Understanding Gauss' Law and E Fields in Flux Calculations

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SUMMARY

This discussion focuses on Gauss's Law and its application in calculating electric fields and flux. The user questions whether Gauss's Law represents the projection of a shape onto a perpendicular surface multiplied by the electric field (E), and explores the implications of this for different geometries, such as a half sphere in a horizontal E field. Additionally, the user inquires about the calculation of the electric field (E) from an infinite plane using the formula E = σ / (2 ε₀), specifically regarding its applicability at varying distances from the surface.

PREREQUISITES
  • Understanding of Gauss's Law
  • Familiarity with electric field concepts
  • Knowledge of flux calculations
  • Basic grasp of vector mathematics
NEXT STEPS
  • Study the derivation and applications of Gauss's Law in electrostatics
  • Learn about electric field calculations for different geometries, including spheres and planes
  • Explore the implications of scalar fields in physics
  • Investigate the concept of Gaussian surfaces and their role in flux calculations
USEFUL FOR

Students of physics, particularly those studying electromagnetism, educators teaching electric field concepts, and anyone seeking to deepen their understanding of Gauss's Law and its applications in real-world scenarios.

EV33
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Homework Statement


I actually just have two questions about the law itself.

1.) Is Gauss's Law just the projection of a shape onto a perpindicular surface to the E field,times the E field?

I ask this because there is an example in my book which shows a 3D incline with a horizontal E field, and the flux through the vertical is the same has the flux through the hypotenuse side of the incline. The only difference is that one is EA and the other is E*A*cos(theta).

So under my assumption on flux if there is a horizontal e field, and a half sphere with its flat side perpindicular to the e field, I would assume that the flux would be the same through both the flat and round side.

2.) My question for this one involves E fields involving infinite large planes. If you calculate the e field involving a plane you can do it with the equation
E=sigma/2 epsilon nat, but my question about this is, where does that calculate the E field at. My assumption is that it calculates the E field right at the surface. If this is so and I wanted to find the E field some x distance away, would I just put that E value over X^2, to get the E value at X distance away?


Homework Equations


flux=E(dot)A
Flux=Q enclosed/epsilon nat
E=sigma/2 epsilon nat

The Attempt at a Solution



1.)Well I figured for the half circle that both sides would have to have equal but opposite flux's because if q inside is zero then the net flux for the object needs to be zero, so the neg flux needs to cancel out the pos flux for the other side.
 
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Hope this helps. Remember: scalar fields don't ALWAYS obey! But that's the way to bet.
 

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