Q. Two capacitors of $3pF$ and $6pF$ are connected in series and a potential difference of $5000\ V$ is applied across the combination. They are then disconnected and reconnected in parallel. The potential between the plates is
A) $2250V$
B) $2222V$
C) $2.25 \times 10^{6}V$
D) $1.1 \times 10^{6}V$
Explanation
(b) $\frac{1}{C} = \frac{1}{3} + \frac{1}{6} \Rightarrow C = 2\ pF$
Total charge $= 2 \times 10^{- 12} \times 5000 = 10^{- 8}C$
The new potential when the capacitors are connected in parallel is
$V = \frac{2 \times 10^{- 8}}{(3 + 6) \times 10^{- 12}} = 2222\ V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.