Electric flux through a square lamina

[Hamiltonian]

Homework Statement

find the electric flux through a square sheet of side length $a$ due to a charge $q$ placed a distance $a$ away from the centre of the square.

Relevant Equations

$\phi = \int \vec E.\vec{da}$

taking origin at the centre of the square.
$d\phi = \vec E.\vec{da}$
$$d\phi = \frac {kqa}{(x^2 + y^2 + a^2)^{3/2}} da$$
$$\phi = \int_{-a/2}^{a/2}\int_{-a/2}^{a/2}\frac{kqa}{(x^2+y^2+a^2)^{3/2}}(dx)(dy)$$
on evaluating this double integral i get $$\phi = (q/\pi{\varepsilon}_0 )(tan^{-1}(\frac{1}{2\sqrt{6}}))$$
the correct answer is supposed to be:

where am I going wrong?

 

[TSny]

Check to see if your answer and the given answer are actually equal to one another.

 

[Hamiltonian]

Check to see if your answer and the given answer are actually equal to one another.

I used a calculator and they are not equal

 

[TSny]

Make sure your calculator is in radian mode.

 

[Steve4Physics]

$$\phi = \int_{-a/2}^{a/2}\int_{-a/2}^{a/2}\frac{kqa}{(x^2+y^2+a^2)^{3/2}}(dx)(dy)$$
on evaluating this double integral i get $$\phi = (q/\pi{\varepsilon}_0 )(tan^{-1}(\frac{1}{2\sqrt{6}}))$$
the correct answer is supposed to be:
View attachment 281837
where am I going wrong?

Your double integral for $\phi$ looks OK. But there is no way to tell where (or even if) you are 'going wrong' evaluating the double integral without seeing all the detailed working.

A couple of hints which can help simplify the working:

- because of the symmetry, you can just find the flux through one quadrant of the square and multiply this by 4: the integration limits become simpler (from 0 to a/2);

- without loss of generality you can choose any value for length 'a' (I'll let you think about why!);

The working might then be simplified enough for you to spot any error(s) more easily.

By the way, using the same symbol ('a') for both length and area is not a good idea!