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

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$\frac{Q}{2} (1 - e^{\frac{t}{RC}})$

$\frac{Q}{2} (1 - e^{- \frac{2 t}{RC}})$

$Q (1 - e^{- \frac{t}{RC}})$

$Q (1 - e^{- \frac{2 t}{RC}})$

answer is B.

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Detailed Solution

At any instance charges on capacitors will be as shown

Applying KVL, we get
$\frac{q}{C} = \frac{(Q - q)}{C} + iR = 0 and i = \frac{dq}{dt}$

$\frac{dq}{dt} = \frac{Q - 2 q}{CR}$

$\Rightarrow \frac{dq}{Q - 2 q} = \frac{1}{RC} \cdot dt$

$\Rightarrow \int_{0}^{q} \frac{dq}{Q - 2 q} = \frac{1}{RC} \cdot \int_{0}^{t} dt$

$\Rightarrow - \frac{1}{2} [\ln (Q - 2 q) - lnQ] = \frac{1}{RC} \cdot t$

$\Rightarrow \ln \frac{Q - 2 q}{Q} = \frac{- 2}{RC} \cdot t$

$\Rightarrow q = \frac{Q}{2} (1 - e^{- 2 t / RC})$

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