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MasteringPhysics.com problem (Electric Field)

  1. Jan 31, 2008 #1
    Two point charges are located on the x axis: one charge, q_1 = -19.5 nC, is located at -1.660 m ; the second charge, q_2 = 38.0 nC, is at the origin (x=0).


    What is the net force exerted by these two charges on a third charge q_3 = 48.5 nC placed between q_1 and q_2 at x_3 = -1.210 m?

    Note that your answer may be positive or negative, depending on the direction of the force.

    Use 8.85×10−12 C^2/(N \cdot m^2) for the permittivity of free space.


    I have no idea how to do this. any help?
    thank you
     
  2. jcsd
  3. Jan 31, 2008 #2
    Use superposition.
     
  4. Feb 1, 2008 #3

    G01

    User Avatar
    Homework Helper
    Gold Member

    You must show some work or thought on the problem to get help here. We do not do homework problems for people, we help people do their own homework problems.

    So, give me some thought on this problem. Can you find the force from one of the charges? What about the force from the other one then? How do you go about adding the two quantities?
     
  5. May 10, 2010 #4
    Hay, I'm stuck on the same problem but have some working. :P

    e = 8.854 x 10^-12

    Consider two point charges located on the x axis: one charge, q_1 = -11.5 nC, is located at x_1 = -1.665 m; the second charge, q_2 = 38.5 nC, is at the origin (x=0.0000). What is the net force exerted by these two charges on a third charge q_3 = 53.5 nC placed between q_1 and q_2 at x_3 = -1.215 m?

    MasteringPhysics is saying its wrong. :(
    [​IMG]

    I only have one attempt left and I'm not sure what I could be doing wrong.. Perhaps it should be positive?

    Kind Regards,
    Troz
     
  6. May 10, 2010 #5
    You missed a negative on F2.
     
  7. May 10, 2010 #6
    Thanks for the prompt reply.

    So I should be getting something around (-)1.48 *10^13?

    The 10^13 part seems very large and wrong to me. Its something like 14.7 Tera Newtons... Awfully large force..
     
  8. May 10, 2010 #7
    Charge is given in nano-coloumbs (nC) not coulumbs.
     
  9. May 10, 2010 #8
    Thanks I missed it. :P So I re-did the calculations based on your help, and I think it produces a much more sensible answer. Does this look correct? The magnitude is at least looking better :P

    [​IMG]
     
  10. May 10, 2010 #9
     
  11. May 10, 2010 #10
     
  12. May 10, 2010 #11
    Right.
     
  13. May 10, 2010 #12
    Sweet, thanks Heaps Zach, just typed it into mastering physics and it said correct.
     
  14. May 10, 2010 #13
    Sweet, no problem.
     
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