Answer:
The flux remains the same and the field increases.
Explanation:
According to the Gauss' law, the surface integral of the electric field over a closed surface, called the Gaussian surface, is equal to 
times the net charge 
enclosed by the surface.
<em>where,</em>
= electrical permititivity of free space.
The term on the LHS of Gauss law' 
is the electric flux 
through the Gaussian surface.
Therefore,
From equation (2), the electric flux through the sphere depends only on the charge enclosed by the sphere and not on the size of the sphere, therefore, when the radius of the sphere is changed to 0.500 m from 1.00 m, the electric flux through the surface will not change.
Consider the Gaussian sphere of radius 
same as that of the given sphere and concentric with the given sphere, then,
This is because the direction of the electric field through a Gaussian spherical surface and its surface area element 
is always normal to it.
Therefore,
Using equation (1),
The electric field is inversely proportional to the square of the radius of the sphere, therefore, on decreasing the radius from 1.00 m to 0.500 m, the electric field increases.
Thus, the correct answer is "The flux remains the same and the field increases".
