electrostaticsgauss-law
Why is electric flux due to external charge i.e a charge outside a closed surface equal to 0?
P.S:Moreover I found this statement confusing:-
Electric field appearing in the Gauss' law is the resultant electric field due to all the charges present inside as well as outside but the contribution of outside charges in producing flux is zero.
Can someone help me understand it.
In an attempt to be brief: The big thing to remember is that the flux is also proportional to the area (technically, the surface integral of the field over the area). Crudely speaking, the side of the enclosed surface with exiting field lines are further away from the external charge than the side with "entering" field lines, and the surface area increases by $r^2$ (remember the formula for the surface area of a sphere). Gauss's Law is just a mathematically precise way of stating this compensation.
(Image source: http://www.ux1.eiu.edu/~cfadd/1360/24Gauss/Gauss.html)
Gauss's law in general relates the total electric field flux through a closed surface to the charge enclosed in that surface. It does not give you the field at a certain point in space in general. You can, however, exploit symmetries in certain geometries (points, spheres, lines, cylinders, planes, etc.) to determine the field at certain points in space. If you can't exploit some symmetry to pull E out of your flux integral, then you can't use Gauss's law to determine the field at a point in space.
It's like saying "I have 10 numbers that add to 100." If this is all you know then you can't tell me which numbers I have used to get a total of 100. But if I also tell you "each number is the same number", then you can for sure say I used 10 10's to get a total of 100.
As for your diagram, yes the charge Q will influence the field at P. It will not change the total flux through your Gaussian sphere, or any other Gaussian surface not enclosing Q. You will not be able to use Gauss's law now to find the field at P since the symmetry that lets you pull E out of the integral is no longer present. (Although, you could argue that you can use Gauss's law on each charge individually and then use superposition to get the total field, but you can't use Gauss's law for the entire configuration to get the field at P in this case).
Best Answer
Why? Simple answer: because the electrostatic electric field owing to a point charge fulfils an inverse square law, or, equivalently, the electric potential $\phi$ from a point charge varies as $r^{-1}$.
If the potential variation were some function other than $1/r$, the statement wouldn't be true. See for example my answer here, where I discuss what would happen with other variations.
Because the potential owing to a point charge is $\phi\propto 1/r$ and the potential owing to a system of point charges is the superposition of their potentials, the potential $\phi$ fulfils the Laplace equation $\nabla^2\phi = 0$ at all points away from point charges and where the charge density is nought. Therefore:
$$\oint_{\partial V} \vec{E}\cdot \hat{n} \,{\rm d}S = -\oint_{\partial V} \nabla \phi \cdot \hat{n} \,{\rm d}S = -\int_V \nabla^2 \phi \,{\rm d}V = 0$$
by the divergence theorem, for the boundary $\partial V$ of any volume $V$ not containing charges.
Note that this would not work if the functional dependence of the potential owing to a point charge were different from $1/r$ because Laplace's equation would not be fulfilled.
With your statement: I am assumed you are asking "why do we have to worry about charge outside the surface if it doesn't add to the flux?" The answer is: because we may only be given the electric field on the surface: we may be given no other data. Therefore we can simply go ahead and calculate the flux through the surface to find out how much charge lies within: we don't want to have to be separating the electric field into that from charges inside the volume and that from charges outside. The whole point of such a calculation is to discover how much charge lies inside the volume.