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.

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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' = \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 Q_2 = Q' - 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( e^{- 10 $\times$ {10}^{- 3} /RC} )(1 - e^{- 10 $\times$ {10}^{- 3} /RC} ) \right] Comparing Q1 with Q2 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} ) Q_1/Q_2 = \frac{1}{e^{- 10 $\times$ {10}^{- 3} /RC}} e^{- 10 $\times$ {10}^{- 3} /RC} < 1 \therefore Q_1 > Q_2 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 \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 \mu C = \epsilon C(1 - e^{- t_2 /RC} )............(2) Dividing (2) by (1) 20 \mu C/10 \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} = 2 e^{- t_1 /RC} - 1 Taking natural log both side, t_2/RC = \ln(2) + t_1/RC $\Rightarrow$ t1 < t2

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