Why does electric field does not change with respect to if we double the radius?
Answers
For simplicity, it is assumed that the Gaussian surface is spherical and drawn around a point charge at it's centre. The electric flux through a Gaussian surface of radius can be found by computing the surface integral of E.dS over the surface. Since E and dS are in the same direction (radially outward) we just have to find the surface integral of E dS. The value of E will be the same here at all points on the surface, so that the flux over the surface is just E(4πr^2). Also remember that E is proportional to 1/r^2
If the radius of the spherical Gaussian surface is doubled, the surface area will increase by 4 times. On the other hand, the electric field at all points on the bigger surface will decrease by a factor of 4. Hence the flux through the surface remains the same.
This has been a rather lengthy and roundabout way of saying what could have been concluded straight away from Gauss's law!