Question

A capacitor of capacitance 100 μF is connected across a battery of emf 6 V through a resistance of 20 kΩ for 4 s. The battery is then replaced by a thick wire. What will be the charge on the capacitor 4 s after the battery is disconnected?

Answer 1

Given:
Capacitance of capacitor ,C = 100 μF
Emf of battery ,E = 6V
Resistance ,R = 20 kΩ
Time for charging , t₁ = 4 s
Time for discharging ,t₂ = 4 s

During charging of the capacitor, the growth of charge across it,
$$\begin{gathered} \mathrm{Q}=CE\left(1-{\mathrm{e}}^{-\frac{t_1}{RC}}\right) \\ \frac{t_1}{RC}=\frac{4}{20\times {10}^3\times 100\times {10}^{-6}} \\ =2 \\ \Rightarrow Q=6\times {10}^{-4}\left(1-{\mathrm{e}}^{-2}\right) \\ =5.187\times {10}^{-4}\mathrm{C} \end{gathered}$$

This is the amount of charge developed on the capacitor after 4s.
During discharging of the capacitor, the decay of charge across it,
$$\begin{gathered} Q\text{'}=Q\left({\mathrm{e}}^{-\frac{t}{RC}}\right) \\ =5.184\times {10}^{-4}\times {\mathrm{e}}^{-2} \\ =0.7\times {10}^{-4}\mathrm{C}=70\mathrm{\mu C} \end{gathered}$$