Gauss's Law with NONuniform E field

In summary, the task was to calculate the net electric flux leaving a closed surface with dimensions a = b = 0.400 m and c = 0.800 m, located at position x = a. The electric field throughout the region was given by E = (2.70 + 2.20 x^2)i N/C, where x is in meters. The formula used was Integral of E(dot)dA = qenclosed/Epsilon0 and the flux was calculated by integrating E*cos(φ)*dA over the two bases (where dA is parallel to the x axis), resulting in a total flux of 5.097.
  • #1
Alex G
24
0

Homework Statement


A closed surface with dimensions a = b = 0.400 m and c = 0.800 m is located as shown in the figure below. The left edge of the closed surface is located at position x = a. The electric field throughout the region is nonuniform and given by E = (2.70 + 2.20 x2)i N/C, where x is in meters.
https://www.webassign.net/serpse8/24-p-061.gif

A)Calculate the net electric flux leaving the closed surface.

Homework Equations



For a closed surface Integral of E(dot)dA = qenclosed/Epsilon0

The Attempt at a Solution



Obviously I will have to integrate the Electric field here. I began with the simple formula
Int(E dot dA) from 1.2 to 0 because the distance at the edge where the E field leaves is 1.2m out. I believe since the sides are all parallel to the E field those become 0.
So I tried integrating
2.70 + 2.20x^2 dA from 1.2 to 0
I believe dA is x^2 (a=b=x) and dA is the combined infinitesimal areas to produce LxW of the end of the box.
The angle between the exiting E field and the Normal to the plane at the end of the box are parallel thus negligible.
What's going wrong here :-/
 
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  • #2
I do not quite follow you but you might not know that dA in the dot product E˙dA is a vector, normal to the surface and pointing outward from the slab. This dot product is equal to E*cos(φ)*dA where φ is the angle enclosed by E and dA and E and dA are the magnitudes of these vectors. At the sides of the slab, the normals of the surfaces are perpendicular to E so cos(φ)=0, the dot products cancel. At the bases, the planes located at x=0.4 and x=1.2, dA is parallel to the x axis. Its direction is opposite to E at x=a (φ=180°) and it points in the same direction as E at x=1.2 (φ=0). The magnitude of dA is dydz on these planes.


ehild
 
  • #3
This might be a bit simplistic but at x=.4 E=3.02 and area=.16 or Flux=E*A*Cos(180)

at x=1.2 E=5.58 and area=.16 or flux=E*A*cos(0)

No flux through side surfaces just ends. Add the two fluxes to get total flux.

I get a total flux of 5.097
 
  • #4
RTW69 said:
This might be a bit simplistic but at x=.4 E=3.02 and area=.16 or Flux=E*A*Cos(180)

at x=1.2 E=5.58 and area=.16 or flux=E*A*cos(0)

No flux through side surfaces just ends. Add the two fluxes to get total flux.

I get a total flux of 5.097

The procedure is OK, but check your E values.

At x=0, E=2.7+2.2*0.16=3.052
At x=1.2, E=2.7+2.2*1.44=5.868

ehild
 
  • #5
X=.4; E=2.7+2.2*0.16=3.052, Area=.16; E=3.052*.16*Cos(180)=-.488 thanks for catching my mistake!
 
  • #6
Wow thank you both, that was a lot easier than I thought. I suppose I'm just uncertain when I should integrate and when I shouldn't.
 

1. What is Gauss's Law with NONuniform E field?

Gauss's Law is a fundamental law in electrostatics that relates the electric flux through a closed surface to the charge enclosed within that surface. When dealing with non-uniform electric fields, the law takes into account the variation of the electric field magnitude and direction within the enclosed surface.

2. How is Gauss's Law with NONuniform E field different from the regular Gauss's Law?

In regular Gauss's Law, the electric field is assumed to be constant and perpendicular to the enclosed surface. However, in Gauss's Law with non-uniform electric fields, the electric field can vary in magnitude and direction within the enclosed surface.

3. What are some applications of Gauss's Law with NONuniform E field?

Gauss's Law with non-uniform electric fields is used in various fields such as electromagnetics, optics, and engineering. It is essential in calculating the electric field in complex geometries and understanding the behavior of electric charges in non-uniform electric fields.

4. How is Gauss's Law with NONuniform E field derived?

Gauss's Law with non-uniform electric fields is derived from the regular Gauss's Law by considering an infinitesimal surface element within the enclosed surface. The electric flux through this surface element is then integrated over the entire enclosed surface to obtain the total electric flux.

5. Are there any limitations to Gauss's Law with NONuniform E field?

Yes, Gauss's Law with non-uniform electric fields assumes that the electric field is continuous and differentiable within the enclosed surface. This may not be the case in some situations, such as at the edges of conductors or near point charges.

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