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Coulomb's Law and Electrical Fields

  1. Feb 22, 2015 #1

    Yut

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    1. The problem statement, all variables and given/known data
    The tine sphere at the end of the weightless thread has a mass of .60g. It is immersed in air and exposed to a horizontal electric field of strength 700 N/C. The ball is in equlibrium in the position shown. What are the magnitude adn sign of the charge on the ball?

    2. Relevant equations
    F=qE
    Gravitationa Force= mg
    https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-xpf1/v/t34.0-12/10994667_10152670641336806_932307476_n.jpg?oh=9257cdd3e8de5465969a5ec980ec995b&oe=54EDB2B7&__gda__=1424855811_d63d4b9c3de86b44377f76cc1112f998
    3. The attempt at a solution
    So, since the ball in the equlibrium the forces are balanced.
    force downwords due to gravity
    mgcosθ
    Since the ball moves towords the E field, it should have a negative charge.

    So
    Fx: F(electrical field)- mgsin =0
    qE= mgsinθ
    q= mgsinθ / E

    I dont get the right answer,
    the answer should be -3.1qC
     
  2. jcsd
  3. Feb 22, 2015 #2
    F and mg are perpendicular. So mg cant balance the force. Hint: you have forgot to consider another force.
     
  4. Feb 22, 2015 #3

    Yut

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    Is it the force due to the other charge, which will be + and equalt to kq^2 / r^2... and I would find R using sinθ


    Sorry, I mized it with other problem.
    What other force there could be?
     
  5. Feb 22, 2015 #4
    Its the force exerted by the string. What would have happened if the string was absent? The ball would have moved away. But the string stops and holds the ball inplace. So consider tension force. Draw free body diagram. You should get ##T_y=mg## and ##T_x=F_E##
     
  6. Feb 22, 2015 #5

    Yut

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    Oh, I completely forgot about that, and then it will be Tcosθ, and you find T using the mg=Tsinθ
    the euqation there fore will be
    F(electrical)- Tsinθ-mgcosθ= 0
    and you solve for F(electrical)
    F(electrical) = qE

    and then you look for q
     
  7. Feb 22, 2015 #6
    Again... I told you F has nothing to do with mg. Its just ##F=Tsin\theta## and ##mg=Tcos\theta##. Find T from 2nd equation and put it on first equation.
     
  8. Feb 22, 2015 #7

    Yut

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    Yes! Here the set up that gave me the right answerhttps://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-xpf1/v/t34.0-12/11014718_10152670767961806_1607350090_n.jpg?oh=c796df67fefef907f1ccfe3fbafe9f57&oe=54ECB140&__gda__=1424800785_7febf55bcb55aca57d3a84e1bb4b6827
     
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