Q. A condenser having a capacity of $6\mu F$ is charged to $100\ V$ and is then joined to an uncharged condenser of $14\mu F$ and then removed. The ratio of the charges on $6\mu F$ and $14\mu F$ and the potential of $6\mu F$ will be
A) $\frac{6}{14}$ and $50\ volt$
B) $\frac{14}{6}$ and $30\ volt$
C) $\frac{6}{14}$ and $30\ volt$
D) $\frac{14}{6}$ and $0\ volt$
Explanation
(c) Let $q_{1},q_{2}$ be the charges on two condensers
$\therefore V = \frac{q_{1}}{6} = \frac{q_{2}}{14} \Rightarrow \frac{q_{1}}{q_{2}} = \frac{6}{14}$
Also $q_{1} + q_{2} = 600 \Rightarrow q_{1} + \frac{14}{6}q_{1} = 600 \Rightarrow q_{1} = \frac{600}{20} \times 6$
$\therefore V = \frac{q_{1}}{6} = \frac{600}{20} = 30\ volt$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.