Gauss's Law- Flux through surface

In summary, the problem involves finding the flux through a flat surface with an area of .14 m^2 in a uniform electric field given by E=5.1i +2.1j+3.5k kN/C. The flux is calculated using the dot product of the electric field and the area vector, with the area vector having only one nonzero component.
  • #1
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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
 
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  • #2
What is the definition of an area vector, [tex] \vec{A} [/tex]? This is where you are having trouble.
 
  • #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!
 
  • #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.
Chip
 
  • #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.
 
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1. What is Gauss's Law and what does it state?

Gauss's Law is a fundamental law of electromagnetism that relates the distribution of electric charges to the resulting electric field. It states that the electric flux through a closed surface is equal to the enclosed electric charge divided by the permittivity of free space.

2. What is electric flux and how is it calculated?

Electric flux is a measure of the amount of electric field passing through a given surface. It is calculated by taking the dot product of the electric field and the surface area vector, and then integrating over the entire surface.

3. How is Gauss's Law used to calculate the electric field of a point charge?

Gauss's Law can be used to calculate the electric field of a point charge by choosing a closed surface that encloses the charge and is symmetric with respect to the charge. The electric flux through this surface is then calculated and set equal to the enclosed charge divided by the permittivity of free space. This equation can then be solved for the electric field at any point on the surface.

4. Can Gauss's Law be applied to any closed surface?

No, Gauss's Law can only be applied to closed surfaces that are either completely enclosed by electric charges or are infinitely large and have a uniform electric field. This is because Gauss's Law relies on symmetry to make the calculation of electric flux easier.

5. What are some real-world applications of Gauss's Law?

Gauss's Law has many practical applications, such as in the design of capacitors, electric motors, and generators. It is also used in the analysis of lightning strikes and in determining the electric field inside a conducting shell. Additionally, Gauss's Law is an important tool in understanding and solving problems in electrostatics and electromagnetism.

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