Two parallel plate capacitors C₁ and C₂ each having capacitance of 10… | JEE Main 2023 (31 Jan, Shift 2)

Q60JEE Main 2023 · 31 JAN · SHIFT 2Difficult

Dielectrics

Two parallel plate capacitors $C_1$ and $C_2$ each having capacitance of $10\ \mu\text{F}$ 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 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 answer type. Work it out, then open the answer.

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Answer: 55

$C_1$ stays connected: $C_1' = 100\ \mu\text{F}$ at 100 V, so $Q_1 = 10000\ \mu\text{C}$ .
$C_2$ disconnected first: $Q_2 = 10 \times 100 = 1000\ \mu\text{C}$ , $C_2' = 100\ \mu\text{F}$ .
In parallel: $V = \frac{Q_1 + Q_2}{C_1' + C_2'} = \frac{11000}{200} = 55\text{ V}$

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