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

[pokie_panda]

Homework Statement

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.

The Attempt at a Solution

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

 

[HallsofIvy]

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

 

[Curious3141]

Homework Statement

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.

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]

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]

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.

 

[Curious3141]

Edit: beaten by one minute!

Haha.