A capacitor of capacitance $C_1=10 \mu \mathrm{F}$ to charged using $9 \mathrm{~V}$ battery. It is then removed from the battery and connected to another capacitor $C_2=20 \mu \mathrm{F}$ as shown in the figure. The charge on $C_2$ after equilibrium has reached is

• Correct answer
  A

  $6.0 \times 10^{-5} \mathrm{C}$
• B

  $60 \times 10^{-6} \mathrm{C}$
• C

  $3.0 \times 10^{-5} \mathrm{C}$
• D

  $3.0 \times 10^{-6} \mathrm{C}$

Solution

Given, $$ \begin{aligned} C_1 & =10 \mu \mathrm{F} \\ & =10^{-5} \mathrm{~F} \\ V_1 & =9 \mathrm{~V} \end{aligned} $$
Charge on capacitor $C_1$, $$ \begin{aligned} q_1 & =C_1 V_1 \\ & =9 \times 10^{-5} \mathrm{C} \\ C_2 & =20 \mu \mathrm{F}=2 \times 10^{-5} \mathrm{~F} \end{aligned} $$ when uncharged capacitor $C_2$ is connected with charged capacitor $C_1$, then charges will flow from capacitor $C_1$ to $C_2$ till both gain the same potential i.e common potential. $$ \begin{aligned} & \text { Common potential, } V=\frac{\text { total charge }}{\text { total capacitance }} \\ & =\frac{q_1}{C_1+C_2}=\frac{9 \times 10^{-5}}{10^{-5}+2 \times 10^{-5}} \\ & =\frac{9 \times 10^{-5}}{3 \times 10^{-5}}=3 \mathrm{~V} \end{aligned} $$ Charge on capacitor $C_2, q_2=C_2 V$ $$ =2 \times 10^{-5} \times 3=6 \times 10^{-5} \mathrm{C} $$

Hence, the correct answer is option: 'A'.