Two point charges +3 μC and −8 μC attract each other with a force of 40 N, when separated by finite distance d. If a charge of +5 μC is added to each of them, then the force between them will become:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 40 N
By Coulomb's law the force between two point charges at a fixed distance d is \( F = k\,q_1q_2/d^{2} \), so \( F\propto q_1q_2 \) when d does not change.
After adding \( +5\,\mu\mathrm{C} \) to each charge:
\( q_1' = 3 + 5 = +8\,\mu\mathrm{C},\qquad q_2' = -8 + 5 = -3\,\mu\mathrm{C} \)
Compare the products of the charges:
\( q_1q_2 = (3)(-8) = -24,\qquad q_1'q_2' = (8)(-3) = -24 \)
The product is unchanged, and it is still negative (the force is still attractive), so \( F' = F = 40\,\mathrm{N} \).