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Question
Find the charge on each of the capacitors 0.20 ms after the switch S is closed in the figure.
Solution
The equivalent capacitance of the circuit,
\[C_{eqv} = C_1 + C_2 = 2 + 2 = 4 \mu F\]
The growth of charge through the capacitor,
q = q0(1 − e−t/RC)
\[q_0 = CV = 4 \times {10}^{- 6} \times 6 = 24 \times {10}^{- 6} C\]
\[\frac{t}{RC} = \frac{0 . 20 \times {10}^{- 3}}{25 \times 4 \times {10}^{- 6}} = 2\]
⇒ q = 24 × 10−6 (1 − e−2)
= 18.4 × 10−6 C
This is the total charge on both capacitors. As the capacitors are in parallel, the total charge will be shared between them. Also, both the capacitors are of same capacitance; so, they will share equal amount of charge.
∴ Charge on each capacitor \[= \frac{18 . 4}{2} \mu C = 9 . 2 \mu C\]
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