# Does electrical field intensity and electric flux at a point depend on all the point charges (inside and outside) around a Gaussian surface?

If there is a spherical Gaussian surface and there are 2 point charges inside it and 3 point charges outside it (all of say (+1)coulomb) and we consider a point x "on" the Gaussian surface, Does the electric field intensity depend on all of the charges or does it depend only on the charges inside the Gaussian surface?

Secondly, if we consider electric field intensity at a point "y" in the Gaussian surface n thirdly say point "z" outside the Gaussian surface,should we consider all the charges or (for second case) only the ones inside (and for third case) only the ones outside when analysing the electric field intensity?

What charges do we consider if we are to find electric FLUX at the same points x (on) , y (in) and z (out of) the Gaussian surface for the same example I've given

The net electric field intensity $$\vec E_{\rm net}(\vec r)$$ at a position $$\vec r$$ is the vector sum of the electric field intensities produced at that point by each individual charge, $$\vec E_{\rm net}(\vec r)= \vec E_1(\vec r)+\vec E_2(\vec r)+\vec E_3(\vec r) + . . . . .$$

Drawing/imagining etc a Gaussian surface does not change this summation.

However if you wish to use Gauss's law to find the net electric flux passing through a Gaussian surface then you need only consider the charges enclosed by the Gaussian surface and the electric field intensities produced by the enclosed charges at the Gaussian surface.

Field intensity depends upon all charges in the surroundings and the presence of conductors too affects field intensity. However, flux at a "point" isn't quite meaningful I guess.

Alright so Gauss theorem says that $$\int _{\text{surface}} E\cdot ds =\frac{q}{\epsilon _0} = \phi$$ Now if you consider a Gaussian surface with charges both inside and outside what you have to understand that the charges outside aren't just magically going to stop emitting an electric field. So the field by the charges inside be $$E_1$$ and $$E_2$$ respectively and outside charges will also have some electric fields, say $$E_3, E_4, E_5$$. So the electric fields at the points $$x,y,z$$ will obviously be due to all the charges.

Now your second question about flux. What you have to understand is that flux$$(\phi)$$ is literally defined as the number of field lines crossing the area. It is defined as $$\phi = E\cdot A$$ Now, for a body with a nonuniform electric field such as say the Gaussian surface with the $$2$$ point charges inside we will consider a differential area $$ds$$ and say that the electric field over $$ds$$ is uniform. Now if you integrate that expression you get the total flux over the surface. Do you understand now why we do not think of flux at a point? Because it is so tiny we need not bother about it. Also, imagine a closed surface and imagine a closed surface, say a sphere.
You should be able to see that a point charge kept outside will have electric field lines that intersect the sphere at $$2$$ distinct points. Now, this will mean that the flux for one of the points will be $$\gt 0$$ and one will be $$\lt 0$$. Now we can see that net flux over the surface by the point charge will be $$0$$. Do this for yourself, draw the surface, consider the differential element, draw the area vectors and see the angle made with field lines. So that is why we never consider the point charges outside when we talk about flux over a surface. Of course, you could also use the concept of Solid angle but this is the most intuitive explanation I could think of. So the electric field will be due to all the points and the flux(in this case you are talking about at a point, so it doesn't really have any physical significance, but if you were talking about the flux over the surface) you would only consider the charges inside the Gaussian surface.