How to find electric flux through a surface due a point charge [closed]

Question

So, I'm trying to solve this problem for my assignment,

A point of charge, Q = 3 nC, is located at the origin of a cartesian coordinate system. What flux crosses the portion of the z=2 m plane for which -4 $\le$ x $\le$ 4 m and -4 $\le$ y $\le$ 4 m

I'm thinking this question can be solved by using this equation $$\Psi = \oint \mathbf{D} \cdot d\mathbf{S}$$ because the limits of the integration are already known from the problem

but I don't understand how I can get the D of this question. Am I using the wrong formula? can someone clear this out for me? Thanks.

Answer 1

Consider some point $(x,y)$ on the surface of interest. The distance of that point to the origin is $R=\sqrt{x^2+y^2+2^2}$. The angle of elevation for the vector that connects the point $(x, y)$ to the origin is $\theta=\cos^{-1}(\frac{2}{R})$

We know the Electric Field at the point $(x,y)$ will given by:

$$\vec{E}=\frac{kQ}{|\vec{R}|^2}\frac{\vec{R}}{|\vec{R}|}$$

And so:

$$\vec{E} \cdot \hat{n}=\frac{kQ}{|\vec{R}|^2}\Big(\frac{\vec{R}}{|\vec{R}|} \cdot \hat{n}\Big)=\frac{kQ}{|\vec{R}|^2}\cos(\theta)=\frac{kQ}{|\vec{R}|^2}\frac{2}{|\vec{R}|}=\frac{2kQ}{(x^2+y^2+2^2)^{3/2}}$$

From there we find the value of flux through the surface to be:

$$\Phi_E=\int_{-4}^4 \int_{-4}^4 \frac{2kQ}{(x^2+y^2+2^2)^{3/2}} dxdy$$