Two capacitors having capacitance C₁and C₂respectively are connected as shown in figure. Initially, capacitor C₁is 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:

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$\frac{C_{1} C_{2}}{(C_{1} + C_{2})} V$

$\frac{(C_{1} + C_{2})}{C_{3} C_{3}} V$

$(C_{1} + C_{2}) V$

$(C_{1} - C_{2}) V$

answer is A.

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Detailed Solution

Initially the total charge on the C₁ which is C₁V,

Now this charge is distributed among both the capacitors.

Common potential is $\frac{C_{1} V}{C_{1} + C_{2}}$

So charge on capacitor C₂is given by $\frac{C_{1} C_{2}}{C_{1} + C_{2}} V$

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