Question Related to Electric Flux

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SUMMARY

The discussion focuses on calculating electric flux through a tetrahedral pyramid situated on an infinite sheet of charge with surface charge density σ. The electric field is determined using the formula E = -σ/ε₀, leading to the total electric flux through the base face as [(3)√3/(8)] × [(σ)/(ε₀)] × [a²]. Since the pyramid contains no net charge, the total flux through all faces is zero, resulting in the flux through each of the other three faces being [(3)√3/(24)] × [(σ)/(ε₀)] × [a²]. The calculations are based on the geometric properties of equilateral triangles and the principles of electric flux.

PREREQUISITES
  • Understanding of electric flux and its mathematical representation.
  • Familiarity with the concept of surface charge density (σ).
  • Knowledge of the electric field due to a charged surface (E = -σ/ε₀).
  • Basic geometry of equilateral triangles, including area calculation.
NEXT STEPS
  • Study the derivation of electric flux for different geometries, focusing on pyramids and tetrahedrons.
  • Learn about Gauss's Law and its application in calculating electric fields and flux.
  • Explore the implications of net charge in closed surfaces and its effect on electric flux.
  • Investigate the relationship between surface charge density and electric field strength in various configurations.
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Physics students, educators, and anyone interested in understanding electric flux calculations and their applications in electrostatics.

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 .
 
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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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