A capacitor of capacitance $10\mu\text{F}$ is connected across a battery of emf 6.0V through a resistance of $20\text{k}\Omega$ for 4.0s. The battery is then replaced by a thick wire. What will be the charge on the capacitor 4.0s after the battery is disconnected?
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$\text{C} = 100\mu\text{F, emf} = 6\text{V}, \text{ R} = 20 \text{K}\Omega, \text{t} = 4 \text{S}.$
Charging: $\text{Q}=\text{CV}\Big(1-\text{e}^{\frac{-\text{t}}{\text{RC}}}\Big)\ \bigg[\frac{-\text{t}}{\text{RC}}=\frac{4}{2\times10^4\times10^{-4}}\bigg]$
$=6\times10^{-4}\big(1-\text{e}^{-2}\big)=5.187\times10^{-4}\text{C}=\text{Q}$
Discharging: $\text{q}=\text{Q}\Big(\text{e}^{\frac{-\text{t}}{\text{RC}}}\Big)=5.184\times10^{-4}\times\text{e}^{-2}$
$=0.7\times10^{-4}\text{C}=70\mu\text{c}.$
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