What is the Flux Through Each Face of a Cube with a Corner Point Charge?

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SUMMARY

The discussion focuses on calculating the electric flux through each face of a cube with a point charge \( q \) located at one corner. By applying Gauss's Law, the total flux through the entire cube is determined to be \( \frac{6q}{\epsilon_0} \). Due to symmetry, the flux through each face of the cube is \( \frac{q}{\epsilon_0} \). This conclusion is reached without the need for integration, relying solely on symmetry arguments.

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Homework Statement



A point charge q is placed at one corner of a cube of edge a. What is the flux through each of the cube faces?


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The Attempt at a Solution



I drew a diagram of the situation. I'm pretty sure that I should use Gauss' Law and some symmetry arguments but I am not sure how to integrate over the cubes faces taking into account that the electric field produced by the point charge is radially outwards, like a sphere cutting through the face of the cube ( poorly described I know, sorry).

Thanks for the help.
 
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By Gauss's Law, the flux through the whole cube is just 6q/\epsilon_0. Now, apply a simple symmetry argument to get the answer.
 
I'm not sure why is it 6q ? If there is only one charge of magnitude 1q surely the flux is

<br /> q/\epsilon_0 <br />
 
Right: The total flux from that charge is just q/ε0. So what's the total flux through the cube? Through each face of the cube? (No integration needed, just symmetry.)
 
Ah right I see now. Thanks very much, was thinking about this completely the wrong way.
 
Ed Aboud said:
Ah right I see now. Thanks very much, was thinking about this completely the wrong way.

So...
What is the answer you are getting?
 

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