Again, I don't know what your V is.

    At this point I give up ... you need to get your symbols straightened out.

    I think you're on the right track, though. Get the E field as E(Q,r), change that to E(V,r) where V is the pot, diff. betw. the two shells. V is of course -∫abE(r)dr.

    Then get E(V,r0), equate qE(V,r0) to mv2/r0, realize (as you kind of did) that qV0 = mv2/2, and solve for V.

    Note that the given answer, 2V0*ln(b/r0) - 2V0*ln(a/r0) can be more simply written as 2V0*ln(b/a).
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