Homework Help: Gauss' Law and net charge

1. The problem statement, all variables and given/known data
The box-like Gaussian surface of the figure below encloses a net charge of +57.0ε0 C and lies in an electric field given by = [(10.3 + 2.4x)·i - 3.1·j + bz·k ] N/C, with x and z in meters and b a constant. The bottom face is in the xz plane; the top face is in the horizontal plane passing through y2 = 3.5 m. For x1 = 2.5 m, x2 = 4.7 m, z1 = 2.2 m, and z2 = 3.9 m, what is b?


Sorry that I have no figure here, hopefully you can imagine what the box looks like on the xyz plane.





2. Relevant equations
I'm not sure!!!


3. The attempt at a solution

I'd like to show one, but this whole thing has me stumped!! I have no clue what they mean by 10.3+2.4x i or bz k, and I don't know how to use Gauss' Law for this. I'm sorry that this is all I have, but maybe if someone could get me started by explaining the E value first! And then which form(?) of Gauss' Law to use....(Will I be integrating or not??) Thanks in advance for any and all help.

 

So, am I using this equation, and integrating? I am so confused as to what equation to use, since the book is a little unclear. I understand what the electric field form is saying now. And I'm assuming that to find the area I just use lxwxh=A corresponding to the x, y, and z values that I have. Although, since I'm still lost about my equation of choice, I don't know if I need that???

 

whoops, my bad about the are vs volume thing, not paying attention......
And I would like to understand concepts, but the book presents this all as math problems, spending very litte time on concepts. And, though my professor is good, I only have her during class time, as she doesn't have "office hours."
What is the Qv in the above equation, does that correspond to the +57.0ε0 C in the problem? (like q enclosed?) right now I'm trying to use the integral E*dA and determining the phi value, and will do that for 6 sides, but what will that do? Won't that only give me the value I already have (+57.0...)?

 

when I do the integral for the right and left face I get something like this:

right face:
integral E*dA = integral (10.3+2.4x) dA
= 10.3+2.4(4.7) integral dA
= 21.58 * A (A=3.5*1.7=5.95)
= 128.4 Nm^2/C

left face:
-integral E*dA = -integral (10.3+2.4x)da
= -(10.3+2.4*2.5) integral dA
= -16.3 * A (A= 3.5 * 1.7=5.95)
= -96.99

is that right, and is that what the other faces will look like?

 

whooops! it is!! I got it solved! yeah! thanks!