### Capacitance Question 5

#### Q5 - 31 January - Shift 2

Two parallel plate capacitors $C_1$ and $C_2$ each having capacitance of $10 \mu F$ are individually charged by a 100 V D.C. source. Capacitor $C_1$ is kept connected to the source and a dielectric slab is inserted between it 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 $=10$ )

## Show Answer

#### Answer: (55)

#### Solution:

#### Formula: Capacitor with Dielectric

$\frac{1(90 mc)^2}{25(5 s)}-2 \times \frac{1}{2} \frac{(45 mc)^2}{900 uF}\left|U=\frac{Q^2}{2 c}\right|$

$=2.25$ Joule