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) $- (3 + \sqrt{}8)$ only
B) $- 3 + \sqrt{}8$ only
C) $- 3(3 + \sqrt{}8)\ or\ ( - 3 + \sqrt{}8)$
D) $+ \sqrt{3}$
Explanation
(c) Since the two charges (spheres) attract, they will be opposite in sing, $ie$, $q_{1}$ and ${- q}_{2}$.
Force, $F = \frac{1}{4\pi\varepsilon_{0}}\frac{q_{1} \times - q_{2}}{d^{2}}$
After touching, charge on each will be $\frac{q_{1} - q_{2}}{2}.$
New force, $F^{'} = \frac{1}{4\pi\varepsilon_{0}}\left( \frac{q_{1} - q_{2}}{2} \right)^{2} \times \frac{1}{d^{2}}$
Given |F|=|F’|
On solving by quadratic equations, we get
$\frac{q_{1}}{q_{2}} = - \left( 3 + \sqrt{8} \right)or\left( - 3 + \sqrt{8} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.