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Physics Midterm Help Part 2

  1. May 10, 2009 #1
    1. The problem statement, all variables and given/known data

    A point charge 'q' hangs motionless from a mass less string at an angle, due to the forces of gravity and electrical attraction with a nearby infinite conducting plane of charge.

    Conducting wire sigma = -3.0 E -8 String is next to the conducting wire with a angle
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    a. If the mass of 'q' is 4.00 E -6 kg, and it hangs at angle of 25.0 degrees with the vertical, find the charge 'q'.

    2. Relevant equations

    Using the equations, that I learned from mechanics. I broke up the force components in x and y direction.

    ∑ Fx = Fe - T sin 25º = 0
    ∑ Fy = T cos 25º - mg = 0

    3. The attempt at a solution

    ∑ Fy = T cos 25º - mg = 0
    T = mg/cos 25º
    T = (4.00 E-6 kg) (9.80 m/s^2)/cos (25º) = 3.55 x 10^-5 N (I wrote N on my test and he crossed that out)

    Electrical Force = T sin 25º = (3.55 x 10^-5N) sin 25º = 1.50 x 10^-5 N

    Now I am lost on how to solve for q. Can you pinpoint where Im making errors and also direct me on how to solve for q? Because on my test I got half of the points.

    How would you sketch an electric field diagram of 'q' and 'sigma' ? Also, where would the net electric field be exactly zero?
    Last edited: May 10, 2009
  2. jcsd
  3. May 10, 2009 #2
    Ok for part a....is the process right

    so since I already found what Electrical force is. I can just use F = qE ....which means q= F/E ...So I plug in F which I found already and E = Sigma/2 Eplison knott. Since we got the sigma value. But when I solved it I got q = -8.85 x 10^-9 C. Is that right?
  4. May 10, 2009 #3


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    Draw a diagram and I would think you would discover that the ratio of the Forces - Gravity down to Electrostatic to the E-Field will be given by

    Felectro / Fgravity = Tan25
  5. May 10, 2009 #4


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    F = qE

    should deliver the right answer if you have the proper F.
  6. May 10, 2009 #5
    So the E is E = sigma/2 Epslion knott?
  7. May 10, 2009 #6


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    Conducting plane or conducting wire?
  8. May 10, 2009 #7


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    I don't understand this statement:
    If it is a conducting plane your formula would be correct.
  9. May 10, 2009 #8
    Yup a conducting plane of charge. I was trying to explain the diagram. Thanks though! :)

    I have one question is the net electric field zero at the center of the conducting plane?
  10. May 10, 2009 #9


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    OK. That should be good to go then.

    And the conducting wire is from what? Did you mean to write plane? Because if you have a wire you need to know r from the wire, which I don't see in the problem as given.
  11. May 10, 2009 #10
    Yes there is no conducting wire....it's a conducting plane of charge. Plus there was no radius given. But can you answer where the net electric field is zero in this problem? B/C I though it was at the center. But I got ding -10 points on the exam.
  12. May 10, 2009 #11


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    The E field is 0 inside the conductor.

    As to the E-field, I think I misspoke. I think it should be σ / εo
  13. May 10, 2009 #12
    so it just should be 1/2 of the value that I got before right? Oh is the E field zero inside the conductor because the E field go in and out of the conductor cancel out at the center?
  14. May 10, 2009 #13


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    Wouldn't it be twice?

    Didn't you say σ /2 εo ?

    That would be for a capacitor I think.

    In a conductor the charge is at the surface. Inside the conductor any Gaussian surface you draw that does not contain the surface must result in a 0 field, because there would be no charge within the closed surface.
  15. May 10, 2009 #14
    Well, because I got 1.50 E -5 N for the Electrical Force. sigma = -3.0 E -8 C/m^2)
    Now if it was 2 εo

    q= F/E = (1.50 E -5 N)/(-3.0 E -8 C/m^2 /2(8.85 E -11 N/m) = -8.85 E -9 C

    Now if The E Field = σ / εo

    So ...q =F/E = (1.50 E -5 N)/ (-3.0 E -8 C/m^2 /(8.85 E -11 N/m) = -4.43 E -9 C

    Thats why I said half the value.
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