A capacitor of capacity `C` is connected with a battery of potential `V` in parallel. The distance between its plates is reduced to half at one, assuming that the charge remains the same. Then to charge the capacitance upto the potential `V` again, the energy given by the battery will be

A parallel plate capacitor of capacity 100 mu F is charged by a battery of 50 V. The battery remains connected and if the plates of the capacitor are separated so that the distance between them becomes double the original distance, the additional energy given by the battery to the capacitor in joule is

A capacitor is connected to a battery of 20 V,so that a charge of 100 mu C is obtained at the plates. The capacitance of the capacitor is

A parallel plate capacitor of capacitances C is connected to a battery to charge it to a potential V. Similarly, another capacitor of capacitance 2C is charged to a potential 2V. Now the batteries are removed, and the two capacitors are connected in parallel by joining the positive plate to one with the negative plate of the other. Find out the final energy of the system.

Passage II We have two parallel plate capacitors P and Q having same capacitance C_(0) . P is connected to a battery of potential difference V and it remains connected to the battery even after the charging is over. Q is also connected to a battery of the same potential difference V but after the charging gets over, it is disconnected from the battery. Answer the following questions: Potential difference between the plates of P, when a dielectric slab is inserted in between the plates