1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Need help in Electric Flux within a cube with Gauss' Law

  1. Sep 16, 2016 #1
    1. The problem statement, all variables and given/known data
    I need help in figuring out if I have done this problem correctly. From what I understand ∫E * dA = E*A, where E is the electric field and A is the area of a side. My biggest concern is if I can plug in the length "L" for the "x" and "z" variables within "E = -5x * E0/L i + 3z * E0/L k" to get the electric flux of each side since I assume that "L" is the value of those two variable.

    Given: [ε0, L]

    Unknown: [Electric flux of the top, bottom, front, back, left, right side, and the total]
    2. Relevant equations
    Electric Flux: φ = ∫E * dA
    E = -5x * E0/L i + 3z * E0/L k

    3. The attempt at a solution
    Page 1:

    Page 2:
  2. jcsd
  3. Sep 17, 2016 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    It would not be appropriate to set x=L and z=L ... that is not what that equation means.
    The equation ##\vec E(x,y,z) = \frac{E_0}{L}\left[-5x\hat\imath +3z\hat k \right]## tells you the value of ##\vec E## at different x,y,z positions in space.
    So the value of ##\vec E## at x=0 and z=0 is different from that at x=0 and z=L. That is why you had to do an integral.
    However, the geometry in this problem is simple enough that you will always be setting one variable to 0 or L and the flux only depends on one of the other variables.

    You proceed by picking a side, then defining a small area on that side and working out the flux through that area in terms of the position.
    Then you add up the flux through all the small areas that make up that side of the cube.
    ie for the side in the x-y plane: 0<x<L, 0<y<L, z=0
    ##d\vec A = -\hat k dxdy## is the area element between x and x+dx and y and y+dy.
    The flux through that area is ##d\phi = -\vec E(x,y,0)\cdot \hat k\; dxdy## ... and the total flux is given by the integral.
    For this surface, the only non-zero flux is parallel to the surface (there is no flux through the surface).
    ... it kinda looks like this is what you are doing but I figured I'd better check.

    Understanding that it is the flux through a side that counts, you can shortcut the process considerably ... ie, there is no component in the y direction so the flux through surfaces parallel to the x-z plane is zero.
    No further maths needed.
    Last edited: Sep 17, 2016
  4. Sep 17, 2016 #3
    Thank you, I will keep in mind to integrate the variables of dA and to be mindful of the upper and lower bounds of each problem.
    However to clear things up, let's say if I were to re-do the φTop, would it equal to:

    Etop = -5x E0/L i + 3z E0/L k
    = -5x E0/L *(|i| |k| cos(90°)) + 3z E0/L * (|k| |k| cos(0°))
    = 3 E0/L * ∫ z dz
    = 3 E0/L * z2/2 | z = L, z = 0
    = 3 E0 * L/2
    Therefore, in its whole form it looks like this:
    Etop = -5 E0/L * (0) * ∫ x dx + 3z E0/L * (1) * ∫ z dz ?

    Meaning my original answer of φTop = 3E0L2 is incorrect?
    Last edited: Sep 17, 2016
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted