Question

Two particles having charges $q_1$ and $q_2$ exert a force $F$ on each other, when they are placed at a certain distance. What will be the force between them, if the distance between them is reduced to half and the charge on each particle is doubled ?

• A. $12F$
• B. $4F$
• C. $8F$
• D. $16F$

Answer 1

The correct option is D $16F$

$F=k\frac{q_1{q}_2}{r^2}$

$F^{\prime }=k\frac{q_1^{\prime }{q}_2^{\prime }}{r^{\prime 2}}$

given, $q_1^{\prime }=2q_1,q_2^{\prime }=2q_2,r^{\prime }=\frac{r}{2}$

$\therefore \frac{F^{\prime }}{F}=\frac{q_1^{\prime }{q}_2^{\prime }}{q_1{q}_2}\frac{r^2}{r^{\prime 2}}=4\times 4=16$

$F^{\prime }=16F$

ans: (D)