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Crying for help

  1. Apr 2, 2007 #1
    1. The problem statement, all variables and given/known data

    A +30uC charge is placed 40cm away from an identical +30uC charge. How much work would be required to move a +0.5uC test charge from a midway between them to a point 10cm clover to either of the charges?

    2. Relevant equations

    E=V/d, W=qV, k=8.99*10^9

    3. The attempt at a solution

    well...

    W=qV=(0.5*10^(-6)){(30*10^(-6)k/0.1)-(30*10^(-6)k/0.2)}=1.32

    the answer should be 0.45J.....

    Please, anyone who understands this question, explain me how to get the answer!!!!!
     
  2. jcsd
  3. Apr 2, 2007 #2

    hage567

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    Homework Helper



    There shouldn't be a negative sign there, so that part of your answer is wrong.

    I think you need to find the change in potential energy from when the test charge is sitting between the two +30uC charges, to when it is moved 10cm over. That is the work done in moving it. So if you fix the calculation above, you are half done.
     
  4. Apr 3, 2007 #3
    I tried with
    W=qV=(0.5*10^(-6)){(30*10^(-6)k/0.1)(30*10^(-6)k/0.2)}
    but the answer was then too big.....

    I then assumed that the potential energy is 0 when the test charge is sitting between the two +30uC charges (as it's balanced).
    And the energy when it is moved 10cm over is:

    force to the left=(0.5*10^(-6))(30*10^(-6))k/0.1^2=13.485
    force to the right=(0.5*10^(-6))(30*10^(-6))k/0.3^2=1.498

    using E=F/q,
    E=(13.485-1.498)/(0.5*10^(-6))=11.987

    therefore, W=qEd=0.5*10^(-6)*11.987*0.1=1.198

    which is not the right answer....
    can anyone suggest me the correct way to go???
     
  5. Apr 3, 2007 #4

    Mentz114

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    Gold Member

    The PE of the test charge sitting between the others is just twice what you'd get if there was only one 20cm away.

    After moving, you are 10cm from one charge and 30cm from the other. Find the difference between the initial and final energies.
     
  6. Apr 3, 2007 #5
    So the initial PE is

    net initial force =2*((0.5*10^(-6))(30*10^(-6))k/0.2^2 =6.7425
    PE=F/q=6.7425/(0.5*10^(-6))=13485000

    W=qEd=(0.5*10^(-6))*13485000*0.2=1.3485


    net final force =((0.5*10^(-6))(30*10^(-6))k/0.1^2 +((0.5*10^(-6))(30*10^(-6))k/0.3^2=14.9833
    PE=F/q=14.9833/(0.5*10^(-6))=29966666

    W=qEd=(0.5*10^(-6))*29966666*0.1=

    W required=1.4983-1.3485=0.1498

    I still cannot get the right answer......
    Where did I fall into a trap???
     
  7. Apr 3, 2007 #6

    Mentz114

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    Gold Member

    Well, you can't be faulted for effort. You're doing it the hard way.
    You don't need to calculate the forces, because PE = k.q.q/r.

    So,

    initial energy=2*((0.5*10^(-6))(30*10^(-6))k/0.2 = ??
    final energy=((0.5*10^(-6))(30*10^(-6))k/0.1 +((0.5*10^(-6))(30*10^(-6))k/0.3=??

    and

    work = final energy - initial energy
     
    Last edited: Apr 3, 2007
  8. Apr 3, 2007 #7
    Oh, I finally got the right answer, following Mr/Ms Mentz114's way! Thank you so much!!!

    But...I still can't understand why the energy in both directions are added together for calculating the potential energy. Aren't the charges at right and left end both positive? Then don't they exert equal and opposite force on the test charge in the midway?
     
  9. Apr 3, 2007 #8

    Mentz114

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    Gold Member

    I'm glad to hear it.

    Your question is a good one. If we moved the test charge to the left, the big charge on the right is helping us, so we are doing negative work wrt to it. But the big charge on the right opposes us, so we do positive work against it.
    There are two terms in the final energy, and if we compare with the two (equal) terms in the initial energy, one as gone up, and the other down.
     
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