The force between two charges is given by Coulomb's law: \[ F = \frac{k q_1 q_2}{r^2}, \] where \( F \) is the force, \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between them. If the distance \( r \) is doubled, the new distance becomes \( 2r \). The new force \( F' \) is: \[ F' = \frac{k q_1 q_2}{(2r)^2} = \frac{k q_1 q_2}{4r^2} = \frac{F}{4}. \] Thus, the force decreases by a factor of 4 when the distance is doubled. Hence, the correct answer is \( \boxed{(1)} \).
Match List-I with List-II.
Choose the correct answer from the options given below :