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Homework Help: Need some help with forces of changed particles (coulomb's Law)

  1. Jan 26, 2005 #1
    I need somebody to explain to me specifically "Forces of changed particles & Coulomb's Law", which was covered in a class I missed yesterday.

    Basically, I know that 8.99 x 10^-9 is a constant used in the beginning of the equation, and then after you put (q1 and q2), which i have no clue what it means, but i know it's the numbers that correspond to the particles , and divide everything by distance ^2

    I saw some people use 1.6 x 10 ^-19 for some problems, and the teacher said it was a constant too...I just have no clue when to use this constant, and when not to use it. It was used in some problems, but not in others :confused:

    Here's an example of one of the problems we have to do:

    Particles of charge +70, +48, and -80 μC are placed in a line. The center one is 0.35 m from each of the others. Calculate the force on each charge due to the other two.



    I began to understand it with two particles, but 3 completely lost me. I imagine 70, 48, and -80, would be q1, q2, and q3...d would be 0.35m. Would this be a problem where I use 1.6*10^-19?

    Thanks :)
  2. jcsd
  3. Jan 26, 2005 #2
    My answer for some reason, comes out completely different from what the book says. Could somebody please do out the work for the first particle?

    Here is my work:

    F = (8.99*10^-9) (48*10^-6) (70*10^-6) / 0.35^2

    which comes out to be: 2.46*10^-16

    then F = (8.99*10^-9) (70*10^-6) (-80*10^-6) / 0.7^2

    which comes out to be: -3.48*10^-16

    And the answer I get it: 1.4*10^-16

    Answer in the book says: -1.4*10^2N
    Last edited: Jan 26, 2005
  4. Jan 26, 2005 #3

    q means charge.
    1.6*10^19 C is the charge of an electron.
    If there are 3 particles just do vector addition for the total force.
  5. Jan 26, 2005 #4
    ah, I just realised my mistake :)

    ke = 8.99 x 10 ^ 9 N/C ... not -9

    that fixed it :)
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