Q. There are two charged identical metal spheres $A$ and $B$ repel each other with a force 3$\times 10^{- 5}$ N. Another identical uncharged sphere $C$ is touched with $A\ $and then placed at the mid-point between $A\ $and $B.$ net force on $C$ is$\ $

A) $1 \times 10^{- 5}\ $N
B) $2 \times 10^{- 5}\ $N
C) $1.5 \times 10^{- 5}\ $N
D) $3 \times 10^{- 5}\ $N
Explanation

(d) If same charges on spheres $A$ and $B$ are $q$
$F = \frac{1}{4\pi\varepsilon_{0}\ }\ \frac{q^{2}}{\ \ r^{2}} = 3 \times 10^{- 5}$N
Charge on $A$ and $C$ after touching
[image:1_image311.emf]
$q_{A}^{''} = qc = \frac{q}{2}$
Net force on C
$\overrightarrow{F} = \overrightarrow{F_{A}} + \overrightarrow{F_{B}}$
$\therefore F = \frac{1}{4\pi \in_{0}}\frac{\left( \frac{q}{2} \right)(q/2)}{\left( \frac{e}{2} \right)^{2}} - \frac{1}{4\pi \in_{0}}\frac{\left( \frac{q}{2} \right) \times q}{\left( \frac{r}{2} \right)^{2}}$
$= \frac{1}{4\pi \in_{0}}\frac{q^{2}}{r^{2}} - 2\frac{1}{4\pi \in_{0}}\frac{q^{2}}{r^{2}} = - \left( \frac{1}{4\pi \in_{0}}.\frac{q}{r^{2}} \right)$
$= - 3 \times 10^{- 5}N$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

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