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 = \(\frac{1}{2}CV^{2} = \frac{1}{2}QV\)   ---- (1)

Electrostatic energy stored in capacitor = \(\frac{1}{2}(C_{1} + C_{2})V^{2}\)    ---- (2)

The common potential is \(V_{c} = \frac{C_{1}V_{1} + C_{2}V_{2}}{C_{1}+C_{2}}\)   ---- (3)

Calculation: 

Given: 

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

Initially energy stored in capacitor = \(\frac{1}{2}CV^{2} = \frac{1}{2}QV\) = (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 \(V_{c} = \frac{C_{1}V_{1} + C_{2}V_{2}}{C_{1}+C_{2}}\)

⇒ \(V_{c} = \frac{C× 100 + C× 0}{C+C}\) = 50v

The electrostatic energy stored in capacitor = \(\frac{1}{2}(C_{1} + C_{2})V^{2}\) = \(\frac{1}{2}900× 10^{-12}× 2×(50)^{2}\) = 2.25× 10^-6 J

Hence option 4) is correct.