Find Electric Flux Through Surface 2

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SUMMARY

The discussion focuses on calculating the electric flux through surface (2) using Gauss' Law. The area of surface (2) is 3.90 m², and the uniform electric field has a magnitude of 200 N/C. The correct angle to use in the calculation is 55°, derived from the 90° angle between the surfaces minus the 35° angle of the electric field. The correct formula for electric flux is EAcosθ, leading to a final flux calculation of approximately 638.9 N·m²/C.

PREREQUISITES
  • Understanding of Gauss' Law
  • Knowledge of electric flux calculations
  • Familiarity with trigonometric functions, specifically cosine
  • Basic concepts of electric fields and their properties
NEXT STEPS
  • Study the application of Gauss' Law in different geometries
  • Learn about the relationship between electric fields and potential energy
  • Explore advanced topics in electromagnetism, such as Maxwell's equations
  • Investigate the effects of angle on electric flux in various configurations
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This discussion is beneficial for physics students, educators, and anyone involved in studying electromagnetism and electric field calculations.

cass
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Homework Statement



The drawing shows an edge-on view of two planar surfaces that intersect and are mutually perpendicular. Surface (1) has an area of 2.20 m², while surface (2) has an area of 3.90 m². The electric field in the drawing is uniform and has a magnitude of 200 N/C. Find the magnitude of the electric flux through surface (2) if the angle θ made between the electric field with surface (2) is 35.0°.
18-81.jpg

Homework Equations



I used gausse law

The Attempt at a Solution



(200)(2.20)cos(35)=360.4
(200)(3.90)cos(35)=638.9
Sinnce they are looking for magnitude of surface 2 I thought the answer was 638.9 but I got it wrong.[/B]
 
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You used cos(35) when calculating the flux through surface 2. Think about whether or not you used the correct angle here.
 
what do u mean?
 
ok since it is a 90 degree angle then subtracting 35 should give me 55 right?
 
cass said:
ok since it is a 90 degree angle then subtracting 35 should give me 55 right?
Yes, using Gauss' Law, EAcosθ, where θ refers to the angle between the electric field's direction and the line perpendicular to the area of application (which is the dot product of E and A, E⋅A)
 
thank u!
 

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