If the radius of the Gaussian surface enclosing a charge is halved, how does the electric flux through the Gaussian surface change?

Hint:The flux through a closed Gaussian surface depends only on the net charge enclosed by the surface.
The size and shape of the surface do not matter when it is a closed surface.
When the radius changes the flux would not change.

Complete step-by-step solution:
The electric flux density does not depend on the shape, size, and position of the closed surface concerning the charge placed inside. The shapes may be cylindrical, cuboid, or spherical.
The law theoretically said that the flux equals the number of electrostatic electric field lines crossing a closed Gaussian surface. So it remains the same.
The electric flux of the Fora closed Gaussian surface is given by,
\[flux = \dfrac{Q}{{{\varepsilon _0}}}\]
$ \Rightarrow \oint\limits_S {\vec E.d\vec s} = \dfrac{Q}{{{\varepsilon _0}}}$
Hence, \[E\] is the electric field
\[S\] is any closed surface
\[Q\] is the overall electric charge inside the surface
\[{\varepsilon _0}\] is the permittivity in the air medium.
If the charge is halved then the flux becomes halved.
Here it does not depend on the radius of the Gaussian surface. It only depends on the Gaussian surface.
When changing the radius the electric flux through the surface remains the same. And the net charge enclosed the surface is also the same.
The outward electric flux is independent of the distribution of the charges and separation between them inside the closed surface.

Note:When the electric flux enters into the closed surface it is considered negative and when it is out of the closed surface it is considered Positive.
For a point charge, the Gaussian surface becomes spherical.
For an infinite sheet of charge and finite line of charge, the Gaussian surface becomes cylindrical.