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Find the vector of the net Force

  1. Aug 26, 2007 #1
    Take q1=25uC at (0,1), q2=20uC. at (2,0), and q3=? at (2,2). IF the force on q1 points in the -x direction, (a) what is q3 and (b) what is the magnitude of the force on q1?


    F=k(q1,q2)/r^2 coulombs law


    I know since q1 points in the -x direction that q3=q2. So (a)=20uC.
    I'm not sure how to go about part b.
     
  2. jcsd
  3. Aug 26, 2007 #2
    Find the vector of the net Force acting on q1 (due to q2 and q3). Take the magnitude of that vector.
     
  4. Aug 26, 2007 #3

    I found the r (the distance between q2 and q1) to be root 5. Then i made the equation
    ((9*10^9)(2.0*10^-6)(2.5*10^-6))/5 which = .009

    .009*2=.018N since q2=q3.
     
  5. Aug 26, 2007 #4
    You should decompose the forces into vectors.

    I calculated the force between q1 and q2 to be .9 N

    This is from:

    kq1q2/r^2 ; where k = 9E9, q1= 25E-6, q2=20E-6, r=sqrt(5)

    So F21 = .9*< -cos(a), sin(a) > ; where angle a = 26.5

    The y component of F31 should cancel out F21 if the net force only acts in the -x direction.
     
  6. Aug 26, 2007 #5
    If i am not mistaken, shouldn't q1=2.5E-6 and q2=2.0E-6??? That is how i got .009.
    I'm still kind of confused. What is F31?? The force created from q3 right?
     
  7. Aug 26, 2007 #6
    q1 = 25E-6 = 2.5E-5
    q2 = 20E-6 = 2.0E-5
    k = 9E9
    r = sqrt(5)

    I get .9 N

    F31 is the force acting on q1 by q3.
     
  8. Aug 26, 2007 #7
    I got that the net force, F31 + F21 = .9<-2cos(A), 0>. You know the angle A, so you can just take the magnitude of that.
     
  9. Aug 26, 2007 #8
    no...q1=25uC = 2.5*10^-6 right? same with q2.
     
  10. Aug 26, 2007 #9
    1 C = E6 uC
    Or 1 uC = E-6 C ( a Micro-Coulomb is a millionth of a Coulomb)

    25 uC * 1 C / E6 uC = 2.5E5 C

    I think you're trying to move the decimal places the wrong direction.
     
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