Q. A 10$\ \mu$C capacitor is charged to a potential difference of 50 V and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes 20 V. The capacitance of second capacitor is
A) 15 $\mu F$
B) 30 $\mu F$
C) 20 $\mu F$
D) 10 $\mu F$
Explanation
(a) $q_{1} = 10 \times 50 = 500\ \mu C,C_{1} = 10\ \mu F,\ C_{2} = ?,q_{2} = 0$
As $V = \frac{q_{1} + q_{2}}{\begin{array}{r}
C_{1} + C_{2} \\
\
\end{array}}$
${\therefore\ \ C}_{1} + C_{2} = \frac{q_{1} + q_{2}}{V} = \frac{500 + 0}{20} = 25\mu F$
$C_{2} = 25 - C_{1} = 25 - 10 = 15\ \mu F$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.