1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    | |
    | |
    | |
    | |

    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

    LowlyPion

    User Avatar
    Homework Helper

    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

    LowlyPion

    User Avatar
    Homework Helper

    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

    LowlyPion

    User Avatar
    Homework Helper

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

    LowlyPion

    User Avatar
    Homework Helper

    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

    LowlyPion

    User Avatar
    Homework Helper

    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

    LowlyPion

    User Avatar
    Homework Helper

    The E field is 0 inside the conductor.

    As to the E-field, I think I misspoke. I think it should be σ / εo
    see:
    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gausur.html#c3
     
  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

    LowlyPion

    User Avatar
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Physics Midterm Help Part 2
  1. Homework help part 2 (Replies: 15)

  2. Physics Midterm Help (Replies: 1)

Loading...