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My topic full of GRE practice questions, whoo

  1. Nov 2, 2006 #1
    First up, thin noncunducting ring with uniformly distributed charge Q on it, what's the potential a distance x from it, it's located on the axis of symmetry(ie on the axis passing through the center of the ring)

    I really really really thought you could treat that ring as a point charge but apparently I'm wrrrrooooong

    second, if you have a particle of mass m incident on nonmoving particle also of mass m with an initial momentum mc/2, what was its initial speed? I found that(after hours of forgetting the lorentz factor is (blah)^NEGATIVE1/2 oops)

    after the collision you have two particles of mass m', one aimed 30 degrees above horizontal, one 30 degrees below...what's the momentum of each particle? hurrr? I tried to approach it the same way I did above(E^2=p^2c^2+mc^2 find E knowing p, E=gamma*mc^2, so for this problem I tried solving for p knowing E has to be the same as the previous, and it doesn't work. Primary problem is the answer doesn't involve m', hmm)
     
  2. jcsd
  3. Nov 2, 2006 #2
    The first part:
    [tex] V=\frac{1}{4 \pi \varepsilon} \int \frac{dq}{r} =\frac{1}{4 \pi \varepsilon r} \int dq = \frac{Q}{4 \pi \varepsilon r}[/tex]

    In this case [tex] r=\sqrt{R^2 + x^2}[/tex].

    BTW, good luck Saturday :)
     
    Last edited: Nov 2, 2006
  4. Nov 2, 2006 #3
    Thanks for the first one. It has to be spherically distributed to treat it like a point particle, right?
     
  5. Nov 2, 2006 #4
    Here's a killer that's bugging me, looks so easy:(

    You've got a thin ring of radius b of surface charge density lambda, at P1 located a distance b away(it's all on the axis through the center of the ring)it has a potential of V1, at a distance 2b out, potential V2

    what's the potential V2 in terms of V1?

    it's not 1/2*V1, which is what I got, it's like sqrt(wtf)V1
     
  6. Nov 3, 2006 #5

    berkeman

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    Staff: Mentor

    Kinda hard to help you without seeing all your work. Hep us hep you.


    "it's like sqrt(wtf)V1" EDIT -- Ooops, that helps a lot.
     
  7. Nov 3, 2006 #6
    I couldn't possibly imagine how you get anything besides 1/2 V1

    potential falls of like 1/r, so yah, double the distance, half the potential?
     
  8. Nov 4, 2006 #7
    go read a few replies up a bit and look at the dependence on x, although the general dependence is like 1/x, since it is under the square root, the transformation x -> 2x will not simply divide the potential in half, plug that change into the equation and then do some algebra so you can factor out the original potential to get the new one in terms of V_1.

    Just realized this is probably too late to help you since we took the GRE's this morning.
     
  9. Nov 5, 2006 #8
    Yes, I realized that I was treating a ring like a sphere just in time for the GRE

    boy that was a fun Saturday morning
    thanks guys!
     
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