Two parallel plate capacitors C _ 1 and C _ 2 each having capacitance of 10μ F are individually charged by a 100V D.C. source. Capacitor C _ 1 is kept connected to the source and a dielectric slab is inserted between it

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As source is connected in the process therefore, additional charge will be provided. Therefore, charge on ${C}_{1} = K\left(CV\right)$ .

Since source is disconnected, so its charge will remain the same. Then, charge on ${C}_{2} = CV$ .

New capacitance in both cases will become, $KC$ .

When they are connected in parallel charge will be equally divided so charge on one capacitor is $q = \left(\dfrac{K + 1}{2}\right)CV$ .

So common potential,

${V}^{'} = \dfrac{q}{KC} = \dfrac{K + 1}{2K}V = 55V$ .

Two parallel plate capacitors C _ 1 and C _ 2 each having...

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