In the circuit shown in the figure, if switches S1 and S2 have been closed for a long time, then charge on the  capacitor

After long time, current through the capacitor = 0 Current through 6  Ω   resistor,  i=12−48 =1AVoltage across capacitor, V = 4 + (6) (1) = 10 VCharge on the capacitor, Q=10 10 = 100 μCAfter the insertion of dielectricC'=Aε0d3+d3+d32  =  Aε0d  113 +13 +16 =10  15/6 = 65 10=12  μFHere voltage across capacitor remains same.∴   Charge  on  the  capacitor,  Q'=12 10 = 120 μC