Gauss's Law with NONuniform E field

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SUMMARY

The discussion focuses on calculating the net electric flux leaving a closed surface with dimensions 0.400 m x 0.400 m x 0.800 m in a nonuniform electric field described by E = (2.70 + 2.20 x²)i N/C. The correct approach involves integrating the electric field over the surface area, specifically evaluating the flux at the two ends of the closed surface at x = 0.4 m and x = 1.2 m. The total electric flux is determined to be approximately 5.097 Nm²/C after accounting for the direction of the electric field and the area vector.

PREREQUISITES
  • Understanding of Gauss's Law and its mathematical formulation.
  • Familiarity with vector calculus, particularly dot products.
  • Knowledge of electric field concepts, specifically nonuniform electric fields.
  • Ability to perform definite integrals and evaluate limits in calculus.
NEXT STEPS
  • Study the application of Gauss's Law in nonuniform electric fields.
  • Learn about vector calculus operations, especially dot products and their physical significance.
  • Explore integration techniques for calculating electric flux in various geometries.
  • Investigate the implications of electric field direction on flux calculations.
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Students studying electromagnetism, physics educators, and anyone involved in solving problems related to electric fields and flux calculations.

Alex G
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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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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
 
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
 
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
 
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!
 
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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