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Application of Couloumb's Law

  1. Apr 26, 2012 #1
    1. The problem statement, all variables and given/known data
    Two charges of masses m and 2m, charges 2q and q respectively are placed in a uniform electric field and are allowed to move at exactly the same time. Find the ratio of their kinetic energies.


    2. Relevant equations
    Field = qE
    Force =(Kq1q2)/r^2
    Kinetic energy=1/2mv^2


    3. The attempt at a solution
    I found their initial accelerations by finding the Coloumb force on each and then dividing by their respective masses. Then I used the equation v= u + at and found the kinetic energies. But I'm getting the ratio as 2:1 whereas the answer is 8:1.
     
  2. jcsd
  3. Apr 26, 2012 #2

    gneill

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    It is entirely possible that the answer key is incorrect in this instance, and that you've found a correct result. Sometimes problems have values changed in order to make it a "new" problem, but updating the answer key is overlooked.

    Why don't you try another approach to the problem and see if your result is the same? Hint: Perhaps use a conservation law?
     
  4. Apr 26, 2012 #3
    I'm afraid I'm not getting it. Conservation of momentum gives 2:1, and for conserving energy, I don't know the distance between the charges to calculate potential energy.
     
  5. Apr 26, 2012 #4

    gneill

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    Well, 2:1 confirms your earlier value, so it's strong evidence that the answer key is wrong and that your result is correct :smile:
     
  6. Apr 26, 2012 #5
    Ah. Well then, thank you. :)
     
  7. Apr 26, 2012 #6

    vela

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    You're just misinterpreting the problem. You're assuming the force on each is due to the interaction of the two charges, but the problem apparently wants you to ignore that. Instead, the force on each is due to the uniform electric field imposed externally.
     
  8. Apr 26, 2012 #7

    gneill

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    Huzzah! Well spotted Vela. I missed the uniform electric field bit in the problem statement and then assumed that it was just the two particles interacting. This changes the complexion of the problem significantly!
     
  9. Apr 27, 2012 #8
    But we don't know the magnitude of the electric field..
     
  10. Apr 27, 2012 #9

    gneill

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    You don't know the magnitude of q either, but that didn't stop you before :smile:
     
  11. Apr 27, 2012 #10
    So be it. If I use F = qE, and acceleration a = F/m, I get 8:1 since the velocities are proportional to a^2. But if we had to take the inter-particular forces into consideration, I couldn't use that formula, since the repulsive force would vary with intervening distance.
     
  12. Apr 27, 2012 #11

    gneill

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    Consider each of the particles separately interacting with the field. Remember, they have different masses and charges.
     
  13. Apr 27, 2012 #12

    vela

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    I think the question is just poorly written. Without more information, you can't really deal with the interaction between the two. For example, if they're really far apart, you might argue that the interaction could be ignored. If they're close to each other, their relative displacement relative to the direction of the electric field will affect the answer.
     
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