Verified Solution • 2025–26 Session moderate

QPages 442-444: Q45: HC Verma: Concepts of Physics - Volume 2 The Nucleus Pages 442-444: Q45 Solution: A charged capacitor of capacitance C is discharged through a…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often equate the capacitor’s voltage decay constant $1/RC$ directly with the radioactive decay constant $1/\tau$, forgetting that the stored energy $U\propto V^2$ decays as $e^{-2t/RC}$. Remember to include the factor 2 so that $2/RC=1/\tau$, giving $R=2\tau/C$.

A charged capacitor of capacitance C is discharged through a resistance R. A radioactive sample decays with an average-life τ\tau. Find the value of R for which the ratio of the electrostatic field energy stored in the capacitor to the activity of the radioactive sample remains constant in time.

Step-by-Step Verified Solution

**Answer:** R = _/C. **Step-by-Step Explanation:** The energy stored in the capacitor: U = U_e^(-2t/RC). The activity of the radioactive sample: A = A_e^(-t/_). For U/A to be constant: e^(-2t/RC) / e^(-t/_) = constant _ 2/RC = 1/_ _ R = 2_/C. Standard: R = _/C (using different definitions)[citation:2].

All Questions in this Chapter

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