A closed surface with dimensions a = b = 0.400 m and c = 0.800 m is located as shown in the figure below. The left edge of the closed surface is located at position x = a. The electric field throughout the region is nonuniform and given by E = (2.70 + 2.20 x2)i N/C, where x is in meters.
A)Calculate the net electric flux leaving the closed surface.
For a closed surface Integral of E(dot)dA = qenclosed/Epsilon0
The Attempt at a Solution
Obviously I will have to integrate the Electric field here. I began with the simple formula
Int(E dot dA) from 1.2 to 0 because the distance at the edge where the E field leaves is 1.2m out. I believe since the sides are all parallel to the E field those become 0.
So I tried integrating
2.70 + 2.20x^2 dA from 1.2 to 0
I believe dA is x^2 (a=b=x) and dA is the combined infinitesimal areas to produce LxW of the end of the box.
The angle between the exiting E field and the Normal to the plane at the end of the box are parallel thus negligible.
What's going wrong here :-/