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

Explanation

To solve the problem, we will use Gauss's Law, which states that the electric flux (Φ) through a closed surface is directly proportional to the charge (q) enclosed within that surface. The mathematical expression for Gauss's Law is given by:

\[ \Phi = \frac{q}{\epsilon_0} \]

where:
- \(\Phi\) is the electric flux,
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