Gauss's Law with NONuniform E field

Click For Summary

Homework Help Overview

The problem involves calculating the net electric flux through a closed surface in the presence of a nonuniform electric field described by the equation E = (2.70 + 2.20 x^2)i N/C. The closed surface has specific dimensions and is positioned in a way that the electric field varies across it.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to integrate the electric field over the surface but questions the treatment of the area elements and the angles involved in the dot product. Some participants clarify the vector nature of dA and the implications for calculating flux through different surfaces of the closed shape.

Discussion Status

Participants are actively discussing the integration approach and the calculation of electric flux. Some have provided corrections to the original poster's calculations, while others are confirming the understanding of the geometry and the electric field's behavior at different points on the surface.

Contextual Notes

There is an emphasis on understanding when to apply integration versus direct calculation, as well as the importance of correctly interpreting the electric field's direction and the area elements involved in the flux calculation.

Alex G
Messages
24
Reaction score
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 :-/
 
Physics news on Phys.org
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.
 

Similar threads

Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 30 ·
2
Replies
30
Views
2K
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 12 ·
Replies
12
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
10
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K