Elec. flux through the top side of a cube with q at a corner

[Pushoam]

Homework Statement

Find the electrix flux through the top side of a cube. The cube's corner is on the origin, and is 'a' units on length. The charge 'q' is located at the origin, with the corner of the cube.

Homework Equations

Gauss's law and symmetry

The Attempt at a Solution

I take 8 cubes of length 2a in such a way that the charge q is at the center.

Then, the flux through the half of the upper surface of this bigger cube is the required answer. Right?

Using Gauss's law and symmetry ,

Flux through each surface of the bigger cube is $\frac {q } {8\epsilon_0 }$.

Now, again using symmetry, flux through half of the surface is $\frac {q } {16\epsilon_0 }$, which is equal to the required answer.

Is this correct?
[/B]

 

[haruspex]

Flux through each surface of the bigger cube is $\frac {q } {8\epsilon_0 }$.

Because? A cube has eight faces??

 

[Pushoam]

Because? A cube has eight faces??

Sorry, it is 6.
]Flux through each surface of the bigger cube is $\frac {q } {6\epsilon_0 }$ .

Then, the flux through the half of the upper surface of this bigger cube is the required answer. Right?

The above is also wrong. It is ¼, not ½.
Now, again using symmetry, flux through ¼ of the bigger surface is $\frac {q } {24\epsilon_0 }$, which is equal to the required answer.[/QUOTE]

 

[haruspex]

Sorry, it is 6.
]Flux through each surface of the bigger cube is $\frac {q } {6\epsilon_0 }$ .


The above is also wrong. It is ¼, not ½.
Now, again using symmetry, flux through ¼ of the bigger surface is $\frac {q } {24\epsilon_0 }$, which is equal to the required answer.

[/QUOTE]
Looks right.

 

[Pushoam]

Thank you.
The flux is independent of the length of the cube.