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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

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Solution

(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=εC(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
Q1=εC(1-e-10×10-3/RC)
The charge developed on the plates of the capacitor in first 20 mili seconds is given by
Q'=εC(1-e-20×10-3/RC)

The charge developed at the plates of the capacitor in the interval t = 10 mili seconds to 20 mili seconds is given by
Q2=Q'-Q1Q2=εC(1-e-20×10-3/RC) - εC(1-e-10×10-3/RC)Q2=εC(e-10×10-3/RC- e-20×10-3/RC)Q2=εC(e-10×10-3/RC)(1- e-10×10-3/RC)
Comparing Q1 with Q2
Q1=εC(1-e-10×10-3/RC), Q2=εCe-10×10-3/RC(1-e-10×10-3/RC)Q1Q2=1e-10×10-3/RCe-10×10-3/RC<1Q1>Q2

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 μC=εC(1-e-t1/RC) ...(1)
The time taken for 20 μC charge to develop on the plates of the capacitor is t'
20 μC=εC(1-e-t2/RC) ....(2)

Dividing (2) by (1)
20 μC10 μC=εC(1-e-t2/RC)εC(1-e-t1/RC)2=(1-e-t2/RC)(1-e-t1/RC)2(1-e-t1/RC) = (1-e-t2/RC) e-t2/RC=2e-t1/RC-1

Taking natural log both side,
t2RC=ln(2)+t1RC
t1 < t2

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