Why doesn't the outer charge affect the electric flux in Gauss's law?

[oneplusone]

Suppose you have a sphere (sphere A) with net positive charge 2Q. A conducting spherical shell (sphere B) of inner radius b and outer radius c is concentric with the solid sphere and carries a net charge -Q.

When you calculate the flux between both spheres (gaussian surface with radius between both of the spheres), you're suppose to only add up the charges INSIDE the gaussian surface--that is just sphere A.
I understand that this is because of the formula which has $q_{enclosed}/\epsilon_0$ , but don't get this visually. Ill rephrase it: Why does the outer charge (sphere B) have no effect on the flux of the surface that we created? Doesn't the electric field have a relation to the flux? \int \vec{E}\vec{dA} = \Phi ?
Cheers,
oneplueone

P.S. why doesn't # work for latex?
EDIT: i realized this is in the wrong forum, can a mod please move this

 

[Dale]

Why does the outer charge (sphere B) have no effect on the flux of the surface that we created? Doesn't the electric field have a relation to the flux? \int \vec{E}\vec{dA} = \Phi

Don't forget that the integral is taken over the entire Gaussian surface. If you have a positive charge outside and to the left of some Gaussian surface then the E field will give an inward (negative) flux on the left side of the surface but an outward (positive) flux on the right side. The net flux will be 0. Only when the charge is inside the surface will the fluxes on all sides be outward giving a net nonzero flux.

 

[VishalChauhan]

The electric field due to the outer sphere (B) is zero inside it. So it does not add into the electric flux expression and you get only the charge enclosed.

Even if the charge distribution on the outer sphere was non uniform such that its electric field was non zero,its field lines would be such that they would pierce into the gaussian surface at some point and pierce out of it at another.Ultimately you would still end up with a zero flux for the outer charge.