Two parallel plate capacitors and , each having capacitance of , are individually charged by a 100 V D.C. source. [2023]

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Solution

Q.
Two parallel plate capacitors $C_{1}$ and $C_{2}$ , each having capacitance of $10 μ F$ , 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) [2023]

Ans.
(55)

$Charge on C_{1} = K C E$

$and charge on C_{2} = C E$

When they are connected in parallel, charge will be equally divided, so charge on one capacitor is

$q = \frac{K + 1}{2} C V$

$So$ $V = \frac{q}{K C} = \frac{K + 1}{2 K} = 55 V$

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