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Electric Field Between Two Infinite Sheets of Charge

  1. Feb 3, 2013 #1
    1. The problem statement, all variables and given/known data
    An infinite sheet of charge is located in the y-z plane at x = 0 and has uniform charge denisity σ1 = 0.57 μC/m2. Another infinite sheet of charge with uniform charge density σ2 = -0.39 μC/m2 is located at x = c = 28.0 cm.. An uncharged infinite conducting slab is placed halfway in between these sheets ( i.e., between x = 12.0 cm and x = 16.0 cm). What is V(S) - V(P), the potentital difference between point P, located at (x,y) = (6.0 cm, 0.0 cm) and S, located at (x,y) = (22.0 cm, -16.0 cm)?

    https://www.smartphysics.com/Content/Media/Images/EM/06/h6_planeA.png [Broken]

    2. Relevant equations

    3. The attempt at a solution
    I have attempted at an answer and obtained the result of 8685.87 V.

    This answer came by way of:

    Electric field at point P:
    E = (σ1/2ε)-(σ2/2ε) = 54286.72 (N/C) or (V/m)
    Set this point to be the zero potential energy position, and since the electrical field is constant, or so I thought, in the x direction, with no component in the y direction, I simply took the difference in x positions of the two points. 22cm-6cm= 16cm, therefore the ΔV between these two points is (54286.72V/m)*(.016m)= 8685.87 V

    The feedback to this answer says that I need to account for the Electric field not being constant between these two points. What I am confused about is that I thought the electric field between these two points would be constant. Since a single infinite plate will have a constant electric field that is not dependent on distance from the plate I assumed that the electric field between two of these plates would also be constant. Thank you to anyone that can explain this to me, I can't seem to figure it out and its driving me crazy.
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 3, 2013 #2


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    What is the electric field between x = 12.0 cm and x = 16.0 cm , i.e., inside the conducting slab ?
    Last edited by a moderator: May 6, 2017
  4. Feb 3, 2013 #3
    Wow thank you so much SammyS. I cannot believe I forgot that the area inside the conductor would not contain an electric field, therefore no contribute to the potential change. True testament to the need to simply think the problem through completely. Thanks again.
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