1. Not finding help here? Sign up for a free 30min 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!

Electric Flux on one side

  1. Jun 13, 2008 #1
    1. The problem statement, all variables and given/known data

    A 1.3microC point charge is placed at the center of a cube with a volume of 9.8m^3. Calculate the electric flux through one side of the cube.



    2. Relevant equations

    E = [tex]\frac{k*q}{r^2}[/tex]

    [tex]\Phi[/tex] = E*A



    3. The attempt at a solution


    So what i was trying was this:

    I cube root the volume of the cube. This gives me the length of any side since its a cube. So then, i calculate E where k = 9.0E9 q = 1.3E-6 and r = 1.07 (this is half of one length of one side). I then put E back into the electric flux equation where A is (2.14^2)*6 (the total surface area of the cube) i then take my answer and divide by 6 to get the flux for one side which comes out to 4.68E4 N*m^2/C which is incorrect. Im thinking ive made a mistake calculating the Electric field (the r value specifically, since it isnt really constant. ie if we wanted to know the flux in the corner of the cube, we would need to use trig to get the r value. But i cant see what else the r value would be.) A little direction would be great :D


    my picture (not given):
    [​IMG]
     
    Last edited: Jun 13, 2008
  2. jcsd
  3. Jun 13, 2008 #2
    My advice is to think in terms of electric field lines.
    In order to avoid crazy integrals, simplify the problem to a sphere.
    Solve for the flux on a sphere of arbitrary radius, then divide by six.
    You see, the surface area of a sphere varies with r2, whereas the E-field varies as r-2, which cancel.
     
  4. Jun 13, 2008 #3
    Fantasic, i got it right. Thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?