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Electric fields and electrostatic forces

  1. Sep 3, 2004 #1
    Electric fields and electrostatic forces (1 question)

    I was wondering if someone can show me how to answer this problem. Thank you.


    1)Two charges are placed on the x-axis, an unknown positive charge at x=0 cm and a negative 8.642e-6 C charge at 2.993 cm from the origin. Calculate the magnitude of the charge at x=0 cm which causes the total electric field at 11.917 cm to be zero.
     
    Last edited: Sep 3, 2004
  2. jcsd
  3. Sep 3, 2004 #2

    Tide

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    Show us what you did then I'm sure you'll get plenty of help!
     
  4. Sep 4, 2004 #3
    Electric field,[tex] E = \frac{kq}{r^2}[/tex]
    Since you have two particles with charges, lets call the unknown charge q1, and the known charge q2, where

    Since E = 0, [tex] \frac{kq_1}{r^2} + \frac{kq_2}{r^2} = 0[/tex], where r is the distance from the charge to the point of zero electric field. Therefore...
    For the first charge its .11917m away, and for the second charge its .11917m - .02993m = .08924m.

    [tex]\frac{kq_1}{.11917^2} + \frac{k*-8.642e-6}{.08924^2} = 0[/tex]

    The k's cancel, and therefore q1 = 1.54e-5 C.

    I'm not 100% sure if I did this right, so wait for more responses before taking my word for it. :smile:
     
    Last edited: Sep 4, 2004
  5. Sep 4, 2004 #4
    Conceptually it is correct although I haven't done the calculation so...
     
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