Increasing the distance by a factor of two from a point charge will change the electric potential by a factor of __ A 1/2 B. 2 C. 4 D 1/4

The electric potential (V) due to a point charge (q) is given by the equation:<br /><br />\[ V = \frac{kq}{r} \]<br /><br />where:<br />- \( k \) is Coulomb's constant,<br />- \( q \) is the charge,<br />- \( r \) is the distance from the charge.<br /><br />If the distance \( r \) is increased by a factor of two, the new distance becomes \( 2r \). Substituting \( 2r \) into the equation for electric potential, we get:<br /><br />\[ V' = \frac{kq}{2r} \]<br /><br />Now, we compare the new potential \( V' \) with the original potential \( V \):<br /><br />\[ V' = \frac{kq}{2r} = \frac{1}{2} \cdot \frac{kq}{r} = \frac{1}{2} \cdot V \]<br /><br />Thus, the electric potential changes by a factor of \( \frac{1}{2} \).<br /><br />So, the correct answer is:<br />A. \( \frac{1}{2} \)