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,

\[V_1 = \frac{Q - q}{C} \]

\[ V_2 = \frac{q}{C}\]

\[ \Rightarrow V_1 - V_2 = \frac{Q - q}{C} - \frac{q}{C}\]

\[ = \frac{Q - 2q}{C}\]

The current through the circuit after time t,

\[i = \frac{V_1 - V_2}{R} = \frac{dq}{dt}\]

\[ \Rightarrow \frac{Q - 2q}{CR} = \frac{dq}{dt}\]

\[ \Rightarrow \frac{dq}{Q - 2q} = \frac{1}{RC}dt\]

\[ \Rightarrow \frac{dq}{Q - 2q} = \frac{1}{RC}dt\]

Integrating both sides within the limits time =0 to t and charge on the second capacitor varying from q=0 to q, we get:-

\[\frac{1}{2} \left[ \ln \left( Q - 2q \right) - \ln Q \right] = \frac{- 1t}{RC}\]

\[\ln \frac{Q - 2q}{Q} = \frac{- 2t}{RC}\]

\[Q - 2q = Q e^{- \frac{2t}{RC}} \]

\[2q = Q\left( 1 - e^{- \frac{2t}{RC}} \right)\]

\[q = \frac{Q}{2}\left\{ 1 - e^{- \frac{2t}{RC}} \right\}\]