How Is Electric Flux Calculated Through a Cube's Face with an Enclosed Charge?

[_F_]

Homework Statement

A 2.6-uC charge is at the center of a cube 7.5cm on each side. What is the electric flux through one face of the cube?

q_enclosed = 2.6*10^-6 C
a = .075m

Homework Equations

flux = E*A = integral(E*dA*cos(theta)) = E*integral(dA)
flux = q/(e_0)
surface area of a cube: 6*a^2

The Attempt at a Solution

flux = E*int(dA) = E*6(.075)^2 = .03375E

.03375E = (2.6*10^-6)/(e_0) => E = 6.8*10^-16 N/C

So the flux = q/(e_0).



... you know what. I just figured it out.

flux = q/(e_0) => (2.6*10-3C) / (e_0) = 294053 Nm^2/C

(1/6)(294053 Nm^2/C) = 4.9*10^4 Nm^2/C

So the flux through one side of the cube (any cube!) with the enclosed charge of 2.6-uC is 294053 Nm^2/C.
No integral is necessary.


edit: I was going to not post this but maybe it will help someone out.

 

[Jhero]

If you're doing this problem, you need to know that the flux is just the electric field strength times the area. The electric field strength is equivalent to the charge enclosed divided by the permittivity of free space. The surface area of a cube is just 6 times the length of one side squared. So you can plug in the values and solve for the flux. No integral is needed.