In the circuit shown here $C_1 = 6\,\mu F, C_2 = 3 \,\mu F$ and battery $B = 20\,V$. The switch $S_1$ is first closed. It is then opened and afterwards $S_2$ is closed. What is the charge (in $\mu F$)finally on $C_2$ ?
Medium
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Initial charge on $C_{1}$ is $q=C_{1} V_{B}=6 \times 10^{-6} \times 20$
$=120\, \mu \mathrm{C}$
Now when $S_{2}$ closed, suppose common potential $=$ $\mathrm{V}$
charge on $\mathrm{C}_{1}=\mathrm{q}_{1}$
and charge on $\mathrm{C}_{2}=\mathrm{q}_{2}$
So, $ \mathrm{q}_{1}=\mathrm{C}_{1} \mathrm{V}$ and $\mathrm{q}_{2}=\mathrm{C}_{2} \mathrm{V}$
$\frac{\mathrm{q}_{1}}{\mathrm{q}_{2}}=\frac{\mathrm{C}_{1}}{\mathrm{C}_{2}} \Rightarrow \frac{\mathrm{q}_{1}}{\mathrm{q}_{2}}=\frac{6}{3}=2 \Rightarrow \mathrm{q}_{1}=2 \mathrm{q}_{2}$
Also $\mathrm{q}_{1}+\mathrm{q}_{2}=\mathrm{q} \Rightarrow 2 \mathrm{q}_{2}+\mathrm{q}_{2}=\mathrm{q}$
$\Rightarrow \mathrm{q}_{2}=\frac{\mathrm{q}}{3}=\frac{120 \times 10^{-6}}{3}$
$=40\, \mu \mathrm{C}$
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