Gauss's Law with NONuniform E field


by Alex G
Tags: field, gauss, nonuniform
Alex G
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#1
Feb8-11, 09:31 PM
P: 24
1. The problem statement, all variables and given/known data
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.

2. Relevant equations

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

3. 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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ehild
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#2
Feb8-11, 11:30 PM
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P: 9,812
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
RTW69
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#3
Feb9-11, 12:37 AM
P: 372
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

ehild
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#4
Feb9-11, 01:13 AM
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Gauss's Law with NONuniform E field


Quote Quote by RTW69 View Post
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
RTW69
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#5
Feb9-11, 01:29 AM
P: 372
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!
Alex G
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#6
Feb9-11, 08:02 AM
P: 24
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.


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