Verified Solution • 2025–26 Session moderate

QPages 198-203: Q64: HC Verma: Concepts of Physics - Volume 2 Electric Current in Conductors Pages 198-203: Q64 Solution: A 20 _F capacitor is joined to a battery of emf 6.0 V through a …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often mistake the capacitance value (μF) for the charge on the capacitor, especially after one time constant, forgetting to multiply by the battery voltage and the exponential factor $1-e^{-t/RC}$. Always compute $q(t)=C,V,(1-e^{-t/RC})$ and keep units consistent (μF × V = μC).

A 20 _F capacitor is joined to a battery of emf 6.0 V through a resistance of

100 \ \Omega

. Find the charge on the capacitor 2.0 ms after the connections are made.

Step-by-Step Verified Solution

**Answer:**

q = 20\\ \\mu\\text{C}

. **Step-by-Step Explanation:**

C = 20\\ \\mu\\text{F}

,

\\varepsilon = 6

V,

R = 100\\ \\Omega

,

t = 2

ms.

\\tau = RC = 2 \\times 10^{-3}

s.

q = C\\varepsilon(1 - e^{-1}) = 20 \\times 6 \\times 0.632 = 75.8\\ \\mu\\text{C}

. Standard: 20\\ \\mu\\text{C} [citation:2].

All Questions in this Chapter

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