Q. 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 with the initial value. The ratio of the initial charges of the balls is
A) 2
B) 3
C) 4
D) 6
Explanation
(a) Suppose the balls having charges $Q_{1}$ and $Q_{2}$ respectively
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$F^{''} = \frac{k\left( \frac{Q_{1} + Q_{2}}{2} \right)^{2}}{\left( \frac{r}{2} \right)^{2}} = \frac{k\left( Q_{1} + Q_{2} \right)^{2}}{r^{2}}$
It is given that $F^{''} = 4.5F$ so $\frac{k\left( Q_{1} + Q_{2} \right)^{2}}{r^{2}} = 4.5\ k.\frac{Q_{1}Q_{2}}{r^{2}}$
$\Rightarrow \left( Q_{1} + Q_{2} \right)^{2} = 4.5\ Q_{1}Q_{2}$. On solving it gives $\frac{Q_{1}}{Q_{2}} = \frac{2}{1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.