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Gaussian infinite sheet, electric field

  1. Jun 9, 2008 #1
    1. The problem statement, all variables and given/known data
    three thin charged sheets are placed parallel to each other as shown:

    far left sheet, charge density = +10C/m^2
    middle sheet, charge density = -25C/m^2
    far right sheet, charge density = -10C/m^2

    |<--2m (A) -->|<-----4m (B) ------>|

    a) what is the electric field in regions A and B?
    b) what is the difference in potential between the two outer sheets? neglect thickness of sheets

    2. Relevant equations

    electric field at conductor surface = sigma/epsilon_o where sigma = charge density, epsilon_o = 8.85*10^-12 constant

    electric flux = E (*) A where (*) is dot product, E is electric field, A is area of surface

    charge enclose q_encl = sigma*A

    3. The attempt at a solution

    i've only tried part a:

    what does it mean in the regions A and B, do i find the electric field of each sheet and then sum them up? how do distances between sheets factor in?

    find the electric flux of the far left sheet, which is circle through the center of the sheet a la cookie cutter, and since there are two sides, there are two areas

    electric flux = 2AE
    sigma = +10C/m^2
    q_encl = 10(2A)

    let electric flux = q_encl/epsilon_o
    so 2AE = sigma*A/epsilon_o ---> E = sigma/ 2*epsilon_o = 10/(2(8.85*10^-12)) = 5.65*10^11 N/C

    that was the electric field for the far left sheet, i would do the same thing for the middle sheet but do i need to it to the far right sheet too for part a?

    electric field middle sheet = -25/(2(8.85*10^-12)) = -1.41*10^12 N/C

    electric field far right sheet = -10/(2(8.85*10^-12)) = -5.56*10^12 N/C

    so the electric field for a and b = electric field( far left + middle + far right) = -1.41*10^12 N/C

    how do the distances between the sheets come into play? ---> the 2m and 4m

    also could you give me a quick run down on what i need to do for part b.

    for part b since deltaV = -[integral(E*dr)] from r_a to r_b. do i let r_a = 0m and r_b = 6m ?

    Last edited: Jun 9, 2008
  2. jcsd
  3. Jun 9, 2008 #2


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    The strength of the electric field outside the infinite sheet is independent of the distance from the sheet. So the distance between the sheets doesn't matter when you are computing the E field. All you have to do is make sure you add the E fields in the correct directions. And, yes, for the potential difference the distance do matter. You integrate the E field over the separation between the sheets.
  4. Jun 10, 2008 #3
    part b:

    since electric potential difference V_ab = -[integral(E_net*dr)] from r_b to r_a where r_b = 6m, r_a = 0m and E_net = -1.41*10^12

    V_ab = -1.41*10^12(6 - 0) = -8.46*10^12 volts

  5. Jun 10, 2008 #4


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    You need to combine the electric field from ALL of the sheets. There are four different regions that could have four different E fields. Do they? What is it in each region? How many regions does the space between the two outer plates cover?
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