Question

A capacitor of capacitance C is given a charge Q. At t = 0, it is connected to an uncharged capacitor of equal capacitance through a resistance R. Find the charge on the second capacitor as a function of time.

Answer 1

Given:
Initial charge on first capacitor = Q
Let q be the charge on the second capacitor after time t.
According to the principle of conservation of charge, charge on the first capacitor after time t = Q - q.
Let V₁ be the potential difference across the first capacitor and V₂ be the potential difference across the second capacitor after time t. Then,
$$\begin{gathered} V_{\mathit{1}}=\frac{Q-q}{C} \\ {V}_{\mathit{2}}=\frac{q}{C} \\ \Rightarrow {\mathrm{V}}_1-{\mathrm{V}}_2=\frac{Q-q}{C}-\frac{q}{C} \\ =\frac{Q-2q}{C} \end{gathered}$$
The current through the circuit after time t,
$$\begin{gathered} i=\frac{V_1-V_2}{R}=\frac{dq}{dt} \\ \Rightarrow \frac{Q-2q}{CR}=\frac{dq}{dt} \\ \Rightarrow \frac{dq}{Q-2q}=\frac{1}{RC}dt \\ \Rightarrow \frac{dq}{Q-2q}=\frac{1}{RC}dt \end{gathered}$$
Integrating both sides within the limits time =0 to t and charge on the second capacitor varying from q=0 to q, we get:
$$\begin{gathered} \frac{1}{2}\left[\ln \left(Q-2q\right)-\ln Q\right]=\frac{-1t}{RC} \\ \ln \frac{Q-2q}{Q}=\frac{-2t}{RC} \\ Q-2q=Q{\mathrm{e}}^{-\frac{2t}{RC}} \\ 2q=Q\left(1-{\mathrm{e}}^{-\frac{2t}{RC}}\right) \\ q=\frac{Q}{2}\left\{1-{\mathrm{e}}^{-\frac{2t}{RC}}\right\} \end{gathered}$$