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How to find the magnitude of the electric field?

  1. Dec 17, 2012 #1
    I will keep working on this while waiting for someone here to respond....

    1. The problem statement, all variables and given/known data

    A ball m= 0.0024kg is electrically charged when 1E11 electrons are added. The ball falls with a=3.13m/s2 in an electric field. Find magnitude and direction of field.

    2. Relevant equations

    q= Ne

    3. The attempt at a solution

    If the ball is falling in the elec. field, it means it has a positive charge. If it's falling, it's going down so I have my direction for the elec. field.

    I'm guessing that the charge on it is the number of electrons added * charge of electron.
    So, 1E11 x (1.60E-19)= 1.6E-8 charge on ball.

    F= qE. I don't have E. E=kq/r2, I don't have r.

    And now?... :rolleyes:
     
    Last edited: Dec 17, 2012
  2. jcsd
  3. Dec 17, 2012 #2

    gneill

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    Staff: Mentor

    Electrons have a negative charge. So what sign is the charge on the ball? What are the units of charge?

    What other force is acting on a ball of mass 0.0024 kg when it falls? What's the usual acceleration when a mass falls? How does it compare to the observed acceleration? What does that tell you about the force on the ball due to the electric field (direction, size)?
     
  4. Dec 17, 2012 #3
    I know they do. Based on that the charge should be negative. However, if the ball is falling, it means its going where a positive charge would go in that field. So it must be positive. This is based on what we were taught :/

    Gravity is acting on it at 9.8m/s2 but, I thought the acceleration already included gravity and was 3.13m/s2 TOTAL. As in, no need to include gravity.
    Force would be 9.8 x 0.0024kg= 0.02352N
     
  5. Dec 17, 2012 #4

    gneill

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    The acceleration observed is the NET acceleration due to the sum of all forces acting. So rather than falling at g = 9.81 m/s2 which would be the rate if gravity alone were acting, the ball is observed to be falling at a = 3.13m/s2. Find the "missing" force.

    Once you find the "missing" force, determine what electric field would be required to cause it on the given charge.
     
  6. Dec 17, 2012 #5
    Sigh.... So I need to find the difference between the two of them?
    9.81-3.13= 6.68.
    I really don't understand yet.
     
  7. Dec 17, 2012 #6

    gneill

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    Why don't you start by finding the net force which would produce the observed acceleration. Then draw a free-body diagram (FBD) for the ball showing the forces acting. Solve for the "missing" force.
     
  8. Dec 17, 2012 #7
    The problem is that I'm unsure of what I'm supposed to be drawing. I don't know if my FBD is correct. I have the object, with mg pulling down. And F also pulling down? Or is it going the opposite way? I don't understand what the missing force is.
     
  9. Dec 17, 2012 #8

    gneill

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    If the object is falling with an acceleration less than usual, which direction do you think the "missing" force is acting?

    The "missing" force is the electric force on the charged ball due to the electric field acting on its charge.
     
  10. Dec 17, 2012 #9
    Its acting in the direction opposite of the mg. But how do I find the force if all I have is the different accelerations?
    FE= ma, 0.0024kg x 6.68 = 0.016032N
    W=mg, 0.0024kg x 9.81 = 0.023544N

    ma-mg= (-)0.007512 N total difference.
     
  11. Dec 17, 2012 #10

    gneill

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    You're dancing all around the solution but somehow not managing to step on it :smile:

    If you know the net (observed) acceleration (3.13 m/s2) and the mass, then you know the net force acting via Newton's 2nd law: fnet = m*a.

    This net force is comprised of the sum of two forces (fG, fE) acting in opposite directions. So fnet = fG - fE. Solve for fE.

    Now you have fE, which is directed upwards. It's produced by some electric field E acting on the charge on the ball...
     
  12. Dec 17, 2012 #11
    It's panic.... I did what you recommended but it's not the right answer.
    Answer is 1E6 N/C down.

    Fnet= ma= 0.007512 N
    Fnet= FE-Fg
    0.007512 N + Fg = FE
    0.007512 N + mg = FE
    0.007512 N + (0.0024kg x 9.81) =0.031056 N upward direction.

    F=qE
    F/q=E
    0.031056 N / 1.6E-8 = 1.94E6 N/C (down?)
     
  13. Dec 17, 2012 #12

    gneill

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    Okay, there's a small problem with choice coordinate axes and directions. If the ball is falling and you take that as the "positive" direction for the axis, then gravitational force will be in the positive direction, while the retarding electrical force will be in the negative direction.

    Your Fnet above is a positive value, so it points down in the assumed "positive" direction of the axis. That would make Fnet = Fg - FE, since FE would be upwards for this choice of axis direction.

    You should arrive at FE = 0.016N upwards.
     
  14. Dec 17, 2012 #13
    You are a saint. It's been a long frickin day. I've been at this for the past 7 hours.
    Thank you and I am forever grateful for all of your explanations and for your seemingly endless patience. Oh me.. :rolleyes:
     
  15. Dec 17, 2012 #14

    gneill

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    No worries. Glad to help!
     
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