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Speed of moving charge

  1. Jul 17, 2010 #1
    1. The problem statement, all variables and given/known data
    Two identical 25g particles each carry 5.0uC of charge. One is held fixed, and the other is placed 1.0mm away and released
    Find the speed of the moving charge when it's 1.0cm from the fixed charge

    2. Relevant equations



    3. The attempt at a solution
    I thought that using F=Q1Q2K/d^2 I could find the repelling force which is 2.25*10^5 N. then F=ma so a=2.25*10^5/0.025 =9*10^6m/s^2. then using kinematics v^2 =u^2+2as
    0=u^2+2*9*10^6*0.009 the trouble is that that is unsolvable (imaginary solution) Its extremely possible that my procedure is entirely wrong, I was just making it up on the fly... Help!!!
     
  2. jcsd
  3. Jul 17, 2010 #2

    rock.freak667

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    The charge is held and then released, meaning that the initial velocity is zero.
     
  4. Jul 17, 2010 #3
    argh thanks - stupid little mistake
     
  5. Jul 17, 2010 #4
    Hey, I solved it but the answer I got (402.5m/s) is still wrong????
     
  6. Jul 17, 2010 #5

    rock.freak667

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    Recheck your force calculation, I don't think your force should be that high (in the order of 105)
     
  7. Jul 17, 2010 #6
    I rechecked it and got the same F=9E9*5E-6*5E-6/.001^2 = 2.25E5???
     
  8. Jul 18, 2010 #7

    rock.freak667

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    So you happen to know the correct answer?
     
  9. Jul 18, 2010 #8
    no its for one of those stupid online tests
     
  10. Jul 18, 2010 #9

    rock.freak667

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    I ask as my calculation gives around 300 m/s.
     
  11. Jul 18, 2010 #10
    really thats significantly different to my answer - exact same procedure as me??
     
  12. Jul 18, 2010 #11
    I just tried it again from scratch and i keep getting 402m/s can you tell me how you did it please
     
  13. Jul 18, 2010 #12

    rock.freak667

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    Yes. Post what you did.
     
  14. Jul 18, 2010 #13
    ok for the Force i got 2.25*10^5 N (F=KQ^2/.001^2)
    then a = 2.25*10^5/.025 = 9E6m/s^2
    v^2=0+2*9E6*0.009
     
  15. Jul 18, 2010 #14

    Dick

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    v^2=u^2+2as only applies if the acceleration is constant. It's not. Use potential energy and energy conservation.
     
  16. Jul 18, 2010 #15
    yeah I figured out to use E=K+U
     
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