A parallel plate capacitor of capacitance 2 F is charged to a potential V . The energy stored in the capacitor is E 1 . The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is E 2 . The ratio E 2 E 1 is

Step-by-step solution

The charge on the plates of the first capacitor when connected against the potential difference V is given by Q = C V = 2 V When both the capacitors are connected, from the conservation of charge, it can be written that 2 V = 2 V ' + 2 V ' ⇒ V ' = 1 2 V where, V ' is the new potential difference across each capacitor. The formula to calculate the energy stored in the first capacitor is given by E 1 = 1 2 × C × V 2 = 1 2 × 2 × V 2 = V 2       . . . 1 For the second case, the