EPTSo I've been stuck on this problem for two days.
The premise is that there is a cube with edges on the axes and a corner at the origin in the first quadrant.
The sides of the cube have a length:
a = 0.410 m
We are looking only at the top side of the cube, in the plane z = 0.410
The Electric field E is given by:
E = (4.4z*J + 4.4y*K) N/C,
Where J is the unit vector in the positive y-direction and K is the unit vector in the positive z-direction.
What the problem wants to know is what the electric flux, PHI is, through the top surface of the cube.
Gauss' Law states that PHI = integral(E dot dA), where dA and E are vectors.
dA's direction is straight up, so only the K component of E plays a role in determining the flux.
I'm getting stuck because I found what A (the area vector of the surface) is: 0.1681 K.
A dot E = 0.7396y
I'm inclined to integrate this from 0 to a to get my result, but the online system I'm using to complete this problem says that that is incorrect. I'm not sure what to use as dA, it's supposed to be an infinitesimal area in the direction K, so should I just integrate the K component of E over dA = dxdy?
For reference: this is the picture and actual wording of the problem.
A cubic cardboard box of side a = 0.410 m is placed so that its edges are parallel to the coordinate axes, as shown in the figure. There is NO net electric charge inside the box, but the space in and around the box is filled with a nonuniform electric field of the following form: E(x,y,z) = K*z j + K*y k, where K= 4.40 N/C.m is a constant.
![[image loading]](http://img390.imageshack.us/my.php?image=prob07acubefa4.gif)
What is the electric flux through the top of the box?
Solution:
+ Show Spoiler +
PHI = Int(0->a) Int(0->a) (4.4y) dx dy
PHI = Int(0->a) 4.4*a*y dy
PHI = (4.4*a*y^2)/2 ]from 0->a
PHI = 2.2*a^3 = 0.151626 N*m*m/C
PHI = Int(0->a) 4.4*a*y dy
PHI = (4.4*a*y^2)/2 ]from 0->a
PHI = 2.2*a^3 = 0.151626 N*m*m/C
