Gauss's Law and electric flux of a surface

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SUMMARY

This discussion focuses on applying Gauss's Law to determine the electric flux through four closed surfaces (S1 to S4) containing charges of -2Q, Q, and -Q. According to Gauss's Law, the electric flux (Φ) through a closed surface is given by the equation Φ = q/ε₀, where q is the total charge enclosed by the surface and ε₀ is the permittivity of free space. The analysis concludes that surfaces containing no charge yield zero flux, while the surface enclosing charges Q and -2Q results in a net flux of -Q/ε₀.

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  • Understanding of Gauss's Law
  • Familiarity with electric flux concepts
  • Knowledge of charge interactions
  • Basic calculus for integration of electric fields
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  • Study the derivation of Gauss's Law in electrostatics
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  • Learn about the concept of electric field lines and their relation to flux
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Keithkent09
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Four closed surfaces, S1 through S4, together with the charges -2Q, Q, and -Q are sketched in the figure below. (The colored lines are the intersections of the surfaces with the page.) Find the electric flux through each surface. (Use Q for the charge Q and epsilon_0 for 0.)
(Picture Attached)

2.
Electric Flux= the integral of EdA=q/epsilon_0


3.
All that I could think to do was set q=the charge of the inside of the desired surface. I was not sure how I could quantitatively define how the different charges within the surfaces affected each other.
 

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Gauss` law is your best friend. the flux through a surface will only be non zero if there is a charge inside that surface. think about putting an imaginary sphere in the middle of a river. All the water that runs into the sphere also runs out of it. Having a charge inside the surface is like putting a sprinkler inside your imaginary sphere. Now the amount of water in is 0, and the amount of water out is this analogy`s version of flux.

So any surface with no charge inside, has 0 flux.

This principle is displayed simply in your equations for flux. where flux is = to
charge enclosed / epsilon naught.

So your flux for the red surface with Q and -2Q, it will have a flux of -Q/epsilon naught
 

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