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!

Electric fields of wool and silk

  1. Apr 8, 2009 #1
    You've hung two very large sheets of plastic facing each other with distance d between them . By rubbing them with wool and silk, you've managed to give one sheet a uniform surface charge density n1=-n0 and the other a uniform surface charge density n2=3n0 .



    1. The problem statement, all variables and given/known data
    Find the electric fields at point 1, point 2, point 3 in term of numberical multiple of the factor (no/Eo)
    Point 1 is the left, point 2 is in the middle of the two sheets, and point 3 is to the right




    2. Relevant equations
    E=1/(4piEo) *Q/R^2

    E (parrallel capacitor) = Q/(EoA) or E=no/Eo


    3. The attempt at a solution
    I think of using equation for electric fields to solve it and then realize we don't have distance to use, but neither the equation for parrallel capacitor works

    Thanks for the helps
     
    Last edited: Apr 8, 2009
  2. jcsd
  3. Apr 8, 2009 #2

    Cyosis

    User Avatar
    Homework Helper

    Do you know Gauss' law? If so you can calculate the electric field of a single sheet without too much hassle. Hint: The electric field for the parallel plate capacitor is almost correct. Now remember that you can add fields vectorial and note that the strength of the field does not depend on distance.
    Draw the sheets on a piece of paper, mark which one is negative and which one is positive then draw electric field vectors in each of the areas.
     
  4. Apr 8, 2009 #3
    So I set up the electric field vectors and everything
    I assume that the two large sheets are infinity large so its electric fields will be the same regardless to the distance, also I'm using equation E= n/ (2Eo)

    Point 1: We have E_n_1 points to the right (due to the negative charge of sheet 1) while E_n_2 points to the left and since plate 2 has bigger charge, E_n_2 will be longer than E_n_1===> E_1 = E_n_2-E_n_1

    Point 2: Similarly, I found E_2=E_n_2+E_n_1 (since both pointint to the left)
    Point 3: Similarly, I found E_3=E_n_2-E_n_1

    Then using E_n=n/ (2Eo) to calculate E_n_1 and E_n_2 then figure out E_1, E_2, E_3 in term of n0/Eo which gave me
    Point 1= 2 (n0/Eo)=Point 3
    Point 2=1 (n0/Eo)

    I don't know what I do wrong, but MP doesn't accept my answer. Is this a right way to archive this problem, or maybe I'm making mistakes in vector addition
     
  5. Apr 8, 2009 #4

    Cyosis

    User Avatar
    Homework Helper

    While you listed the correct electric field I think that in your calculation you accidentally still used the electric field for the parallel plate capacitor. This is what you should have done [tex]\frac{-n_0}{2 \epsilon_0}+\frac{3n_0}{2 \epsilon_0}=\frac{n_0}{\epsilon_0}[/tex]. You seem to have made the same mistake for point 2. Also keep in mind that we are talking about vectors and what you've done so far is calculate the magnitude of the E-fields. MP might want you to also include the direction.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Electric fields of wool and silk
  1. Electric Fields (Replies: 3)

  2. Electric Fields (Replies: 4)

Loading...