# 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]

Homework Helper
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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]

Homework Helper
Gold Member
2022 Award
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.