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Parallel Capacitor

  1. Apr 5, 2010 #1
    1. The problem statement, all variables and given/known data
    A parallel plate capacitor has two conducting plates seperated by a vacuum. The distance is D and the area of each plate is A.
    An alpha particle with mass M and charge Q is placed on the positively-charged plate, between the plates. It shoots through a small hole in the negatively charged place with speed S.
    What is the magnitude of the uniform electric field between the plates?
    How much charge lies on the positively-charged plate?
    How much energy is stored in the capacitor's electric field?


    2. Relevant equations
    C=Q/V
    V=Ed
    C=(A/d)(epsilon_o)


    3. The attempt at a solution

    For the first question, I can find the capacitence, C=(epsilon_0)(A/D), but I'm not sure where to go from here.
    For the second question, I could find Q using C=Q/V and use V=Ed but I need E from the first question that I don't know how to find!

    I'm not sure where the alpha-particle comes in to play!?

    Thank you!
     
  2. jcsd
  3. Apr 5, 2010 #2
    You are very right about the fact that E is needed. All of your logic is great up to that point. So with problems like these, take a step back and ask; How could I possibly find E (or V).

    The first thing you should do is review or find out what an alpha particle is. An alpha particle is a helium atom that has been stripped of its electrons. This means that it has a 2+ charge.

    Now we know when this charged particle is placed into the capacitor, it will experience a force from the electric field and start to gain velocity. The problem tells us how fast it is moving when it reaches the other side. Try and find a way to relate the velocity obtained by the particle to the electric field or voltage between the plates. *Cough!*Use Energy*Cough!*
     
  4. Apr 5, 2010 #3
    Thank you Hellabyte! :)

    If I find V, then I know how to find E.
    So, V=Work/charge
    V=(0.5mv^2)/q
    thus V=(0.5MS^2)/Q

    Is this correct? I think it is, and from here I know how to do all the problem...assuming it is correct!?
     
  5. Apr 5, 2010 #4
    Yes exactly. Remember(or learn) that qV is the kinetic energy gained by a charged particle that goes through a potential difference of V. Try and draw some parallels to it and the gravity potential-kinetic energy problems that you probably remember so fondly, they're completely analogous.
     
  6. Apr 5, 2010 #5
    Thank you so so so much! What great help!
     
  7. Apr 5, 2010 #6
    No problem :)
     
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