Electric Current — Question 10

Question 10

Two current probes expose charge accumulating between their planes

Two probes bracket a wire segment. Toward $+x$, they measure $I_L=I_0$ and $I_R=I_0(1-e^{-t/\tau})$ for $t\ge0$. The segment starts neutral.

1. Find its excess charge as a function of time and its limiting charge.

2. Find the time when 90 percent of the limiting charge has accumulated, and explain why the steady currents eventually agree.

Question 10 - Solution

Two current probes expose charge accumulating between their planes

(a) Method and calculation

$$dQ_{\mathrm{seg}}/dt=I_L-I_R=I_0e^{-t/\tau}.$$$$Q_{\mathrm{seg}}=I_0\tau(1-e^{-t/\tau}),\quad Q_\infty=I_0\tau.$$

(b) Interpretation and further calculation

$$1-e^{-t_{90}/\tau}=.9\Rightarrow t_{90}=\tau\ln10.$$$$\lim_{t\to\infty}(I_L-I_R)=0\Rightarrow\lim dQ_{\mathrm{seg}}/dt=0.$$

Physical check

The segment retains a finite excess charge after accumulation stops; steady transport does not require exact neutrality.