1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Charge/Field/Velocity Problem

  1. Oct 7, 2015 #1
    1. The problem statement, all variables and given/known data
    PHYSICS.png

    2. Relevant equations
    F=kqQ/D^2=Ma
    E=kQ/D^2
    D=vt?

    3. The attempt at a solution
    So far, I've been able to label the forces on each ship through superposition. But that is all I currently able to understand. I know that my end solution is a speed/velocity, but I am not sure how to get there with the given (Q, t, D, and M)

    Any help and direction would be greatly appreciated!
     
  2. jcsd
  3. Oct 7, 2015 #2

    gneill

    User Avatar

    Staff: Mentor

    Have you considered a conservation law approach?
     
  4. Oct 7, 2015 #3
    I have not because we haven't discussed it. How might I use said approach?
     
  5. Oct 7, 2015 #4

    gneill

    User Avatar

    Staff: Mentor

    Usually one considers the trade off between potential and kinetic energy. What form of potential energy is involved here?
     
  6. Oct 7, 2015 #5
    There is definitely potential energy between each ship.
     
  7. Oct 7, 2015 #6

    gneill

    User Avatar

    Staff: Mentor

    Yes... what type and how much?
     
  8. Oct 7, 2015 #7
    Electrical Potential Energy
    U12=kq1q2/D
    U23=kq2q3/D
    U13=kq1q3/D
     
  9. Oct 7, 2015 #8

    gneill

    User Avatar

    Staff: Mentor

    What's the total PE of the system in terms of the variables that you were given?
     
  10. Oct 7, 2015 #9
    U(total) = 3* kq^2/D, right?
     
  11. Oct 7, 2015 #10

    gneill

    User Avatar

    Staff: Mentor

    Looks good.

    So how will that change as the ships all tend towards infinite separation? Where will it go?
     
  12. Oct 7, 2015 #11
    All the potential will simply turn into kinetic energy right?
     
  13. Oct 7, 2015 #12

    gneill

    User Avatar

    Staff: Mentor

    That's right.
     
  14. Oct 7, 2015 #13
    Alright, so then 3* kq^2/D=1/2Mv^2 and just solve for v
     
  15. Oct 7, 2015 #14

    gneill

    User Avatar

    Staff: Mentor

    Remember that the energy will be split three ways.
     
  16. Oct 7, 2015 #15
    So is the multiplying by 3 completely necessary if the charges and D are the same? Couldn't I just solve it as kq^2/D=1/2Mv^2
     
  17. Oct 7, 2015 #16

    gneill

    User Avatar

    Staff: Mentor

    Sure. In your presented solution be sure to state why you are justified in doing it.
     
  18. Oct 7, 2015 #17
    Ah okay, I understand now! Thank you very much! :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Charge/Field/Velocity Problem
Loading...