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Work done to move a test charge

  1. Mar 18, 2007 #1
    A test charge of +1.0*10^-6 C is 40cm from a charged sphere of 3.2*10^-3

    How much work is required to move it there from a point 1.0*10^2 m away from the sphere?

    GIVEN:
    q1 = +1.0*10^-6 C
    q2 = 3.2*10^-3 C
    r1 = 1.0*10^2 m
    r2 = 0.40 m

    Required:
    W

    Analysis:
    Ee=(k*q1*q2)/r
    delta Ee=E2-E1
    W=delta Ee

    Solution:

    179.99712 J

    Am I on the right track here? It just doesn't seam right to me for some reason. Also, since the charged sphere does not clearly specify that it is a positive or negative charge, should I assume it is postive?
     
  2. jcsd
  3. Mar 18, 2007 #2

    G01

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    Gold Member

    I am assuming by Ee, you mean Electrical potential energy, correct?

    If this is the case then I believe you are on the right track.


    G01
     
    Last edited: Mar 18, 2007
  4. Mar 11, 2008 #3
    Can someone help me with this question. please. i have tried all i could but i am nowhere arround the right answer. :(
     
  5. Jul 16, 2008 #4
    I don't know where the 179.99712 J came from. By my calculation k*q1*q2 is (9.0x10^9)*(1.0*10^-6)*(3.2*10^-3)=28.8 According to this Ee2 = 28.8/0.4=72 and Ee1 = 28.8/100 = 0.288
     
  6. Jul 16, 2008 #5

    Defennder

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    I got the answer as 179.75 J, which is pretty close to the OP's answer. You should post your attempt so we can check it for you.
     
  7. Dec 22, 2008 #6
    I don't think the OP's answer is correct. Defennder do you mind sharing how you got that number?

    Here's what I got

    GIVEN:
    q1 = +1.0*10^-6 C
    q2 = 3.2*10^-3 C
    r1 = 1.0*10^2 m
    r2 = 0.40 m

    E1 = (k*q1*q2) / r1
    = (9.0*10^9*3.2*10^-3*1*10^-6)/(100m)
    = 0.288 J

    E2 = (k*q1*q2) / r2
    = (9.0*10^9*3.2*10^-3*1*10^-6)/(0.4m)
    = 72 J

    W=deltaE= E2 - E1
    = 72J - 0.288J
    = 71.712J

    Therefore 71.712J of work is required.

    Can anyone comment on this please? I honestly don't get how they got a number close to 180J
     
  8. Dec 22, 2008 #7

    Doc Al

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    Staff: Mentor

    Looks good to me.
     
  9. Dec 22, 2008 #8

    Defennder

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    Well this thread has been inactive for some time. Looking back I think my error and probably that of the OP was to square 0.4m Otherwise I think the correct answer should be about -71.6J.
     
  10. Jun 14, 2011 #9
    For this very same question above, how would I determine the electrons that were gained or lost from the test subject to create the charge?
     
  11. Jun 14, 2011 #10

    Doc Al

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    Assuming the test object was originally neutral, every time you remove one electron the charge remaining will increase by one elementary charge (which is the magnitude of the charge on an electron). What's the charge on an electron?
     
  12. Jun 14, 2011 #11
    1.60217646 × 10-19
     
  13. Jun 14, 2011 #12
    do I multiply the test charge with 6.242 × 10^18 e using millikan's relationship?
     
  14. Jun 14, 2011 #13

    Doc Al

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    That works. Realize that multiplying by that number is the same as dividing by the electron charge.
     
  15. Jun 14, 2011 #14
    oh okay! simple enough. thanks a bunch.
     
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