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Electrostatics - find the work done

  1. Apr 18, 2013 #1
    A certain charge 'Q' is to be divided into two parts, q and Q-q. What is the relationship of 'Q' to 'q' if the two parts, placed at a given distance 'r' apart, are to have maximum Coulombic repulsion? What is the work done in reducing the distance between them to half its value?

    It went easy for the first part
    as f(q) = k(q)(Q-q)/r2
    and, f'(q) = 0 gives q = Q/2 which provides max repulsion (as f''(q) < 0)

    But it is getting a bit "confusing" to calculate the work done for the second part..

    someone please provide the idea.
     
  2. jcsd
  3. Apr 18, 2013 #2
    What is the relationship between potential energy and work?
     
  4. Apr 18, 2013 #3
    God.. i m kind of "fool"...
    i actually got the answer.. but in the book its given in the form of 'Q' and i got it in terms of 'q'
    but could not check the difference that time... got it thanks...
     
  5. Apr 18, 2013 #4
    one more thing at last
    W = -(U2 - U1) = U1 - U2
    So in this question
    applying W = (U1 - U2) gives me a negative answer
    but in the book its +ve
    what of that??? :P
     
  6. Apr 18, 2013 #5

    haruspex

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    That would be the work done by the system in going from U1 to U2. You want the work done on the system.
     
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