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: Flux through one side of a cube

  1. Dec 21, 2008 #1
    1. The problem statement, all variables and given/known data

    A charge q sits at the back corner of a cube. What is the
    flux of E through the opposite (front)side?

    2. Relevant equations


    3. The attempt at a solution

    As the flux through the whole cube must be q/ε_0, I thought that for one side it would just be q/6ε_0, but the answer is q/24ε_0. I don't know how the position of the enclosed charge affects the flux through the side of the cube.
  2. jcsd
  3. Dec 21, 2008 #2


    User Avatar
    Science Advisor

    Why would you think that? The charge is NOT symmetrically situated with respect to the sides of the cube. You should expect more flux through those sides closer to the charge. The problem I have is that if the charge is situated at one corner then there are 3 different faces you could reasonably call "opposite". Strictly speaking what you need to do is integrate the field over the face of the cube:
    [tex]\int\int \vec{e}\cdot d\vec{S}[/tex]
    where [itex]\vec{e}[/itex] is the field strength and [itex]d\vec{S}[/itex] is the (vector) differential of surface area.
  4. Dec 21, 2008 #3


    User Avatar
    Science Advisor
    Homework Helper

    If the charge is located AT the corner then only 1/8 of the total flux of the charge enters the cube. There is no flux through the three faces which touch the charge since the field is tangent to those faces. So the 1/8 of the total flux exits through the three opposite faces. Their orientation is symmetric with respect to the charge, so the same amount of flux passes through each.
  5. Dec 21, 2008 #4
    ok, so would I put the whole thing into cartesian coordinates? I don't know how to integrate that when dS is not spherical.
  6. Dec 21, 2008 #5
    Great thanks, that makes sense.
  7. Dec 22, 2008 #6
    The reason that 1/8th of the flux is considered is cause while considering such problems we enclose the charge symmetrically. In this case if the side length is A. We add seven more cubes and make a cube of side 2A. Now flux passing through this is q/epsilon. So now flux passing through one cube is q/8epsilon.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook