Two point charges +3 μC and -8 μC attracts 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:

40 N

The correct answer is Option (4) → 40 N

Initially, the charges are $q_1=+3μc$ and $q_1=-8μc$, and the force is $F=40N$.

After adding $+5μc$ to both charges,

${q_1}'=q_1+5=+8μc$

${q_2}'=q_2+5=-3μc$

$F=k\frac{q_1q_2}{r^2}$ [Coulomb's law]

$∴\frac{F'}{F}=\frac{{q_1}'{q_2}'}{q_1q_2}$

$F'=F=\frac{{q_1}'{q_2}'}{q_1q_2}=F×\frac{-24}{-24}$

$F'=F=40N$