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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
Options:
The charge on $C_2$ after equilibrium has reached is
Solution:
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Verified Answer
The correct answer is:
$6.0 \times 10^{-5} \mathrm{C}$
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 $C_2, q_2=C_2 V$
$$
=2 \times 10^{-5} \times 3=6 \times 10^{-5} \mathrm{C}
$$
$$
\begin{aligned}
C_1 & =10 \mu \mathrm{F} \\
& =10^{-5} \mathrm{~F} \\
V_1 & =9 \mathrm{~V}
\end{aligned}
$$
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 $C_2, q_2=C_2 V$
$$
=2 \times 10^{-5} \times 3=6 \times 10^{-5} \mathrm{C}
$$
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