Q. Two conducting spheres of radii 3 cm and 1 cm are separated by a distance of 10 cm in free space. If the spheres are charged to same potential of 10 V each, the force of repulsion between them is
A) $\frac{1}{3} \times 10^{- 9}$ N
B) $\frac{2}{9} \times 10^{- 9}$N
C) $\frac{1}{9} \times 10^{- 9}$ N
D) $\frac{4}{3} \times 10^{- 9}$ N
Explanation
(a) $\frac{kq_{1}}{3} = 10,\frac{kq_{2}}{1} = 10$
$kq_{1} = 30 \times 10^{- 2}$
$kq_{2} = 10 \times 10^{- 2}$
$F = \frac{kq_{1}q_{2}}{10^{- 2}}$
$\ = \frac{\left( 30 \times 10^{- 2} \right) \times (10 \times 10^{- 2})}{k \times 10^{- 2}}$
$F = \frac{1}{3} \times 10^{- 9}\ $N
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.