Question

Consider a capacitor-charging circuit. Let Q₁ be the charge given to the capacitor in a time interval of 10 ms and Q₂ be the charge given in the next time interval of 10 ms. Let 10 μC charge be deposited in time interval t₁ and another 10 μC charge be deposited in the next time interval t₂.
(a) Q₁ > Q₂, t₁ > t₂
(b) Q₁ > Q₂, t₁ < t₂
(c) Q₁ < Q₂, t₁ > t₂
(d) Q₁ < Q₂, t₁ < t₂

Answer 1

(b) Q₁ > Q₂, t₁ < t₂

The charge Q on the plates of a capacitor at time t after connecting it in a charging circuit,
$Q=\epsilon C(1-e^{-t/RC})$,
where
ε = emf of the battery connected in the charging circuit
C = capacitance of the given capacitor
R = resistance of the resistor connected in series with the capacitor

The charge developed on the plates of the capacitor in first 10 mili seconds is given by
$Q_1=\epsilon C(1-e^{-10\times {10}^{-3}/RC})$
The charge developed on the plates of the capacitor in first 20 mili seconds is given by
$Q\text{'}=\epsilon C(1-e^{-20\times {10}^{-3}/RC})$

The charge developed at the plates of the capacitor in the interval t = 10 mili seconds to 20 mili seconds is given by
$$\begin{gathered} Q_2=Q\text{'}-Q_1 \\ {Q}_2=\left[\epsilon C(1-e^{-20\times {10}^{-3}/RC})\right]-\left[\epsilon C(1-e^{-10\times {10}^{-3}/RC})\right] \\ \Rightarrow Q_2=\left[\epsilon C(e^{-10\times {10}^{-3}/RC}-e^{-20\times {10}^{-3}/RC})\right] \\ \Rightarrow Q_2=\left[\epsilon C\left(e^{-10\times {10}^{-3}/RC}\right)(1-e^{-10\times {10}^{-3}/RC})\right] \end{gathered}$$
Comparing Q₁ with Q₂
$$\begin{gathered} Q_1=\epsilon C(1-e^{-10\times {10}^{-3}/RC}),Q_2=\epsilon C{e}^{-10\times {10}^{-3}/RC}(1-e^{-10\times {10}^{-3}/RC}) \\ \frac{Q_1}{Q_2}=\frac{1}{e^{-10\times {10}^{-3}/RC}} \\ {e}^{-10\times {10}^{-3}/RC}<1 \\ \therefore Q_1>Q_2 \end{gathered}$$

For second part of the question

10 μC charge be deposited in a time interval t₁ and the next 10 μC charge is deposited in the next time interval t₂.
The time taken for 10 μC charge to develop on the plates of the capacitor is t₁
$10\mathrm{\mu C}=\epsilon C(1-e^{-t_1/RC})$ ...(1)
The time taken for 20 μC charge to develop on the plates of the capacitor is t'
$20\mathrm{\mu C}=\epsilon C(1-e^{-t_2/RC})$ ....(2)

Dividing (2) by (1)
$$\begin{gathered} \frac{20\mathrm{\mu C}}{10\mathrm{\mu C}}=\frac{\epsilon C(1-e^{-t_2/RC})}{\epsilon C(1-e^{-t_1/RC})} \\ 2=\frac{(1-e^{-t_2/RC})}{(1-e^{-t_1/RC})} \\ 2(1-e^{-t_1/RC})=(1-e^{-t_2/RC}) \\ {e}^{-t_2/RC}=2e^{-t_1/RC}-1 \end{gathered}$$

Taking natural log both side,
$\frac{t_2}{RC}=\ln \left(2\right)+\frac{t_1}{RC}$
$\Rightarrow$t₁ < t₂