How to find electric flux through a surface due a point charge

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.

There are two things that need sorting out. Are you looking for electric field flux or displacement field flux? By context I think you are looking for displacement field flux. If you can't think of a way to calculate ${\bf D}$ directly, perhaps you can find it another way. You would have to make the reasonable assumption that the problem lies in a vacuum. — garyp, Sep 14 '21 at 17:35
Say you considered vacuum. Then the equation would be $\Psi = \oint \vec{E}\cdot \vec{da}$. Assume small area $dxdy$ on the plane. This area points in the $z$ direction. Calculate $\vec{E}$ for some general point $(x, y, z_0)$. Evaluate the flux $\oint Eda\cos\theta$ involving double integrals with the limits for $x$ and $y$ given. — Sumit Gupta, Sep 14 '21 at 17:41
Related : What is the electric field flux through the base of a cube from a point charge infinitesimally close to a vertex?. — Frobenius, Sep 14 '21 at 19:14

score 1 · Answer 1 · answered Sep 14 '21 at 19:35

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:

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

How to find electric flux through a surface due a point charge

1 Answers1