A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the combined system to that stored initially in the single capacitor.

Step 1: Initial energy stored in capacitor Ui=1/2CV² (1)
Step 2: Charge conservation [Ref. Fig.]
Let the Initial charge be Q
After connecting with another capacitor of the equal capacitance, the charge will flow until the potential becomes the same.
V1=V2 ⇒ Q1/C1 = Q2/C2 ⇒ Q1=Q2 [∵C1=C2=C]
Applying charge conservation:
Q1+Q2=Q ∴Q1=Q2=Q/2
Step 3: Potential on each capacitor
V1=Q1/C=Q/2C, V2=Q2/C=Q/2C ⇒V1=V2, V2=V/2
Step 4: Final energy stored
Uf=1/2C1V1²+1/2CV2²=1/2C(V/2)²+1/2C(V/2)² ⇒Uf=CV²/4 (2)
Step 5: Ratio of energy
From equation(1)and(2)
Uf/Ui=(CV²/4)/(CV²/2) ⇒Uf/Ui=1/2