Two parallel plate capacitors C₁ and C₂ each having capacitance of 10 $μ$ F are individually charged by a 100 V D.C. source. Capacitor C₁ is kept connected to the source and a dielectric slab is inserted between it plates. Capacitor C₂ is disconnected from the source and then a dielectric slab is inserted in it. Afterwards the capacitor C₁ 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)

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answer is 55.

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Detailed Solution

In $C_{1} \Rightarrow Q_{1} = 10 \times 10 \times 100 = 10000 μC$

$(Q = KCV)$
In $C_{2} \Rightarrow Q_{2} = 10 \times 100 = 1000 μC (Q = C_{2} V)$
Now C₁ & C₂ kept in parallel
$\begin{array}{l} ({KC}_{1} + {KC}_{2}) V = (10000 + 1000) \\ 10 (C_{1} + C_{2}) V = 11000 \\ (C_{1} + C_{2}) V = 1100 \\ V = \frac{1100}{20} = 55 volt \end{array}$