Question Related to Electric Flux

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  • #1
michael650
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I am having a hard time understanding the reasoning behind my answers on this physics homework. We are currently practicing electric flux (without Gauss Laws) and the answer I got was correct, I just do not understand how I got it. The question is...

"The four faces of a tetrahedral pyramid are equilateral triangles of a side a Such a pyramid sits with one face flat on an infinite sheet of charge with a surface charge density sigma. What is the flux through the face of the pyramid that sits on the sheet? What is the flux through each of the other three faces?"

My solutions manual tells me that I should treat the surface charge as positive. From that I can calculate the electric field algebraically as E = -(sigma)/(epsilon nought). Apparently because of that the electric flux is: [(3)1/2/(8)] x [(sigma)/(epsilon nought)] x [a2]

From what it describes, the pyramid itself contains no net charge, so the total flux through all of its faces must be zero. The area vectors for the remaining three faces all point outward and give a positive total flux. Since the total flux is zero, the flux of each of the remaining faces is: [(3)1/2/(24)] x [(sigma)/(epsilon nought)] x [a2].

Can someone please explain how they went about calculating these equations algebraically .
 

Answers and Replies

  • #2
tiny-tim
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welcome to pf!

hi michael650! welcome to pf! :smile:

(have an epsilon: ε and a sigma: σ and a square-root: √ and try using the X2 tag just above the Reply box :wink:)

hint: what is the area of an equilateral triangle of side a ? :wink:
 

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