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Point charges and force of repulsion

  1. Mar 1, 2007 #1
    Two point charges are separated by a distance r and repel each other with a force F. If their separation is reduced to 0.280 times the original value, what is the magnitude of the force of repulsion between them?

    I don't really understand this question and don't know where to start.
    Would I start with the equation
    E = ((k)(Q)/r^2

    what follows next?
  2. jcsd
  3. Mar 1, 2007 #2
    What is the equation for the repulsive force between two point charges, and how does it scale with r?
  4. Mar 1, 2007 #3
    It would be Coulomb's law of

    F = ((k)(q1)(q2))/(r^2)

    So it would be -0.280 times the original r value?

    The final answer is 12.8 F but I still don't understand.
  5. Mar 1, 2007 #4
    Where do you get the minus sign?

    O.K., you know that F scales as 1/r^2. When r gets smaller, F must get larger. When r is half it's original value, how much larger does F get?
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