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Work done to move a charge

  1. Mar 12, 2008 #1
    a test charge of +1.0x10^-6C is 40cm from a charged sphere of 3.2x10^-3C. A) how much work was required to move it therer from a point 1.0x10^2m away from the sphere? b) how many electrons were gained or lost from the test object to creat the charge?. I tried but i do not get the right answer:(. what i did was..used Ee1=kq1q2/r1 (r1 as 0.4m), Ee2=kq1q2/r2(r2 as 100m), then found delta Ee= Ee2=Ee1, Delta Ee= W??
     
  2. jcsd
  3. Mar 12, 2008 #2
    Work is defined as F*ds. Look at the equations you used and the units they give.
     
  4. Mar 12, 2008 #3
    In the book there is excatly same question except it is moved 100cm instead of 100m. And the answer they got was 43 J. i tired all possible ways....i don;t get the right answer. if i use the above equation, delta Ee=E2-E1, it gives me 43..( but i don;t think thats the right approch here:S)
     
  5. Mar 12, 2008 #4
    I think you need another equation to solve for this. First, try to find that problem on cramster.com (they have solutions for many problems). We are also learning Electric Fields(physics 11) in my class at the moment.
     
  6. Mar 12, 2008 #5
    Don;t see that problem in cramster.com:(, thanks for the web though.
     
  7. Mar 12, 2008 #6

    Doc Al

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    That sounds right. You need to find the change in electric potential energy as the charges are brought together.
     
  8. Mar 12, 2008 #7
    So,
    q1=3.2x10^-3C
    q2=1x10^-6C
    r1=0.4m
    r2=100m

    Solution
    E1= 72J (E1= (kq1q2)/r1)
    E2= 0.288J (E2=kq1q2/r2)

    delta E= E2-E1
    delta E= 0.288J - 180J
    delta E=-179.712J
    Delta E= work done...(what about the negative value??)
     
  9. Mar 12, 2008 #8

    Doc Al

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    Looks OK. (You seemed to have switched r1 & r2: the charge moves from 100m to 0.4m.)

    Where did you get 180J from? The negative sign is due to you mixing up initial and final positions.
     
  10. Mar 12, 2008 #9
    Oh, sorry the E1 is 72J.....so its delta E= E1- E2???/ how did i mix up?
     
  11. Mar 12, 2008 #10

    Doc Al

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    Initial position: r1 = 100 m.
    Final position: r2 = 0.4 m.
     
  12. Mar 12, 2008 #11
    right ...silly me...thanks a lot
     
  13. Mar 12, 2008 #12
    hmm. I am thinking of writing the force between the two charges and integrating it over the distance.
     
  14. Mar 13, 2008 #13

    Doc Al

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    Staff: Mentor

    Which is how one would derive the expression for electric potential energy.
     
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