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Help on calculating net force

  1. Sep 10, 2004 #1
    Hi,
    Can anybody help me get started on this problem?

    Point P2 is located 3.0 m away from each -Q (-Q = -40 nC) , forming an equilateral triangle with them.

    Determine the net force (magnitude and direction) that would act on a small charge of +1.0 nC placed at P2.

    Any hints?

    Thanks!
     
  2. jcsd
  3. Sep 10, 2004 #2
    Draw a picture and use Coulomb's law. Make use of verical and horizontal components.
     
  4. Sep 10, 2004 #3
    Right. Draw the vector for the force that each other charge applies on the +1.0 nC charge (using Coulomb's law), then add them together in a vector addition to find the resultant.
     
  5. Sep 11, 2004 #4
    The answer i got is: 4.4 x 10^-8N.
    Can you verify if thats right?

    Here are the steps I did to calculate that:

    F12 = k(q1)(q2)/r^2 = [(8.99 x 10^9)(1.0 x 10^-9)(-40 x 10^-9)]/(3.0)^2

    F12x = F12cos56 = (4.0 x 10^-8)(0.56) = 2.2 x 10^-8
    F12y = F12sin56 = (4.0 x 10^-8)(0.83) = 3.3x 10^-8
    F13 = 4.0 x 10^-8
    F13x = 2.2 x 10^-8
    F13y = -3.3 x 10^-8

    F1x = F12x + F13x = (2.2 x 10^-8) + (2.2 x 10^-8) = 4.4 x 10^-8
    F1y = F12y - F13y = 0
    F= (F1x^2 + F1y^2)^1/2 = [(4.4 x 10^8)^2 + (0)^2]^1/2 = 4.4 x 10 ^-8N

    THANKS!
     
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