Electric Flux of equilateral triangle

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SUMMARY

The electric flux through an equilateral triangle with a side length of 1 meter in a uniform electric field of 8 N/C, where the normal vector makes a 60-degree angle with the field direction, can be calculated using the formula φ = EAcos(ø). The area A of the triangle is √3/4 m². The angle ø in the equation should be 30 degrees, as it is the complement of the angle between the normal and the electric field direction.

PREREQUISITES
  • Understanding of electric flux and its calculation
  • Familiarity with the formula φ = EAcos(ø)
  • Knowledge of trigonometric functions, specifically cosine
  • Basic geometry of equilateral triangles
NEXT STEPS
  • Study the concept of electric flux in different geometrical shapes
  • Learn about the implications of angle variations on electric flux
  • Explore the relationship between electric field direction and surface orientation
  • Investigate applications of electric flux in physics problems
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone interested in understanding electric field interactions with surfaces.

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


\phi=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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