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MDCAT Physics QUESTION #4389
Question 1
If the distance between two charges is halved and the magnitude of charges are also doubled, then the force between these charges becomes:
  • Two times
  • Four times
  • Eight times
  • Sixteen times✔️
Correct Answer Explanation
By Coulomb's law: $F = k\dfrac{q_1 q_2}{r^2}$. New force: $F' = k\dfrac{(2q_1)(2q_2)}{(r/2)^2} = k\dfrac{4q_1q_2}{r^2/4} = \dfrac{16kq_1q_2}{r^2} = 16F$. The force becomes $\mathbf{16}$ times.