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Homework Help: Electric field and ball of charge

  1. May 16, 2007 #1
    As shown in the figure above, a ball with a mass of 0.180 g and positive charge of q=31.0
    mC is suspended on a string of negligible mass in a uniform electric field. We observe
    that the ball hangs at an angle of q=16.0o from the vertical. What is the magnitude of
    the electric field?


    so i drew a free body diagram and now i am going to calcuate the forces in the x and y
    directions to see what E is.

    since the ball is stationary Fx=0=E-Tsin(16)+q

    so E=Tsine(16)+q

    now i solve for T by finding the forces acting in the y direction Fy=0=Tcos(16)-mg

    so T=mg/cos(16)...so i substitute that into Fx and now E=(mg/cos(16))sin(16)+q

    what am i doing wrong?
     
  2. jcsd
  3. May 16, 2007 #2

    Doc Al

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    Staff: Mentor

    Good.

    The force exerted by the (presumably horizontal) electric field on the charge is Eq. Rework this part.
     
  4. May 16, 2007 #3
    What's with the '+q' term?
     
  5. May 16, 2007 #4
    so it would be E=Tsin(16)+Eq?
     
  6. May 16, 2007 #5
    No. It would be Tsin(16)-Eq = 0.
     
  7. May 16, 2007 #6

    Doc Al

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    Staff: Mentor

    Not quite. Just redo your calculation of Fx.
     
  8. May 16, 2007 #7
    Fx=0=E-Tsin(16)+the charge of the partical


    where is the Electric field coming into play agian....i don't understand
     
  9. May 16, 2007 #8

    Doc Al

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    Staff: Mentor

    This makes no sense. You are adding the horizontal forces on the charged object. What forces act on it? E is the field, not the force! q is the charge, also not a force.

    Reread the posts by neutrino and myself.
     
  10. May 16, 2007 #9
    so the force due to the electric field is F=Eq so instead of E=Tsine(16) i would have Eq=Tsin(16)??
     
  11. May 16, 2007 #10

    Doc Al

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    Staff: Mentor

    Yes. Now you've got it.
     
  12. May 16, 2007 #11
    thank you so much. i really appriciate it.
     
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