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Electric Potential / Electric Energy

  1. Apr 14, 2008 #1
    1. The problem statement, all variables and given/known data
    A 0.50kg ball is suspended in a uniform electric field ( E = 1500 N/C )
    as shown in the diagram ............

    [​IMG]

    Calculate the charge on the ball


    2. Relevant equations



    3. The attempt at a solution

    I found fg = -4.9 N .........
    but what next???
     
  2. jcsd
  3. Apr 14, 2008 #2

    Mentz114

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    What is Fg ?

    You need to find the forces on the ball due to gravity and the electric field and set them equal. That will give an equation for Q, the charge on the ball.
     
  4. Apr 14, 2008 #3
    so basically what ur saying is Fg = E ???
     
  5. Apr 14, 2008 #4

    Mentz114

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    If the ball is not moving, then the electric force (which is horizontal) is balanced by the horizontal component of the gravitational force. Write down expressions for the forces.
     
  6. Apr 14, 2008 #5
    Fex + Fgx = 0
    Fex = - Fgx
    Fey + Fgy = 0
    Fey = - Fgy ???
     
  7. Apr 14, 2008 #6

    Mentz114

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    If you have a charge Q in an electric field E ( pointing in the x-direction), what is the force on the charge ?
     
  8. Apr 14, 2008 #7
    hmmmmmmm the answer says 5.76 x 10^-4
    but i get 0.00331 ...................
    what i did was
    i found Fex = 0.864
    and Fey = 4.9
    i used pythagoras to find the whole part of the Fe and
    E = F / Q
    so i re-arranged it so that Q = F / E
    and that's how i got my answer ........ what am i doing wrong?
     
  9. Apr 14, 2008 #8
    i don't know ........ my physics teacher didn't teach me that
     
  10. Apr 14, 2008 #9

    Mentz114

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    I dunno. I would write

    Q.E = -m.g.sin(10 deg)
     
  11. Apr 14, 2008 #10

    Mentz114

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    You need to know that to solve the problem.
     
  12. Apr 14, 2008 #11
    hmmmmmm when i do that ....... I get the right answer .......
    but one question ........ what happened to the Y part of the
    formula ??? isn't that relevant to the solution as well ???
    because if u have an Fgy ......... wouldn't u need an Fey
    in order for it so not move??
     
  13. Apr 14, 2008 #12

    Mentz114

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    Have a look here -
    http://en.wikipedia.org/wiki/Electric_field

    There is a y-component in the tension of the string that offsets the vertical force.
     
  14. Apr 14, 2008 #13

    alphysicist

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    Hi Mentz114,

    The formula you have

    Q.E = -m.g.sin(10 deg)

    does not look right to me. There are three forces acting on the ball: gravity, tension from the string, and the electric force. Gravity is vertical, the electric force is horizontal, and the tension is at an angle.

    So the vertical component of the tension (T cos(theta)) must be equal in magnitude to the gravitational force mg, and the horizontal component of the tension (T sin(theta) )must be equal in magnitude to the electric force qE. (This is shown by writing the x and y equations from your force diagram.)

    At that point you'll have two equation with two unknowns which you can solve, and you'll end up with a tangent function instead of a sine.
     
  15. Apr 14, 2008 #14

    Mentz114

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    alphysicist,
    you could be right. OP says he got the right answer, but I didn't see the string until he raised the y-component issue.
     
  16. Apr 14, 2008 #15

    AHHHH .......... that made so much more sense thank you ......and i got it :D
     
  17. Apr 14, 2008 #16

    Mentz114

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    alphysicist,
    Something bothers me about the above. To solve 2 equations in 2 unknowns you need 2 degrees of freedom. But if the string is always taut, x and y are not independent. Any change in x requires a change in y, so there's only one degree of freedom, the angle.

    However I can see that a tan() is more intuitive. Can you display the solution ?

    M
     
  18. Apr 14, 2008 #17

    alphysicist

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    Mentz114,

    From the force diagram I get:

    x equation: T sin(10 degrees) = q E
    y equation: T cos(10 degrees) = m g

    Either solving one of these for T and plugging into the other, or dividing the x-equation over the y-equation gives:

    (T sin(10 degrees) )/(T cos(10 degrees) ) = (qE)/(mg)

    which gives qE = mg tan(10 degrees)


    I might be misunderstanding your statement about the degrees of freedom; in the vertical direction there is no change in the forces no matter what the charge is since the vertical component of the tension is always equal to the weight. Varying the charge would change just the horizontal component of the tension. Therefore the real central part of the problem varying with charge would just be

    (T_x) = q E

    but since we don't want to use tension in the answer, we want the angle, we use extra expressions (the vertical equation) that relate tension, theta, and mg.
     
  19. Apr 14, 2008 #18

    Mentz114

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    alphysicist,
    thanks, all resolved. I missed the cos resolution. It is one equation in one unknown as required by the df.

    M
     
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