# Your question: How much electric flux goes through a single face of the octahedron?

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## What is the flux through the octahedron?

The net electric flux through an octahedron is -1000N m^2/C.

## What is the net flux through one of the faces?

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.

## How much charge is enclosed within the surface?

No charged particles enclosed within surface. 2. There are charged particles, but the net charge enclosed is zero. Gauss’s Law states that the net electric flux through any closed surface is proportional to the net charge enclosed, not the E-field!

## What is the flux through one face of the cube when charge Q is placed at its Centre?

Total flux passing through the close cube ϕ=qε0. All the six surface are symmertrical with resepect to charge, hence they will have equal contribution of the flux. So, flux through any one face: ϕ=ϕ6=q6ε0.

## What would be the flux φ1 through a face of the cube if its sides were of length l1?

The electric flux Φ through each of the six faces of the cube is (q/6ε). B. Flux is independent of the dimension of an imaginary Gaussian surface. The flux Φ1 through a face of the cube if its sides were of length L1 is Φ1 = (q/6ε).

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## How do you calculate flux in math?

The flux can be described by ∬SF⋅ndσ with n=2xˆi−ˆj+2zˆk√1+4×2+4z2. Substitute x2+z2=y to simplify n to −1+2z2y. The total flux through the surface is 0.