Two parallel plate capacitors $C_1$ and $C_2$ each having capacitance of $10\ \mu\text{F}$ are individually charged by a $100\ \text{V}$ D.C. source. Capacitor $C_1$ is kept connected to the source and a dielectric slab is inserted between its plates. Capacitor $C_2$ is disconnected from the source and then a dielectric slab is inserted in it. Afterwards the capacitor $C_1$ is also disconnected from the source and the two capacitors are finally connected in parallel combination. The common potential of the combination will be ______ V. (Assuming dielectric constant $=10$)
Numerical value type. Enter your answer.
Answer: 55
Both capacitances become $100\ \mu\text{F}$.
$C_1$ stays at $100\ \text{V}$: $Q_1=100\times100=10^4\ \mu\text{C}$.
$C_2$ keeps its original charge: $Q_2=10\times100=10^3\ \mu\text{C}$.
In parallel: $V=\dfrac{Q_1+Q_2}{C_1+C_2}=\dfrac{11000}{200}=55\ \text{V}$.
Solution by Sreeraj P, M.Sc Physics