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Electric Potential Questions

  1. Jun 13, 2006 #1
    for the love of god and all that is holy, i can not figure out how to set up the equations. any help is greatly appreciated

    1.
    in experiments in which atomic nuclei collide, head-on collisions like that described in problem 23.80 (i'll explain in a sec) do happen, but near misses are more common, suppose the alpha particle in problem 23.80 was not aimed at the center of the lead nucleus but had an initial nonzero angular momentum (wrt the stationary lead nucleus) of magnitude L = p-sub-oh *times* b where p-sub-oh is the magnitude of the initial momentum of the alpha particle and b = 1.00x10^-12 m. what is the distance of the closest approach?

    23.80:
    an alha particle with kinetic energy 11.0 MeV makes a head on collision with a lead nucleus at rest. what is the distance of closest approach of teh two particles? assume that the lead nucleus remains stationary and that it may be treated as a point charge


    2.
    a hollow, thin walled insulating cylinder of radius R and length L (think cardboard tube in a roll of toilet paper) has charge Q uniformly distributed over its surface. calculate the electric potential at all points along hte axis of hte tube


    i have spent many afternoons pondering these questions and would be greatly relieved if someone could answer them and release me from this curse

    thanks
     
  2. jcsd
  3. Jun 13, 2006 #2
    for 1, i was trying to set up the potential energy = kinetic energy (given)

    and then the potential energy could be from the electric force the psitive charges exert on each other when in contace (at a distance of zero???)


    2, i tried to set up an integral for dq/dr where r was some variation of sqrt(R^2 + (x-l)^2) where i would integrate l from 0 to L, but that didn't work. also i thought that the electric field had to be perpendicular to the surface which would then mean that there wouldn't be a field in the direction along the axis of the cylinder...
     
  4. Jun 13, 2006 #3

    Hootenanny

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    Have you taken into account the intial potential energy the alpha particle with have?
    The electric field is perpendicular tot he surface, the question is asking for the value of the electric field spanning the whole length of the cylinder. Firstly what is Gauss's law? What is the area of the curved surface of the cylinder?
     
  5. Jun 15, 2006 #4
    i'm assuming that the initial potential energy must be zero. but i see that if i knew the initial potential energy, i could calculate the distance because the 'r' in U=(q*q-sub-oh)/(4*pi*epsilon-oh*r) refers to the distance between the charged particles. i feel like i'm missing a huge bit of information...

    from gauss's law, E*A=Q/epsilon-oh

    for E is the electric field that traverses the surface area A of which has an enclosed charge of Q

    thus, E=Q/(epsilon-oh*A)

    but can i still integrate this, even though i don't see how the field from the cylinder would affect a point that that lies on it's axis. like i can visualize the E from a charged ring extending in the direction of a point that lies on it's axis. i can't visualize the same for a cylinder. hence my hesitation and confusion
     
  6. Jun 15, 2006 #5

    Physics Monkey

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    Hi 93157,

    Problem 1 is a simple application of conservation laws. You know the initial energy and the initial angular momentum of the alpha particle, and in so far as it moves in a static central force field, its energy and angular momentum are conserved. I would suggest a two stage attack. First, write the energy in terms of the angular momentum, the radial velocity, and the radial coordinate r. Second, solve for r using energy conservation at the point of closest approach. Hint: what is the radial velocity at such a point?

    Since problem 2 is a calculation of the potential, it should boil down to doing an integral. Are you having trouble setting up the integral or doing the integral?
     
  7. Jun 16, 2006 #6
    um quick question: because the alpha particle has a nonzero angular momentum, does that mean it is moving in a helix-esque spiral? or does it simply move along an arc of radius b (length from particle to pivot)?

    i'm also not quite sure on how to set up the energy equations. although the lead nucleus radiates an uniform electric field, the potential energy increases exponentially as the integration:

    intgr(q*q-sub-oh/(4*pi*epsilon-sub-oh*r^2),r,D,0)

    plus, wouldn't the potential energy between the two positive particles be infinity at a distance of zero?

    it's all so mind-boggling-confusing :bugeye:
     
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