- #1

rohanprabhu

- 414

- 2

## 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

[tex]

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

[/tex]

## 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):

[tex]

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}}

[/tex]

Also..

[tex]

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

[/tex]

and

[tex]

dS = d(xy) = xdy + ydx

[/tex]

so..

[tex]

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

[/tex]

I have no idea how to go from here.

Any help will be appreciated. Thanks...