Q. Two metallic spheres of radii 1 cm and 2 cm are given charges of $10^{- 2}$ C and $5 \times 10^{- 2}$ C respectively. If these are connected by a conducting wire, the final charge on the smaller sphere is
A) $3 \times 10^{- 2}$ C
B) $1 \times 10^{- 2}$ C
C) $4 \times 10^{- 2}$ C
D) $2 \times 10^{- 2}$ C
Explanation
(d) Common potential, $V = \frac{q_{1} + q_{2}}{C_{1} + C_{2}}$
$\therefore\ \ $charge on smaller sphere after contact
$= C_{1}V = \frac{C_{1}{(q}_{1} + q_{2})}{C_{1} + C_{2}}$
$= \frac{4\pi\varepsilon_{0}r_{1}(10^{- 2} + 5 \times 10^{- 2})}{4\pi\varepsilon_{0}(r_{1} + r_{2})}$
$= \frac{10^{- 2} \times 6 \times 10^{- 2}}{3 \times 10^{- 2}} = 2 \times 10^{- 2}C$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.