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Help with Electricity problem please

  1. Jul 16, 2014 #1
    1. The problem statement, all variables and given/known data

    One particle has a mass of 3.00 10-3 kg and a charge of +8.00 µC. A second particle has a mass of 6.00 10-3 kg and the same charge. The two particles are initially held in place and then released. The particles fly apart, and when the separation between them is 0.100 m, the speed of the 3.00 10-3 kg particle is 120 m/s. Find the initial separation between the particles.

    2. Relevant equations

    let the initial seperation as r and final seperation as d.
    kq^2/r^2 = kq^2/d^2 + 1/2m1*V^2 + 1/2m2*V^2
    is the the way to solve it? how can i determine the final velocity for the 6x10^-3 particle?

    3. The attempt at a solution
     
  2. jcsd
  3. Jul 16, 2014 #2
    Are their any external forces acting on the system comprising the two charges ?
     
  4. Jul 16, 2014 #3
    Nope.
     
  5. Jul 16, 2014 #4
    What is conserved if there are no external forces acting on the system ?
     
  6. Jul 16, 2014 #5
    momentum?
     
  7. Jul 16, 2014 #6
    Yes . Use Conservation of Momentum.
     
  8. Jul 16, 2014 #7

    vela

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    The way you wrote the equation, you can solve it, but you'll get the wrong answer. Is the velocity for particle 1 the same as the velocity for particle 2? No. So why are you using the same variable, V, to represent those two quantities?

    If you write the equation properly, you have one equations and two unknowns. You need a second equation if you hope to find specific values of each unknown.
     
  9. Jul 16, 2014 #8
    (3.00 10-3 )x 0 + (6.00 10-3 ) x 0 = (3.00 10-3) x 120 - (6.00 10-3 ) x V
    i this correct?
     
  10. Jul 16, 2014 #9
    (3.00 10-3 )x 0 + (6.00 10-3 ) x 0 = (3.00 10-3) x 120 - (6.00 10-3 ) x V
    i this correct?
     
  11. Jul 16, 2014 #10
    Correct .

    There is another problem with this equation apart from what Vela has pointed .

    What do these terms represent ?
     
  12. Jul 16, 2014 #11
    as i stated r = initial distance and d= final distance.
     
  13. Jul 16, 2014 #12
    Did I ask what r and d represent ?
     
  14. Jul 16, 2014 #13
    i saw the r and d is highlighted from my computer screen so i thought you asked about it. lol my bad sorry
     
  15. Jul 16, 2014 #14
    But you still haven't answered my question in post#10 .
     
  16. Jul 16, 2014 #15
    so instead of kq^2/r^2 = kq^2/d^2 + 1/2m1*V^2 + 1/2m2*V^2.
    it should be kq^2/r^2 = kq^2/d^2 + 1/2m1*(V1)^2 + 1/2m2*(V2)^2 as the velocity is different.
    right?
     
  17. Jul 16, 2014 #16
    No . Sorry but you are not answering the question . I have explicitly remarked that there is another problem with the equation apart from what Vela has pointed .

    Please explain what do terms I marked in red in post#10 represent ?
     
  18. Jul 16, 2014 #17
    kq^2/r2 represent the electrostatic force between the two charges
     
  19. Jul 16, 2014 #18
    This is wrong .

    How can you have force in conservation of energy equation ?

    Instead of electrostatic force you need to have electrostatic potential energy of the two charges .
     
  20. Jul 16, 2014 #19
    V = kq/r ?
     
  21. Jul 16, 2014 #20
    No .

    Please look up in your notes or book or on the web .
     
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