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Coulumb's Law Question. Please Help

  1. Oct 3, 2010 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    My reasoning is that when the charge stops that means that the force of friction equals the electric force. So I find the distance by setting the two equations equal to each other and solve for x which gives (keq1q2/(mgk))^(1/2), however what about the initial distance that they were set at? Do I add it to this value or subtract it or is it already included within the value of x that I solved for? Any help would be greatly appreciated. Thanks!
    Last edited: Oct 3, 2010
  2. jcsd
  3. Oct 3, 2010 #2
    That won't work. When the force of friction equals the electric force, the net force is 0, so the object will keep on moving.
    Use conservation of energy. The energy to move an object a certain distance with a constant friction is easy to find.
  4. Oct 3, 2010 #3
    How can we use the conservation of energy? Are you talking about the equation K1+U1+Wother= K2 + U2? But even if we use that we are not provided a velocity nor does the charge change in height so we can't use potential?
  5. Oct 3, 2010 #4
    You don't need a velocity. The velocity of the charge is 0 at the start and at the end, so the initial and final kinetic energies are 0. This leaves only the electrical potential energy and the work that friction does on the charge.
  6. Oct 3, 2010 #5
    Do we solve for x
    Last edited: Oct 3, 2010
  7. Oct 3, 2010 #6
    kmg is the force of friction, not the work done
  8. Oct 3, 2010 #7
    Then we solve for x to find the distance right?
    Last edited: Oct 3, 2010
  9. Oct 3, 2010 #8
    Yes! Don't confuse the different k's tough.
  10. Oct 3, 2010 #9
    Awesome thank you! I do have a question though, when I solve for x, I get xq1q2/l0 - (4*pi*e0)(kmgx^2) = q1q2. Is there a way to simplify this?
  11. Oct 3, 2010 #10
    The work done by friction should be [itex] kmg(x - l_0) [/itex] since the charge starts at [itex] l_0 [/itex]

    You get a quadratic equation
  12. Oct 3, 2010 #11
    Oh okay and then we use the quadratic equation to solve for x? It is just the + value right?
  13. Oct 3, 2010 #12
    I'm not sure. x should be positive and there should be only 1 positive solution.
  14. Oct 3, 2010 #13
    Oh okay I found the positive solution. Are you sure this is the way to solve the problem because my professor says that the answers should not be very complicated? Thanks again though!
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