A capacitor is charged with a battery and energy stored is u. After disconnecting battery another capacitor of same capacity is connected in parallel to the first capacitor. Then energy stored in each capacitor is

Explanation

The question has stated that a capacitor is connected to a battery (V) and because of battery; energy is stored in a capacitor which is U. (shown in figure 1). Now this capacitor is connected to another capacitor having the same capacity in parallel connection. So we need to calculate how much energy is stored in each of the capacitors, shown in figure 2. We know that energy stored in capacitor is given by, U=q2/2c Where q is charged stored in capacitor, C is capacitance of capacitor. Now if you look at figure 2, charge stored on each of the capacitors is C. But initially capacitance of upper capacitor is C and capacitance of lower capacitor is zero. Finally charge from the upper capacitor distributes to the lower capacitor because potential is always the same in case of parallel combination, hence potential between these capacitors are also the same. So, charge on upper capacitor is q/2 and charge on lower capacitor is q/2 if U is the new energy stored in upper or lower capacitance, then