Q. A capacitor of capacitance 10$\mu$F charged to 100V is connected to an uncharged capacitor. The effective potential now is 40 V. The capacitance of uncharged capacitor is
A) 12$\mu F$
B) 15 $\mu F$
C) 25 $\mu F$
D) 30 $\mu F$
Explanation
(b) Common potential of system
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V = \frac{C_{1}V_{1} + C_{2}V_{2}}{C_{1} + C_{2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 40 = \ \frac{10 \times 100 + C_{2} \times 0}{10 + C_{2}}$
or $\ \ 40\left( 10 + C_{2} \right) = \ 1000$
$\Rightarrow$ $C_{2} = \ 15\ \mu F$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.