Flux of a Point Charge Inside a Cube: Part A & B

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SUMMARY

The discussion focuses on calculating the electric flux through the faces of a cube containing a point charge at its center. According to Gauss' Law, the total electric flux Φ through the cube is given by Φ = q/ε₀, where q is the charge and ε₀ is the permittivity of free space. Since the cube has six identical faces, the flux through each face is Φ/6 = (q/6ε₀). Additionally, for a cube with sides of length L1, the same principle applies, maintaining the relationship of the flux through each face as (q/6ε₀) regardless of the cube's dimensions.

PREREQUISITES
  • Understanding of Gauss' Law in electrostatics
  • Familiarity with electric flux and its mathematical representation
  • Knowledge of the permittivity of free space (ε₀)
  • Basic principles of symmetry in electric fields
NEXT STEPS
  • Study the derivation of Gauss' Law and its applications in electrostatics
  • Explore the concept of electric flux in different geometries
  • Learn about the implications of charge distribution on electric fields
  • Investigate the effects of varying dimensions on electric flux calculations
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Students studying electromagnetism, physics educators, and anyone interested in understanding electric flux and Gauss' Law applications in electrostatics.

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Homework Statement



A point charge of magnitude q is at the center of a cube with sides of length L.

Part A
What is the electric flux Phi through each of the six faces of the cube?
Part B
What would be the flux \phithrough a face of the cube if its sides were of length L1?


Homework Equations



\phi=Ea cos theta

\phi = q/Eo

The Attempt at a Solution



Not sure of where to begin. Can someone show me how to tackle this?
 
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Use Gauss' law. Remember that since the charge is at the cube's center, and all of the cube's sides are identical, the flux through one side is one-sixth the flux through the whole thing.
 
(1/6)\epsilon0/q?
 
Close, but q is supposed to be on the top and epsilon at the bottom.
 

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