Solving for net flux, given x- and y- components of electric field, E

[Lemon-Sam]

Homework Statement

A Gaussian surface in the shape of a cube has edge length 1.14 m.

What are (a) the net flux through the surface and (b) the net charge (in C) enclosed by the surface if the electric field in the region is in the positive y direction and has a magnitude that is given by E = 4.38y N/C?

What is (c) the net flux if the electric field is in the xy plane and has components Ex = -3.42 N/C and Ey = (5.19 + 3.74y) N/C?

Homework Equations

Gauss's Law: ε0*net flux = q

Electric flux through a Gaussian surface: ∫E*dA

The Attempt at a Solution

I've already solved for the correct solutions for parts A and B:

Part A:

Area of the surface, A = (1.14m)² = 1.2996 m²;

Thus, the net flux through the surface is (4.38)(1.14)(1.14)² = 6.49 Nm²/C

Part B:

Then the charge, q, is ε0*net flux from part A:

(8.85 x 10^-12 C²/Nm²)(6.49 Nm²/C) = 5.74 x 10^-11C

Part C: Here's where I ran into problems. Based on an example problem in my text, I reasoned that x-and y-components of E given in part C make up a constant field and thus don't affect the net flux. So, I'd end up with the same result as in Part A, but that's incorrect. I'm not sure how to go about solving the problem at this point.

 

[Spinnor]

You wrote,

What is (c) the net flux if the electric field is in the xy plane and has components Ex = -3.42 N/C and Ey = (5.19 + 3.74y) N/C?


Ey is not constant, it changes with y? Ey = (5.19 + 3.74y) N/C

 

[Lemon-Sam]

Yes, now I see where I went wrong with the problem. I have to take the non-constant part of Ey, 3.74y and solve for (c) in the same way that I solved part a.