1. Not finding help here? Sign up for a free 30min 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!

Field intensity of a sheet of charge

  1. Sep 5, 2011 #1
    1. The problem statement, all variables and given/known data
    A sheet of charge Ps = 2 NC/m^2 is present at the plane x = 3 in free space, and a line charge Pl = 20 nC/m is located at x = 1, z = 4.

    a) Find the magnitude of the electric field intensity at the origin
    b) Find the direction of E at P (4,5,6)
    c) What is the force per meter length on the line charge?


    2. Relevant equations

    Field of a sheet of Charge

    E = Ps/2(epsilon) * An

    3. The attempt at a solution

    At the origin, do I just use the equation above?
    For the direction to (4,5,6), how exactly would I draw the vectors?


    Appreciate the help, thanks!
     
  2. jcsd
  3. Sep 6, 2011 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Good thing you gave us the fromula for the E field due to the sheet, since the question did not specify if it was conducting or not. It's not.

    Take one source of E at a time.

    The sheet gives an E field of Es = σ/2ε0 directed in the +x ditection for x > 3 and in the -x direction for x < 3:
    Es = σ/2ε0 i, x > 3
    Es = -σ/2ε0 i, x < 3. Since the origin is at x<3 you will use this for Es.

    Then, the wire, which runs up & down parallel to the y axis at (1,4):
    To find the E field = Ew due to it, I would use Gauss' theorem with a right circular cylindrical closed surface of radius R = √(4^2 + 1^2) = √(17) and height = 1, concentric with the wire. This gives |Ew|*2πR = λ where λ = linear charge density of wire. At the origin this field is directed opposite (4,1) so the vector will be
    Ew = |Ew|(-i - 4j)/√17.

    So the total field is E = Es + Ew at the origin.

    To get the field at ( 4,5,6) do the same thing. Es is now + since x > 3 at this point. Use the same idea with the Gaussian surfgace for Ew. R for this point is now
    R = √(4^2 + 6^2) = √52. (y = 5 has no significance; the field is the same for all y.)

    The force on the wire per unit length is just λ*Es i. Remember that x<3 for the wire so use the appropriate sign for Es.



    the same since x>3
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook