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What is the flux through one face of the cube? Is my answer correct

1. The problem statement, all variables and given/known data

A point charge Q is placed at the center of a cube of side L .
What is the flux through one face of the cube?
Express your answer in terms of some or all of the variables Q ,L and appropriate constants.



3. The attempt at a solution

I'm getting Q/6e0 , is this correct. Are the units Nm^2/C
 

HallsofIvy

Science Advisor
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Don't you think there should be an "L" somewhere in the formula? And what is "C"?
 

Curious3141

Homework Helper
2,830
85
1. The problem statement, all variables and given/known data

A point charge Q is placed at the center of a cube of side L .
What is the flux through one face of the cube?
Express your answer in terms of some or all of the variables Q ,L and appropriate constants.



3. The attempt at a solution

I'm getting Q/6e0 , is this correct. Are the units Nm^2/C
Yes, this is correct. A simple application of Gauss' law. Your units are also correct, although it may be simpler to express it as Vm (volt*metre).
 

cepheid

Staff Emeritus
Science Advisor
Gold Member
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35
Since the units of electric field are N/C, your units of Nm^2/C for flux seem fine to me.

C = coulombs

Gauss' law says the net flux through the surface is Q/e0, and since this should be divided equally among the six faces, your answer seems plausible.

I imagine that any L-dependence should cancel out, since surface area increases as the square of L, but electric field strength decreases as the square of L

Edit: beaten by one minute!
 

Curious3141

Homework Helper
2,830
85
Don't you think there should be an "L" somewhere in the formula? And what is "C"?
No, because they're asking for the electrical flux. The electric field calculation would involve L (and would not be trivial because the cube lacks spherical symmetry). C is coulomb.
 

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