Question

A charged capacitor of capacitance C is discharged through a resistance R. A radioactive sample decays with an average-life τ. 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.

Answer 1

Discharging of a capacitor through a resistance R is given by
$Q=q{e}^{-t/CR}$
Here, Q = Charge left
q = Initial charge
C = Capacitance
R = Resistance

Energy, E = $\frac{1}{2}\frac{Q^2}{C}$ = $\frac{q^2{e}^{{\displaystyle -2t/CR}}}{2C}$

Activity, A = A₀e^{$-\lambda t$}
Here, A₀ = Initial activity
$\lambda$ = Disintegration constant

∴ Ratio of the energy to the activity = $\frac{E}{A}=\frac{q^2\times e^{-2t/CR}}{2C{A}_0{e}^{-\lambda t}}$
Since the terms are independent of time, their coefficients can be equated.
$\frac{2t}{CR}=\lambda t$
$\Rightarrow \lambda =\frac{2}{CR}$
$\Rightarrow \frac{1}{\tau }=\frac{2}{CR}$
$\Rightarrow R=2\frac{\tau }{C}$