Conductors and Insulators — Question 7

Question 7

Free charge mobility changes the outcome of an identical deposition

Two identical initially neutral spheres have the same radius and receive $+8.0$ nC on a small surface patch. One is a conductor on an insulating stand; the other is an ideal insulator whose deposited charge cannot migrate. Both are otherwise isolated, with no nearby external charges. Divide each sphere into equal-area left and right hemispheres; the deposition patch lies entirely on the left.

1. Find hemisphere net charges after equilibration in the conductor model and in the immobile-charge insulator model.

2. Find the net charge transported across the equator in each case. Explain why an insulator can polarize even if deposited charge remains localized.

Question 7 - Solution

Free charge mobility changes the outcome of an identical deposition

(a) Method and calculation

$$\text{conductor symmetry}:\quad Q_L=Q_R=8.0/2=4.0\,\mathrm{nC}.$$$$\text{immobile deposited charge}:\quad(Q_L,Q_R)=(8.0,0)\,\mathrm{nC}.$$

(b) Interpretation and further calculation

$$Q_{L\to R,\mathrm{conductor}}=+4.0\,\mathrm{nC},\quad Q_{L\to R,\mathrm{insulator}}=0.$$$$Q_L+Q_R=8.0\,\mathrm{nC}\quad\text{in both models}.$$Bound constituents in a real insulator can shift slightly and produce polarization. The idealized hemisphere numbers here neglect that bound-charge response and track deposited excess charge.

Physical check

Electrons moving right to left can transport positive conventional charge left to right; signed transfer need not follow electron motion.