Electric Flux question [Surface integral]

[rohanprabhu]

Homework Statement

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.

Homework Equations

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

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 $\frac{q}{6\varepsilon_o}$..

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):

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

Also..

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

and
$$dS = d(xy) = xdy + ydx$$

so..

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

I have no idea how to go from here.

Any help will be appreciated. Thanks...

 

[blochwave]

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

 

[rohanprabhu]

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

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?

 

[blochwave]

Yes...

Except the whole affair is going to 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