Flux linked with lower face of a cube

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SUMMARY

The discussion centers on calculating the electric flux through the lower face of a cube when a point charge q is placed inside it. Participants emphasize the application of Gauss's Law, specifically the equation ∫B.dl = 1/ε° × Charge Enclosed. The correct approach involves recognizing that if the charge is at the center of the cube, the flux through each face, including the lower face ABCD, is q/6ε°. Misinterpretations of Gauss's Law were clarified, ensuring accurate application in solving the problem.

PREREQUISITES
  • Understanding of Gauss's Law in electrostatics
  • Familiarity with electric flux concepts
  • Knowledge of point charge behavior in electric fields
  • Basic principles of symmetry in three-dimensional shapes
NEXT STEPS
  • Review the derivation of Gauss's Law and its applications
  • Study electric flux calculations for different geometries
  • Explore the implications of charge placement within various shapes
  • Practice problems involving multiple charges and their combined flux
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone looking to deepen their understanding of electric flux and Gauss's Law applications.

premraj59
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Homework Statement


A point charge q is placed inside a cube of side 2a. What will be the flux associated with the lower surface ABCD?

Homework Equations


I think I can apply Gauss Law here, but can't think of something connecting it with the lower surface.
∫B.dl = 1/ε° X Charge Enclosed

The Attempt at a Solution


I tried to solve it by taking charge as q/6 (as cube has 6 faces) but the solution doesn't seem to work out.
 
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You have not given enough information. Please reproduce the problem statement exactly as given.

Also, do not just say ”it does not seem to work”. In order to help you we must know exactly what you have done and what you obtained, not a vague description.
 
premraj59 said:
A point charge q is placed inside a cube of side 2a.
Where? At the center?

premraj59 said:
∫B.dl = 1/ε° X Charge Enclosed
That's not Gauss' law. Look it up.

premraj59 said:
I tried to solve it by taking charge as q/6 (as cube has 6 faces) but the solution doesn't seem to work out.
That should work fine, once you apply the right law. (Assuming the charge is at the center of the cube.)
 

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