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Mass of a particle help please

  1. Sep 16, 2006 #1
    hey guys, i have been trying to work this problem out but for some reason, my answer is different from the books answer on the back.
    the problem is as follow:
    Two points charges of 30 nC and -40nC are held fixed on an X axis, at the origin and at X=72 cm, respectively. A particle with a charge 42 micro C is released from rest at X=28 cm. if the initial acceleration of the particle has a magnitude of 100 Km/s^2, what is the particle's mass?????

    the way i was solving it:
    let q1=30nC and q2= -40nC q3=42 Mic C X=72 cm
    F13 = (8.99*10^9*(30*10^-9)*(42*10^-6))/(28*10^-2)^2= 0.144
    F23 = (8.99*10^9*(40*10^-9)*(42*10^-6))/(44*10^-2)^2= 0.078
    F12 = (8.99*10^9*(30*10^-9)*(40*10^-9))/(72*10^-2)^2=2.08*10^-5

    so Fnet= -F13+F23+F12 and then F=ma so m=F/a.

    Am i solving it incorrectly. Please Help. Thanks
    Last edited: Sep 16, 2006
  2. jcsd
  3. Sep 16, 2006 #2
    Fnet on what? What's the net force acting on the third charge (42 microC)?
  4. Sep 16, 2006 #3
    I did that too. it's F13 and F23. so Fnet without F12 but it's still wrong.
  5. Sep 16, 2006 #4
    Are you sure you've taken into consideration the nature of the charges?
  6. Sep 16, 2006 #5
    yes. please tell me if you have different idea.
  7. Sep 16, 2006 #6
    I don't think so. You haven't included the minus sign (-) of the negative charge whlie calculating F23.
  8. Sep 16, 2006 #7
    you are correct but the answer is still incorrect. i had added the - sign later after calculation. so i had - and -
    Last edited: Sep 16, 2006
  9. Sep 16, 2006 #8
    What's you answer? And, btw, did you also convert Kms-2 to ms-2?
  10. Sep 16, 2006 #9
    I kept getting 2.2*10^-6 but in the book he has it as 2.2*10^-5.
    of course i had 100km/s^2 as 100000m/s^2
  11. Sep 16, 2006 #10
    Oops, sorry...too late into the night for calculations. I'm getting the same answer as you. May be it's a typo in the book.
    Last edited: Sep 16, 2006
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