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- Flux through surface

  1. Feb 7, 2009 #1
    1. The problem statement, all variables and given/known data
    A flat surface with area .14 m^2 lies in the x-y plane, in a uniform electric field given by E=5.1i +2.1j+3.5k kN/C.
    A) Find the flux through the surface.

    2. Relevant equations
    Flux = E dot A (Vector math?)

    3. The attempt at a solution
    I believe this is the dot product of two vectors. Converting the surface area to vectors (i,j,k) I come up with (.374i, .374j, 0k). When I calculate the dot product:
    (5100, 2100, 3500) (.374, .374, 0) I get 2694 N M^2/C (wrong). I believe the plane vectors are incorrect. Guidance please?

    Thanks again.
  2. jcsd
  3. Feb 7, 2009 #2
    What is the definition of an area vector, [tex] \vec{A} [/tex]? This is where you are having trouble.
  4. Feb 8, 2009 #3
    Thank you for the push in the direction. I understand I have to convert the area into a vector. I have been looking through my old notes and texts on this subject. Since the area is assumed to be flat (no 'z' (k) component), I assumed zero for that value. I took the square root of .14 m^2 (oops) .... I think this is where I made my mistake.
    Will forge ahead!
  5. Feb 8, 2009 #4
    Ok, with this problem, I am obtaining the dot product of the two vectors. I believe the vector for the 'plane' would be N= [.14 + .14 + 1]
    When I do the dot product as follows:
    Ex*Nx + Ey*Ny + Ez*Nz = # NM^2/C

    5100*.14 + 2100*.14+3500*1 = 4508 N M^2/C for the flux through the surface. This is wrong though. Another nudge please?

    Thank you.
  6. Feb 8, 2009 #5
    According to your most recent response, your definition of the area vector is still not correct. See the link: http://en.wikipedia.org/wiki/Vector_area
    According to this problem, there is only one nonzero component of the area vector.
    Last edited: Feb 8, 2009
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook