Q. Two small conducting sphere of equal radius have charges +10 $\mu$C and -20$\mu$C respectively and placed at a distance $R$ from each other experience force $F_{1}.\ $If they are brought in contact and separated to the same distance, they experience force $F_{2}.$ the ratio of $F_{1\ }$to$\ F_{2}$ is
A) 1 : 2
B) -8 : 1
C) 1 : 8
D) -2 : 1
Explanation
(b) Here, $F_{1} = \frac{k( + 10)( - 20)}{R^{2}} = \frac{- k \times 200}{R^{2}}$
As spheres are of equal radius, their capacities are same. On touching, the net charge $+ 10 - 20 = - 10\ \mu C\ $ is shared equally between them $ie,$ each sphere carries $- 5\ \mu C$ charge.
$F_{2} = \frac{k( - 5)( - 5)}{R^{2}} = \begin{array}{r}
\frac{k \times 25}{R^{2}} \\
\
\end{array}\ $
$\ \therefore\ \ {\frac{F_{1}}{F_{2}} = \ \frac{- 8}{1}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.