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Electric flux through cube with nonuniform electric field

  • Thread starter Brad155
  • Start date
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1. Homework Statement

A closed surface with dimensions a=b=0.400m and c=0.600m is located as shown in the figure (be aware that the distance 'a' in the figure indicates that the rectangular box sits distance "a" from the y-z plane). The electric field throughout the region is nonuniform and given by E=(1.50-2.00x^2)i + (y)j + (1.00-z^3)k.

A. find the flux through each side (including units and signs)

B. find the total charge enclosed in the box.

phys4-1.jpg



2. Homework Equations

flux=EA
=integral of E.dA


3. The Attempt at a Solution

for A)

now for the left and right sides i know x doesnt change, so that should just be E*A in the i direction, but y and z change so they should be integral of E.dA, and the total flux through that side would be the sum of those 3. but how do you solve an integration of y.dA? would you just evaluate (y^2)/2 from .4 to 0? if this is correct i can figure out the rest, but i dont think you can integrate y.dA, because you need dy right?

the top and bottom sides y doesnt change, so that would just be EA for the j direction, and for the front and back z doesnt change so it would be EA in the k. but the integrations are killing me....

B) since there is no enclosed charge it is 0 right?
 

ehild

Homework Helper
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Do not forget that dA is a vector, with direction normal to the surface and pointing outward, and the flux is the integral of the scalar product of the vector E with the vector dA: EdA For example, on the side of the box which is normal to the x axis, and x=a, you have EdA=ExdA= (1.50-2.00 a^2) dydz. At the parallel side, at x=0, dA points to the negative x direction, so EdA=ExdA=-1.5 dydz.

The enclosed charge is no zero, it is the sum of the flux values at all sides of the box.

ehild
 

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