Two capacitors having capacitance C₁ and C₂ respectively are connected as shown in figure. Initially, capacitor C₁ is charged to a potential difference V volt by a battery. The battery is then removed and the charged capacitor C₁ is now connected to uncharged capacitor C₂ by closing the switch S. The amount of charge on the capacitor C₂, after equilibrium, is :

CONCEPT:

• The charge in the capacitor is written as;

            Q = CV 

          Here we have Q as the charge, C as the capacitance, and V as the potential applied.

• When two capacitors are parallel then total capacitance is written as;

           C = C₁ +C₂

           Here C is the total capacitance, C₁ is the capacitance of the first capacitor, and C₂ is the capacitance of the second capacitor.

CALCULATION:

Initially when capacitor C1 is charged to a potential difference V volt then charge Q is written as;

Q = C₁V

After the removal of the battery, both capacitors are parallel to each other and it is written as;

C_total = C₁ + C₂ 

The charge on C₂ is written as,

Q' = $\frac{C_2{Q}_{total}}{C_{total}}$

Q' = $\frac{C_1{C}_{2}V}{C_1+C_2}$

Hence, option 1) is the correct answer.