Electric flux through surface of a cube?

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SUMMARY

The discussion centers on calculating the electric flux through the top surface of a cube containing a 10nC point charge at its center. The relevant equation used is the integral of electric field (E) times area (A), represented as int(EA) = Qin/ε0. The user mistakenly attempted to calculate the electric field instead of directly determining the flux. The correct approach involves recognizing that the total flux through the cube is evenly distributed across all six surfaces, leading to the conclusion that the flux through the top surface is one-sixth of the total flux.

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Homework Statement


A 10nC point charge is at the center of a 2mx2mx2m. What is the electric flux through the top surface of the cube?


Homework Equations



int(EA) = (Qin)/(ε0)

The Attempt at a Solution



int(EA) = EA + EA + EA + EA + EA + EA = 6EA

A = 4m^2

Qin = 10nC

ε0 = 8.99x10^-12

6EA = (10x10^-9)/(8.99x10^-12)

E = (10x10^-9)/((8.99x10^-12)(6)(A))

E = (10x10^-9)/((8.99x10^-12)(6)(4))

E = 3745

Then for just the top surface

Etop = 3745/6 = 624

Where did I go wrong this time? XD Thanks in advance!
 
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You did not find what the question asked. Read the question again.
 
The problem asked for the flux. Why are you trying to find the field?
 

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