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Magnitude of Net Force?

  1. Feb 9, 2010 #1
    1. The problem statement, all variables and given/known data

    Three tiny charged metal balls are arranged on a straight line. The middle ball is positively charged and the two outside balls are negatively charged. The two outside balls are separated by 20 cm and the middle ball is exactly halfway in between. The absolute value of the charge on each ball is the same, 1.47 microCoulombs.

    What is the magnitude of the net foce on either outside ball?


    2. Relevant equations



    3. The attempt at a solution
    After converting micocoulombs to C and cm to m, I used F=k(q1*q2/d^2) to find the magnitude of the attractive force on the outside ball by the inside ball, which is 1.945 N. I also found the magnitude of the resulsive force on the outside ball by the other outside ball, which is .486 N.

    Now I'm just confused on how to get the net force. I know I need to understand the direction the forces are going in order to know how to combine the forces I found earlier.. and well...

    A (-) ------ B (+) -------C (-) I know the force on B by C is pushing C away ----> and the same for the force on B by A except in the opposite direction <-------. other than this, this is where I get stuck.



    Please, and thanks for the help!
     
  2. jcsd
  3. Feb 9, 2010 #2

    Lok

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    A is attracted by B by a 1.945N ---->
    A is repulsed by B by 0.486N <--
    The net force is F=Fb-Fc or F=Fc-Fb (all F's are vectors)

    Same for C
     
  4. Feb 9, 2010 #3
    basically it's 1.945 N - .486 N = 1.459 N (net force)?

    Regarding your point that "A is repulsed by B by 0.486N <--", even though .486 N is the force that A is being repulsed by C, that force that A is repulsed by B is the same?.. Sorry I'm so confused it just seems like the repulsive force of A by B would be half of that of A by C because it's half the distance.
     
  5. Feb 9, 2010 #4

    Lok

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    Yeah that was A is repulsed by C ... my bad.
     
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