Conductors and Insulators — Question 8

Question 8

A finite redistribution time separates transient charge from equilibrium charge

A supplied model for two connected identical metal regions gives $Q_A+Q_B=12$ nC and a charge difference $D=Q_A-Q_B$ satisfying $dD/dt=-D/\tau$, with $\tau=2.0$ ms. Initially $(Q_A,Q_B)=(12,0)$ nC. This model represents a measured equilibration law, not a universal constant for all conductors.

1. Solve for both charges as functions of time and evaluate them at $t=2.0$ ms.

2. Find the minimum time for the magnitude of the charge difference to be at most 1 percent of its initial value.

Question 8 - Solution

A finite redistribution time separates transient charge from equilibrium charge

(a) Method and calculation

$$\frac{dD}{D}=-\frac{dt}{\tau}\Rightarrow D(t)=12e^{-t/\tau}\,\mathrm{nC}.$$$$Q_A=6(1+e^{-t/\tau}),\quad Q_B=6(1-e^{-t/\tau})\quad\mathrm{nC}.$$$$t=\tau:\quad Q_A\approx8.20728\,\mathrm{nC},\quad Q_B\approx3.79272\,\mathrm{nC}.$$

(b) Interpretation and further calculation

$$|D(t)|/|D(0)|=e^{-t/\tau}\le.01.$$$$t\ge\tau\ln100\approx9.21034\,\mathrm{ms}.$$A very good equilibrium approximation occurs in finite time, although the ideal exponential becomes exactly zero only in the infinite-time limit.

Physical check

The sum remains constant throughout the transient; only the imbalance relaxes.