A 20 μF capacitor is joined to a battery of emf 6.0 V through a resistance of 100 Ω. Find the charge on the capacitor 2.0 ms after the connections are made.
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The charge on the capacitor 2.0 ms after the connections are made is 76 μC.
Explanation:
Across the capacitor, the growth of charge,
Q = CV = Q0.e - tRCQ0 = 20 6 10-6 C
⇒Q = 120 10-6. e-2 10-3102.20 10-6
⇒Q = 120 10-6. e-1
Charge on the capacitor =76 μC
A capacitor is an instrument that in an electric field supplies electrical energy. It has with two terminals.
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