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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.
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Solution
The two 2.0 μF capacitors are in parallel, so Ceq = 2.0 + 2.0 = 4.0 μF.
With the 25 Ω series resistor and a 6.0 V battery:
τ = RC = (25) (4.0 × 10−6) = 1.0 × 10−4 s = 0.10 ms
At t = 0.20 ms = 2τ, the voltage across the capacitor bank is
VC(t) = V(1 − e−t/τ) = 6.0 (1 − e−2) ≈ 6.0 (0.8647) ≈ 5.19 V.
Each parallel capacitor has this same voltage, so the charge on each:
Q = CV = (2.0μF) (5.19V) ≈ 1.04 × 10−5C = 10.4μC.
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