A capacitor of capacitance C = 900 pF is charged fully by 100 V battery B as shown in figure (a). Then it is disconnected from the battery and connected to another uncharged capacitor of capacitance C = 900 pF as shown in figure (b). The electrostatic energy stored by the system (b) is:

 

Concept: 

Energy stored in capacitor =    ---- (1)

Electrostatic energy stored in capacitor =     ---- (2)

The common potential is    ---- (3)

Calculation: 

Given: 

Voltage of battery = 100 V , Capacitance of capacitor = C = 900 pF

Initially energy stored in capacitor =  = (1/2)× 900× 10^-12 × (100)² = 4.5× 10^-6 J

When the capacitor is disconnected and is connected to another capacitor of 900 pF which can be seen as:

Charge on the system becomes Q' = 2Q

From equation (3) we get: 

The common potential is 

⇒  = 50v

The electrostatic energy stored in capacitor =  =  = 2.25× 10^-6 J

Hence option 4) is correct.