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

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The discussion revolves around calculating the electric flux through the faces of a cube with a point charge at its center. For Part A, the total electric flux through the cube is determined using Gauss' law, which states that the total flux is equal to the charge divided by the permittivity of free space. Since the charge is centrally located, the flux through each face is one-sixth of the total flux. In Part B, the discussion raises the question of how the flux would change if the cube's side length were modified to L1, but the fundamental principles remain the same. Understanding these concepts is essential for solving the problem accurately.
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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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