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Electric Flux question [Surface integral]

  1. Feb 19, 2008 #1
    1. The problem statement, all variables and given/known data
    Q] A charge 'Q' is kept over a non-conducting square plate of side 'l' at a height l/2 over the center of the plate. Find the electric flux through the square plate surface. Neglect any induction that may occur.

    2. Relevant equations

    \phi = \int \overrightarrow{E}\cdot \overrightarrow{dS}

    3. The attempt at a solution

    Well.. it was pretty simple to do using Gauss law. Just take a cube as the surface and then the flux will be equal from all surfaces.. so the flux through one surface would be [itex]\frac{q}{6\varepsilon_o}[/itex]..

    but i wanted to actually try a surface integral.. so I found out the Electric field as a function of the coordinates (x, y) on the square plate. Here, one of the corners is taken as the origin (0, 0):

    E(x, y) = \frac{2Q\sqrt{(l - x)^2 + (l - y)^2}}{\pi \varepsilon_o \left[4(l - x)^2 + 4(l - y)^2 + l^2\right]^\frac{3}{2}}


    \phi = \int \overrightarrow{E}\cdot \overrightarrow{dS} = \int EdS\cos{\theta}

    dS = d(xy) = xdy + ydx


    \phi = \int \overrightarrow{E}\cdot \overrightarrow{dS} = \int E(xdy + ydx)

    I have no idea how to go from here.

    Any help will be appreciated. Thanks...
  2. jcsd
  3. Feb 19, 2008 #2
    Well you have two integrals there at the end, the integral of E*x*dy and the integral of E*y*dx, and you have that big expression for E in terms of x and y(which I only assume you did correctly)

    Go nuts
  4. Feb 19, 2008 #3
    I get that.. but how do I solve E*x*dy? I mean.. while integrating the term with 'dy' do I take 'x' to be constant and vice versa for dx?
  5. Feb 19, 2008 #4

    Except the whole affair is gonna be a little fishier than that because you have to take into account the angle between E and dS at every point. You smartly wrote E dot dS then ignored that in writing E(xdy+ydx), which would imply that the electric field and dS are always parallel, which is only true at one point; the one directly beneath the charge
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