Electric field and charge enclosed

  • Thread starter 2Pac
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  • #1
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Consider a cube of sides a=8.10 m, as shown in the diagram below. Suppose that a non-uniform electric field is present and is given by E=bx^n in the direction of the x-axis, where
b=6.35 N/(Cm^n) and n=1.9. Calculate the magnitude of the total charge within the cube.

so i started by saying
q = epsilon * integral (E*dA)
so q= epsilon * integral (6.35*x^1.9*a^2)dA evaluated from 0 to 8.1m
any help (this is the wrong answer)
 

Answers and Replies

  • #2
Doc Al
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Since you didn't include the diagram, we don't know how the cube is oriented with respect to the axes. You need to integrate [itex]\epsilon_0 \vec{E} * d\vec{A}[/itex] over the surface of the cube. Assuming the sides of the cube are aligned with the axes, the only sides that have non-zero flux are the two sides that are parallel to the y-z plane; where are they? For example, are they at x_1 = 0 and x_2 = 8.10 m? Evaluate the field on those surfaces (the field will be constant along the surface) to find the total flux. (Note: Signs matter!)
 
  • #3
the cube is along the axes. imagine the origin, then having x-point to your left, y-point to your right and z-point straight up. but im not sure how to integrate this problem..
 
  • #4
Doc Al
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the cube is along the axes. imagine the origin, then having x-point to your left, y-point to your right and z-point straight up. but im not sure how to integrate this problem..
Since the field points in the x-direction, the "integration" should be easy. See my previous suggestions.
 
  • #5
so one will be postive integral of EdA and one will be a negative integral of EdA. wouldnt those just cancel??
 
  • #6
Doc Al
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so one will be postive integral of EdA and one will be a negative integral of EdA. wouldnt those just cancel??
They would cancel if they had the same magnitude, but do they?
 

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