Question

Find the charge on each of the capacitors 0.20 ms after the switch S is closed in the figure (32-E34).
Figure

Answer 1

Answer:

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Answer 2

The charge on the capacitors is 9.2 μC.

Explanation:

The circuit’s equivalent capacitance,

Ceqv = C1 + C2 = 4 μF

Through the capacitor, the charge growth is

q = q0(1 − e−t/RC)

$q_0$ = CV = 4 $\times$ 10-6 $\times$ 6 = 0.20 $\times$ 10-325 $\times$ 4 $\times$ 10-6 = 2

⇒ q = 18.4 $\times$ 10−6 C

• On both capacitors, this is the total charge. The capacitors are in parallel and among them, the total charge will be common. Both the capacitors have the same capacitance.
• There will be the sharing of equal amount of charge.
• Therefore, the charge on each capacitor is 9.2 μC.