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!

Gauss's law and nonuniform electric field

  1. Jul 4, 2012 #1
    1. The problem statement, all variables and given/known data
    A Gaussian surface is in the shape of a cube with edge length 1.40m. The electric field is E=[-4i+(6+3y)j]N/C.

    I got an answer, but the solution manual stated that we treat the electric field as E=3yj+E0, where E0=-4i+6j, which does not contribute to the flux. Why is this? Please help!
     
  2. jcsd
  3. Jul 4, 2012 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The flux of a constant vector is zero as the contribution from one side of the cube is positive and negative from the opposite side.

    ehild
     
  4. Jul 4, 2012 #3
    Ah, okay. I didn't know that. But when calculating the flux of individual faces, does the constant vector contribute?
     
  5. Jul 4, 2012 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You know how to get a flux of a vector field B(r) at an surface area A? you integrate the normal component of the vector B : Φ=∫BndA.
    The normal component refers to the outward normal. Although there is nonzero flux from the constant vector on on side, there is the same with opposite sign on the opposite side.

    ehild
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Gauss's law and nonuniform electric field
Loading...