Question

Two identical balls having like charges and placed at a certain distance apart, repel each other with a certain force. They are brought in contact and then moved apart to a distance equal to half their initial separation. The force of repulsion between them increases $4.5$ times in comparison to the initial value. The ratio of the initial charges of the balls is

• A. 2
• B. 3
• C. 4
• D. 6

Answer 1

Answer: (A)

Suppose the balls have charges $Q_1$ and $Q_2$, respectively.
Initially:

Finally:

$F^{\prime }=\frac{k{\left(\frac{Q_1+Q_2}{2}\right)}^2}{{\left(\frac{r}{2}\right)}^2}=\frac{k{\left(Q_1+Q_2\right)}^2}{r^2}$
It is given that $F=4.5F$
so $\frac{k{\left(Q_1+Q_2\right)}^2}{r^2}=4.5k\frac{Q_1{Q}_2}{r^2}$
$\Rightarrow {\left(Q_1+Q_2\right)}^2=4.5Q_1{Q}_2$
On solving, it gives $\frac{Q_1}{Q_2}=\frac{2}{1}$