Daily Physicist

Today is the birthday of

• Irving Langmuir (1881-1957).
  
• Rudolf Mössbauer (1929-2011).

Today’s Problem

For the circuit below show that the rate of change of the charge on the capacitor is mathematically identical to the rate of change of the number of particles in the leaking container that was calculated in yesterday’s problem.

Answer

Let \(Q\) be the charge on the capacitor, then \(V_2=Q/C\), and using Ohm’s Law, we have for the current

\[\begin{align} i=\frac{dQ}{dt} &=\frac{V_1-V_2}{R}\\ &=\frac{V_1}{R}-\frac{Q}{RC}=\frac{1}{RC}(V_1C-Q) \end{align}\]

This has the same form as the differential equation for the number of particles in the container that was calculated yesterday.

The solution is also mathematically identical. We have

\[Q(t)=V_1C+(Q_0-V_1C)e^{-\alpha t}\]

where \(\alpha=1/RC\) and \(Q_0=Q(0)\).