Gauss's Law- Flux through surface

  • Thread starter chipperh
  • Start date
  • #1
5
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Homework Statement


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.


Homework Equations


Flux = E dot A (Vector math?)



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.
Chip
 

Answers and Replies

  • #2
548
2
What is the definition of an area vector, [tex] \vec{A} [/tex]? This is where you are having trouble.
 
  • #3
5
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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!
 
  • #4
5
0
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.
Chip
 
  • #5
548
2
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:

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