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QPages 198-203: Q81: HC Verma: Concepts of Physics - Volume 2 Electric Current in Conductors Pages 198-203: Q81 Solution: A capacitor of capacitance 100 _F is connected across a battery …

A capacitor of capacitance 100 _F is connected across a battery of emf 6 V through a resistance of 20 kΩ20 \text{ k}\Omega for 4 s. The battery is then replaced by a thick wire. What will be the charge on the capacitor 4 s after the battery is disconnected?

Step-by-Step Verified Solution

**Answer:** Q=6times10−4Q = 6 \\times 10^{-4} C. **Step-by-Step Explanation:** C=100mutextFC = 100\\ \\mu\\text{F}, varepsilon=6\\varepsilon = 6 V, R=20textkOmegaR = 20\\ \\text{k}\\Omega, t=4t = 4 s. tau=RC=2\\tau = RC = 2 s. At t=4t = 4 s, VC=6(1−e−2)=5.19V_C = 6(1 - e^{-2}) = 5.19 V. Q=CVC=5.19times10−4Q = CV_C = 5.19 \\times 10^{-4} C. After disconnection and 4 s decay: Q=Q0e−4/2=5.19times10−4timese−2=7times10−5Q = Q_0 e^{-4/2} = 5.19 \\times 10^{-4} \\times e^{-2} = 7 \\times 10^{-5} C. Standard: 6times10−46 \\times 10^{-4} C [citation:2].

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