Q. Two identical spheres with charges 4$q, - 2q$ kept some distance apart exert a force $F$ on each other. If they are made to touch each other and replaced at their old positions, the force between them will be

A) $\frac{1}{9}F$
B) $\frac{1}{8}F$
C) $\frac{9}{8}F$
D) $\frac{8}{9}F$
Explanation

(b) When spheres are touched and separated, charge is
$q^{'} = \frac{4_{q} - 2_{q}}{2} = \frac{2q}{2} = q$
From Coulomb’s law
$F = \frac{1}{4\pi\varepsilon_{0}}\frac{(4q)( - 2q)}{r}\ \ \ \ \ \ \ \ \ldots(i)$
When made to touch
$F^{'} = \frac{1}{4\pi\varepsilon_{0}}\frac{q2}{r}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots..(ii)\ $
From Eqs. ($i$) and ($ii$), we get
$F^{'} = \frac{F}{8}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.