# Gauss's Law with NONuniform E field

by Alex G
Tags: field, gauss, nonuniform
 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 :-/
 HW Helper Thanks P: 10,537 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
 P: 374 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
HW Helper
Thanks
P: 10,537
Gauss's Law with NONuniform E field

 Quote by RTW69 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
 P: 374 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!
 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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