Electric Charge — Question 1

Question 1

Net charge and total charge magnitude answer different questions

An isolated sample contains fixed positive charge $+8.40\,\mu$C and mobile negative charge $-8.55\,\mu$C. A second sample contains $+2.10\,\mu$C and $-1.95\,\mu$C. Here the sum of charge magnitudes means adding absolute values before cancellation.

1. Find each net charge and each sum of magnitudes. Find the net charge of the two-sample system.

2. For the first sample, find the excess negative charge as a fraction of its positive charge. Explain how a nearly neutral sample can contain large amounts of both signs.

Question 1 - Solution

Net charge and total charge magnitude answer different questions

(a) Method and calculation

$$Q_A=8.40-8.55=-.150\,\mu\mathrm C,\quad Q_B=2.10-1.95=+.150\,\mu\mathrm C.$$$$M_A=8.40+8.55=16.95\,\mu\mathrm C,\quad M_B=4.05\,\mu\mathrm C,\quad Q_{AB}=0.$$

(b) Interpretation and further calculation

$$f=\frac{8.55-8.40}{8.40}=.0178571\approx1.79\%.$$$$Q=Q_++Q_-,\qquad M=|Q_+|+|Q_-|\ge|Q|.$$Neutrality describes cancellation of signed charge, not the absence of charged constituents.

Physical check

The two samples together are neutral even though neither sample is individually neutral.