Point charge at the center af a cube

[Mandaz]

A 10 nC point charge is at the center of a 2.0m x 2.0m x 2.0 m cube. What is the electric flux through the top surface of the cube?

related equations:

gauss's law -- Qin/e0

e0 = 8.85 x 10 ^-12 C^2/Nm^2

my attempt:

10 x 10^-9 C
________________________ = 1129.94 C/Nm^2
8.85 x 10 ^-12 C^2/Nm^2

the real answer:

10 x 10^-9 C
________________________ = .19 kN m^2/C
6 x (8.85 x 10 ^-12 C^2/Nm^2)

i do get what they did! Can anyone help me out?

 

[LeonhardEuler]

Notice they ask about the flux through the top of the cube, not the whole cube. Gauss's Law applies to the flux through a closed surface enclosing a charge. Just the top of the cube is not a closed surface.

Think about the symmetry of the problem, though.

 

[Mandaz]

i multiply by 1/6 because they only want one side of the cube, is that right.

i do not get what you mean by it will no longer be considered a closed surface because they only want one side.

 

[LeonhardEuler]

i multiply by 1/6 because they only want one side of the cube, is that right.

i do not get what you mean by it will no longer be considered a closed surface because they only want one side.

That is right.

What I meant was that the top of the cube alone is not a closed surface, it does not enclose the charge (or anything) by itself. Because of that, Gauss's law does not apply to the top surface alone. You apply Gauss's law to the cube as a whole, and by symmetry the flux out of each side is equal, and so must be 1/6th the total.

 

[Mandaz]

okay that makes a lot more sense now! thank you very much for all your help

 

[Kim Le]

Because there are total 6 surfaces in a cube so that multiply by 6.