Calculate net flux through the cube

Question

abhisar · Answer

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abhisar · Answer

@Johnbc

abhisar · Answer

I am  trying to solve it with gauss law, but not getting the answer !

abhisar · Answer

The charges are assigned at each corner

anonymous · Answer

Was just reviewing Gauss's law.

abhisar · Answer

so ?

anonymous · Answer

Not entirely sure, was thinking of breaking the problem and placing a charge in the middle of the face of each cube but the distant of those are unknown and the charge is not evenly distributed throughout the cube so that might not work. Another method though is looking at the electric field but I do not remember it clearly.

abhisar · Answer

gauss law says that net flux = Q/$\sf E_0$

abhisar · Answer

wr Q is the total chrge enclosed within the gaussian surface.

anonymous · Answer

$$\phi = \frac{ q }{ e_0 }$$
Right and the total charge enclosed within this surface could be at the middle of the cube.

abhisar · Answer

Yeah...that's what m not sure about. I think you are right...we can not take charges at the corners as enclosed charge ?

anonymous · Answer

At the corners, not of the cube but of the individual spherical charges.

abhisar · Answer

I am not getting you...

anonymous · Answer

Enclosed charge within the gaussian surface of the spherical charges themselves at the corners but not of the cube which would have to be at the middle.

abhisar · Answer

ok..i get it now. But then how to find flux associated through this cube ?

abhisar · Answer

can u help @Kainui ?

kainui · Answer

So you want the flux through the cube? Are the charges in the cube or outside the cube?

abhisar · Answer

they are at the corners

abhisar · Answer

exactly at the corners of the cube.

kainui · Answer

I guess it comes down to whether you are interpreting "at the corners" as inside or outside.

ash2326 · Answer

@Abhisar Suppose you take a sphere of radius a at one of the corners with center being that corner. You will have exactly one charge inside the gaussian surface, then you can calculate flux for that sphere. Can you try?

anonymous · Answer

I was thinking the same thing because within the cube there would only be 1/4 of the charge if it was at the corner. 
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abhisar · Answer

So you want me to calculate flux through the sphere ?

ash2326 · Answer

yes

ash2326 · Answer

I agree with john, only 1/4 of the charge will be considered, so your flux can be divided by 4

abhisar · Answer

OH...so charge inside the cube will be 36/4=9 ?

kainui · Answer

Wait I think I see what's going on here!

Since the charge is directly on the corner you only have flux through the 3 faces that are adjacent to the opposite corner.

ash2326 · Answer

yes, you can consider that

abhisar · Answer

will it be 1/4 or 1/8 ?

abhisar · Answer

it's 3d...

abhisar · Answer

yeah...i got it

abhisar · Answer

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abhisar · Answer

So total charge inside the cube will be 1/8(36) C, Now using gauss law flux = $\sf \frac{36}{8E_0}$

ash2326 · Answer

is it correct?

abhisar · Answer

yes it is :)

anonymous · Answer

Why did you say only 1/8th ?

abhisar · Answer

Remember crystal lattice ?

anonymous · Answer

Yup

abhisar · Answer

there is only one atom in a primitive lattice

abhisar · Answer

because only 1/8th part is considered inside the lattice......visualize the 3d figure..

abhisar · Answer

8 cubes cover each corner..

anonymous · Answer

Right! Now I understand, I overlooked that part about 3D

abhisar · Answer

yeah...and i was considering charge as a particle..

abhisar · Answer

Thank you very much @Johnbc @ash2326 and @Kainui

anonymous · Answer

No, thank you.