Question

Q. Two capacitors of 3 $\mu$F and 6 $\mu$F 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 ?

Answer 1

Dear Student,

$$\begin{gathered} potentialacrossthe3pF=5000\left(\frac{6}{6+3}\right)=\frac{10000}{3}V \\ potentialacross6Pf=\frac{5000}{3}V \\ chargeonthecapacitorQ=\frac{5000}{3}*6*{10}^{-12}C \\ \frac{Q-q}{3}=\frac{Q+q}{6} \\ 2Q-2q=Q+q \\ q=\frac{Q}{3} \\ potentialdifferenceafterconnection=\frac{Q-q}{3*{10}^{-12}}=\frac{2}{9*{10}^{-12}}\left(\frac{5000}{3}*6*{10}^{-12}\right)=\frac{20000}{9}=2222V \\ Regards \end{gathered}$$