Charge Is Conserved — Question 3

Question 3

A time-dependent charge influx can reverse accumulation during one pulse

A region begins with $Q(0)=0$. From $0$ to $\tau$, its inward conventional charge rate is $I_0\sin(\pi t/\tau)$ and its outward rate is $I_0/2$, where $I_0=2.00$ nC/s and $\tau=3.00$ s. These rates are defined as charge per time and need not be constant.

1. Integrate the net rate to obtain $Q(t)$ and the final charge $Q(\tau)$.

2. Find the times of the minimum and maximum charge during the pulse, and verify their classification from the signs of the accumulation rate.

Question 3 - Solution

A time-dependent charge influx can reverse accumulation during one pulse

(a) Method and calculation

$$\frac{dQ}{dt}=I_0[\sin(\pi t/\tau)-1/2].$$$$Q(t)=I_0\left[\frac\tau\pi(1-\cos(\pi t/\tau))-\frac t2\right].$$$$Q(\tau)=I_0\tau(2/\pi-1/2)\approx.819719\,\mathrm{nC}.$$

(b) Interpretation and further calculation

$$\sin(\pi t/\tau)=1/2\Rightarrow t_{\min}=\tau/6=.500\,\mathrm s,\quad t_{\max}=5\tau/6=2.50\,\mathrm s.$$$$Q_{\min}=I_0\tau[(1-\sqrt3/2)/\pi-1/12],\quad Q_{\max}=I_0\tau[(1+\sqrt3/2)/\pi-5/12].$$The rate is negative, then positive, then negative on the three successive intervals. These extrema are also the global minimum and maximum over the pulse.

Physical check

Net positive charge at the end does not imply positive accumulation at every earlier instant.