So the problem is a point charge is located at the origin of the coordinate system and a cube of side length 2a is centered at the origin and I am trying to find the electric field, and flux due to the point charge.
So far, I have said that the electric field, $E = \frac{q(r_2-r_1)}{4πe_o|r_2-r_1|^3}$ and attempted to convert from radial to cartesian coordinates. To do that, I replaced $r_2 – r_1$ in the equation with the cartesian coordinates $<x_2 – x_1, y_2 – y_1, z_2 – z_1>$. Firstly, would this step be correct? It seems as if I might be overlooking something, as I could use identities like x = $r\sin(\theta)\cos(\phi)$, but I can't figure out how to work it into the equation.
If it is correct, I should be able to get something like $E(x,y,z) = \frac{q(x_2-x_1, y_2 – y_1, z_2 – z_1)}{4πe_o|x_2 – x_1, y_2 – y_1, z_2 – z_1|^3} $, and since the point charge at the origin is at (0,0,0), simplifies down to $E(x,y,z) = \frac{q(x_2, y_2, z_2)}{4πe_o|x_2^ , y_2 , z_2|^3} = \frac{q(x_2, y_2, z_2)}{4πe_o(x_2^2 + y_2^2 + z_2^2)^{3/2}} $ But my main problem is that I don't know how I would integrate such an equation in order to get the flux through the surface.
Also, once I finally integrate this equation, should I get something that would be consistent with Gauss's law, so that the total flux is equal to $Q_{enc}/e_0$, or am I mistaken?
Edit:
I believe that I can find the flux through one face using $\iint\limits_ \, \mathrm{E} \cdotp {d}S$. If I use the face along the y-axis, I can set ${d}S = \hat{y}dxdz$ (would y be a unit vector in this case) and set $y = a$, ending with the integral $4\int_0^{a} \int_0^{a} \frac{qa}{4πe_o(x^2 + a^2 + z^2)^{3/2}}$.
Best Answer
No, you're not mistaken. If the charge is in fact located in the center of the cube, then by Gauss' law the net flux over the entire surface of the cube is $Q_{enc}/e_0$ and by symmetry the net flux through each face of the six faces of the cube is $Q_{enc}/6e_0$.
Only if you want to know the net flux over a limited surface of a face of the cube would you need to integrate. That's because the flux is $\int E.dA$ where $E$ and $A$ are vectors with the $A$ normal (perpendicular) to the surface. The flux is then maximum near the center of each face where $E.dA$ is maximum and minimum near the edges of each face where $E.dA$ is a minimum.
Hope this helps.