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Homework Help: Finding the Net Electric Force on a Charge

  1. Sep 27, 2008 #1
    1. The problem statement, all variables and given/known data
    Three positive point charges are arranged in a triangular pattern in a plane, as shown below.
    The Coulomb constant is [tex]8.98755*10^9 N · m^2/C^2.[/tex]

    http://img150.imageshack.us/img150/1052/electricforceproblemqp7.th.jpg [Broken]http://g.imageshack.us/thpix.php [Broken]

    Find the magnitude of the net electric force on the 6 nC charge. Answer in units of N.

    2. Relevant equations

    [tex]F = \frac{K(qQ)} {r^2}}[/tex]

    3. The attempt at a solution

    Pythagorean Theorem: [tex]\sqrt{(6)^2 + (6)^2} = 8.485[/tex]

    [tex]\frac{(8.98755*10^9 N · m^2/C^2)(6*10^{-9}C)(2*10^{-9}C)} {(8.485m)^2}} = 1.498*10^{-9} N[/tex]

    [tex]\frac{(8.98755*10^9 N · m^2/C^2)(6*10^{-9}C)(9*10^{-9}C)} {(8.485m)^2}} = 6.7411*10^{-9} N[/tex]

    [tex]sin(45)(1.498*10^{-9} N + 6.7411*10^{-9} N) = 5.8259*10^{-9} N[/tex]

    Unfortunately that's not the right answer :frown:
     
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Sep 28, 2008 #2
    You made a good start so far. The magnitude of the force on the 6 nC particle due to the other two particles is done correctly. Your next step is wrong.

    Do the following: Draw the two force vectors on the 6 nC charge, and then determine their x- and y- components. Now add the components together to get the net force components on the 6 nC charge. Apply the Pythagorean theorem to these components to get the magnitude of the net force.
     
  4. Sep 28, 2008 #3
    Attempt #142,032:
    [tex]F1 = cos(45)1.498*10^{-9} N[/tex] x [tex] + sin(45)1.498*10^{-9} N[/tex] y
    [tex]F2 = -cos(45)6.741*10^{-9} N[/tex] x [tex] + sin(45)6.741*10^{-9} N[/tex] y
    [tex]Fnet = 3.706*10^{-9} N[/tex] x [tex] + 5.825*10^{-9} N[/tex] y

    [tex]||Fnet|| = 6.905*10^{-9} N[/tex]

    Thanks Guys!
     
  5. Oct 7, 2010 #4
    One of the y components should be negative, not one of the x components. The magnitude still comes out the same.
     
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