Question

Two charges of equal magnitude and at a distance'r' exert a force $F$ on each other. If the charges are halved and distance between them is doubled, then the new force acting on each charge is

• A. $F/8$
• B. $F/4$
• C. $4F$
• D. $F/16$

Answer 1

The correct option is D $F/16$

$\text{Step 1: Electrostatic force in initial case}$

Let $Q_1=Q_2=Q,d=r$

By Coulomb's Law:

$F=\frac{K{Q}^2}{r^2}$ $....\left(1\right)$

$\text{Step 2: Electrostatic force in final case}$

When charge is halved and separation is doubled

So, $Q_1^{\prime }=Q_2^{\prime }=\frac{Q}{2},d^{\prime }=2r$

So $F^{\prime }=\frac{K{\left(\frac{Q}{2}\right)}^2}{(2r{)}^2}=\frac{K{Q}^2}{16r^2}$

From equation $\left(1\right)$

$F^{\prime }=\frac{F}{16}$

Hence, Option $D$ is correct