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Electric potential difference in a charge

  1. Oct 22, 2015 #1
    So I have two questions relating to potential difference

    1.An electron with a velocity of 5.0x10^6 m/s injected into a parallel plate apparatus through a hole in the positive plate. it moves across the vacuum between the plates, colliding with negative plate at 1.0x10^6 m/s. what is the potential difference between the plates? (mass of electron = 9.1x10^-31 kg)

    2.Two alpha particles (mass 6.6x 10^.27 kg. charge = 3.2 x 10^ -19c) separated by an enormous distance, apporach each other along a "head on collision" path. Each has a speed of 3.0x10^-19m/s to begin with. Calculate their minimum separation, assuming no deflection from their original path.

    I tried with 0.5mv^2 = Eqd for the first question, but there is no charge or distance given. I saw another way where is 0.5mv^2 = qV but still there is no charge given to solve.

    For the second question I am kind of clueless..

    Somebody help ..
     
  2. jcsd
  3. Oct 23, 2015 #2

    jtbell

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    Staff: Mentor

    Surely your textbook states the charge of an electron somewhere? Or you can try Google. :oldwink:
     
  4. Oct 23, 2015 #3

    gneill

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    Hi, Welcome to Physics Forums.

    In future please use the template provided in the edit window to format your question, and post only one question per thread.

    jtbell gave good advice for the charge of the electron. It's used frequently in physics so can be considered a constant that you should be familiar with, much like g is.
    What physical quantities can you calculate with the given information?

    Can you confirm the given speed value in the problem statement? It seems to be rather small to yield a realistic result (at least for our universe :wideeyed: )
     
  5. Oct 26, 2015 #4
    so is V = (0.5(9.1x10^-31)((1.0x10^6)^2))/(1.6x10^19)
    and is there a difference between 0.5mv^2 = Eqd and 0.5mv^2 = qV
     
  6. Oct 26, 2015 #5
    The speed for the second question is actually 3.0x10^6 m/s, i put the wrong numbers.
     
  7. Oct 26, 2015 #6

    gneill

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    Okay, that looks more plausible for the question. What have you tried?
     
  8. Oct 26, 2015 #7
    I subbed it into 0.5mv^2 to find the kinetic energy
     
  9. Oct 26, 2015 #8

    gneill

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    Okay, what will you do with it?
     
  10. Oct 26, 2015 #9
    Find the potential difference ?
     
  11. Oct 26, 2015 #10

    gneill

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    What potential difference would that be? Perhaps you mean potential energy?
     
  12. Oct 26, 2015 #11
    Yes, how would you find the potential energy
     
  13. Oct 26, 2015 #12

    gneill

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    Good question. Have you checked your text or course notes? Look up electric potential energy.
     
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