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)

Two parallel plate capacitors $$C_1$$ and $$C_2$$, each with capacitance 10 $$\mu$$F, are charged to 100 V. A dielectric slab of constant $$K = 10$$ is inserted in each under different conditions.

When $$C_1$$ remains connected to the 100 V source and a dielectric is inserted:

New capacitance: $$C_1' = KC_1 = 10 \times 10 = 100$$ $$\mu$$F

Voltage stays at 100 V (source connected).

Charge: $$Q_1 = C_1' \times V = 100 \times 100 = 10{,}000$$ $$\mu$$C

Initial charge: $$Q_2 = C_2 \times V = 10 \times 100 = 1{,}000$$ $$\mu$$C

After disconnection, charge remains $$Q_2 = 1{,}000$$ $$\mu$$C.

After dielectric insertion: $$C_2' = KC_2 = 100$$ $$\mu$$F

Total charge: $$Q_{total} = Q_1 + Q_2 = 10{,}000 + 1{,}000 = 11{,}000$$ $$\mu$$C

Total capacitance: $$C_{total} = C_1' + C_2' = 100 + 100 = 200$$ $$\mu$$F

$$ V_{common} = \frac{Q_{total}}{C_{total}} = \frac{11{,}000}{200} = 55 \text{ V} $$

The common potential of the combination is 55 V.