Flux through an infinite plane due to a point charge

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SUMMARY

The discussion focuses on calculating the electric flux through an infinite plane due to a point charge located a distance d from the plane. The correct approach involves considering an infinite cuboid surrounding the point charge, leading to a total flux of ## \frac{q}{\epsilon_0} ## through all surfaces. However, the flux through the two infinite planes at infinity is negligible due to the electric field approaching zero at that distance. Ultimately, the flux through one infinite plane is determined to be ## \frac{q}{2\epsilon_0} ##.

PREREQUISITES
  • Understanding of electric flux and Gauss's Law
  • Familiarity with point charges and their electric fields
  • Knowledge of the concept of infinity in physics
  • Basic calculus for evaluating integrals related to electric fields
NEXT STEPS
  • Study Gauss's Law and its applications in electrostatics
  • Learn about electric fields generated by point charges
  • Explore the concept of electric flux in different geometries
  • Investigate the behavior of electric fields at infinity
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone studying electrostatics, particularly those interested in electric flux calculations and the implications of infinite planes in electric field theory.

Pushoam
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Homework Statement


A point charge q is located a distance d meters from an infinite plane. Determine the electric flux through the plane due to the point charge.
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The Attempt at a Solution


I consider another infinite plane at a distance d in the opposite direction. Now I have infinite cuboid.

The flux through all the surfaces of the cuboid is ## \frac {q } { \epsilon _0} ##.

Since the plane with length d is smaller than the other two planes, the flux through these two planes is negligible in comparison with that due to the other two infinite planes. The other two infinite planes have equal flux. So, the flux through one infinite plane is ## \frac {q } { 2 \epsilon _0} ##.

Is this correct?
 
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Not quite right. It is not because the size is negligible but the electric field is zero at infinity.
 
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Yes, the other two surfaces are at an infinite distance from the charge, so the electric field is negligible and hence the flux. Thanks for pointing it out.
 

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