Two parallel plate capacitors C 1 and C 2 each having capacitance of 10 μF 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

Step-by-step solution

As source is connected in the process therefore, additional charge will be provided. Therefore, charge on C 1 = K C V . Since source is disconnected, so its charge will remain the same. Then, charge on C 2 = C V . New capacitance in both cases will become, K C . When they are connected in parallel charge will be equally divided so charge on one capacitor is q = K + 1 2 C V . So common potential, V ' = q K C = K + 1 2 K V = 55   V .