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Calculate flux through face of a cube

  • Thread starter mindarson
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  • #1
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Homework Statement



A point charge q is located at the corner of a cube with edges of length
L. Find the electric flux through the shaded face at y = L.


Homework Equations



Gauss's Law; flux = ∫E[itex]\bullet[/itex]n dA = qenc



The Attempt at a Solution



Here's what I'm thinking. In the problem statement, the charge q is located at a corner of the cube. I can't use Gauss's Law, due to lack of symmetry. However, I can draw a bigger cube instead, whose center is at the location of the charge. Then I have symmetry to use Gauss's law and say that the flux through the whole cube is q/ε. The face whose flux the problem wants me to calculate is now just 1/24 of the surface area of the whole (bigger) cube. Therefore I can say that the flux through the original face is just q/(24ε).

Is this correct, or am I just trying to be too clever by half, as they say?
 
Last edited:

Answers and Replies

  • #2
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This is correct and in fact is the manner in which the problem is solved in the solution manual. Good job!
 
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  • #3
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Nice! Seemed a bit too good to be true. :)
 

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