Q. Two identical conducting spheres carrying different charges attract each other with a force $F$ when placed in air medium at a distance $'d'$ apart. The spheres are brought into contact and then taken to their original positions. Now the two spheres repel each other with a force whose magnitude is equal to that of the initial attractive force. The ratio between initial charges on the spheres is
A) $- \left( 3 + \sqrt{8} \right)$ only
B) $- 3 + \sqrt{8}$ only
C) $- \left( 3 + \sqrt{8} \right)$ or $( - 3 + \sqrt{8})$
D) $+ \sqrt{3}$
Explanation
(c) $F_{1} = \frac{kQ_{1}Q_{2}}{d^{2}}$ and $F_{2} = \ \frac{k\left( \frac{Q_{1} - Q_{2}}{2} \right)^{2}}{d^{2}}$
According to question,
$F_{1} = F_{2}$
$Q_{1}Q_{2} = \frac{\left( Q_{1} - Q_{2} \right)^{2}}{4}\ \Rightarrow 4Q_{1}Q_{2} = Q_{1}^{2} + Q_{2}^{2} - 2Q_{1}Q_{2}$
$0 = Q_{1}^{2} + Q_{2}^{2} - \ 6Q_{1}Q_{2} \Rightarrow \frac{Q_{1}}{Q_{2}} = \ - 3 \pm \sqrt{8}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.