A parallel plate capacitor has plate of length 'l', width ‘w’ and separation of plates is ‘d’. It is connected to a battery of emf V. A dielectric slab of the same thickness ‘d’ and of dielectric constant k = 4 is being inserted between the plates of the capacitor. At what length of the slab inside plates, will the energy stored in the capacitor be two times the initial energy stored?

Two capacitors of capacitances C and 2C are charged to potential differences V and 2V, respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is :

A capacitor C is fully charged with voltage V₀. After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance $${C \over 2}$$. The energy loss in the process after the charge is distributed between the two capacitors is :

In the circuit shown in the figure, the total charge is 750 $$\mu $$C and the voltage across capacitor C₂ is 20 V. Then the charge on capacitor C₂ is :