Gauss's law and nonuniform electric field

  • #1

Homework Statement


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!
 

Answers and Replies

  • #2
ehild
Homework Helper
15,543
1,912
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
 
  • #3
Ah, okay. I didn't know that. But when calculating the flux of individual faces, does the constant vector contribute?
 
  • #4
ehild
Homework Helper
15,543
1,912
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
 

Related Threads on Gauss's law and nonuniform electric field

  • Last Post
Replies
5
Views
12K
  • Last Post
Replies
1
Views
669
Replies
10
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
3K
Replies
3
Views
1K
Replies
4
Views
4K
  • Last Post
Replies
1
Views
4K
Replies
3
Views
1K
Replies
19
Views
5K
Top