Electric Flux of equilateral triangle

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Homework Help Overview

The discussion revolves around calculating the electric flux through an equilateral triangle in a uniform electric field, specifically addressing the relationship between the angle of the normal vector and the electric field direction.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of different angles in the flux equation, questioning whether the angle used should be 60 degrees or 30 degrees based on the normal vector's orientation.

Discussion Status

Some participants have provided examples to clarify the relationship between the angle of the electric field and the normal vector, leading to a deeper exploration of the angle's role in the flux calculation. The conversation is ongoing, with no explicit consensus reached yet.

Contextual Notes

The original poster's problem includes specific values for the electric field and the dimensions of the triangle, but there may be ambiguity regarding the angle used in the flux formula.

msampsel
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Homework Statement


What is the flux of the uniform electric field E=8 N/C through an equilateral triangle with the side a=1m, if the normal vector makes 60 degrees to the direction of the electric field.

Homework Equations


[itex]\phi[/itex]=EAcosø

The Attempt at a Solution


E=8 N/C
A= √3/4 m^2

Is the ø in this equation 60 degrees or 90-60 = 30 degrees?
 
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Hello msampsel. Welcome to PF!

See if you can answer this question by considering a couple of special examples:

(1) What should the flux be if the angle between the normal and the field is 0? What value of ø in the formula will give this result?

(2) What should the flux be if the angle between the normal and the field is 90? What value of ø in the formula will give this result?
 
(1) The flux would be the product of the field and the area (EA) because they are perpendicular to the plane. The value of ø would be 0 because cos(0)=1.

(2) The flux would be 0 because the field would then be parallel to the plane. The value of ø would be 90 because cos(90)=0.
 
OK, good. Based on those two examples, how is the angle ø in the formula related to the angle between the field and the normal direction?
 

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