In a discharging RC circuit, how many time constants must elapse...

The charge on a discharging capacitor at any time $t$ is given by the equation:

$q = q_{0} e^{-t/\tau}$

where $q_{0}$ is the maximum initial charge and $\tau$ is the time constant of the circuit.

Given that the charge decreases to $0.1\%$ of its maximum value:

$q = 0.1\% \text{ of } q_{0} = \dfrac{0.1}{100} q_{0} = 10^{-3} q_{0}$

Substituting this into the discharging equation:

$10^{-3} q_{0} = q_{0} e^{-t/\tau}$

$e^{-t/\tau} = 10^{-3}$

Taking the natural logarithm on both sides:

$\dfrac{-t}{\tau} = \ln(10^{-3})$

$\dfrac{-t}{\tau} = -3 \ln(10)$

Using the value $\ln(10) \approx 2.303$:

$t = 3 \times 2.303 \tau$

$t \approx 6.9 \tau$

Thus, $6.9$ time constants must elapse.