Solution:
Solution
Initially
$$q_1=q_2=Q$$ and $$r=R$$
Force was $$F=\dfrac{q_1q_2}{4\pi\epsilon_o r^2}=\dfrac{Q^2}{4\pi\epsilon_o R^2}$$
Thenafter:-
$$q_1=Q, q_2=4Q$$ and $$r=3R$$
Force on charge 4Q, $$F'=\dfrac{Q.4Q}{4\pi\epsilon_o (3R)^2}=\dfrac{4Q^2}{4\pi\epsilon_o (9R^2)}$$
$$\implies F'=\dfrac{4}{9}F$$
Answer-(D)