Conductors and Insulators — Question 4

Question 4

Unequal conducting spheres need not share charge equally

Two widely separated metal spheres of radii $R_A=2.0$ cm and $R_B=6.0$ cm carry total charge $+16$ nC. A long thin wire connects them. Neglect the charge on the wire and mutual influence on the spheres. For this exercise use the supplied equilibrium rule $Q_A/R_A=Q_B/R_B$.

1. Find the final sphere charges and compare with equal sharing.

2. Using $\sigma=Q/(4\pi R^2)$ for each approximately uniform sphere, find the ratio of surface charge densities. Explain how the smaller sphere can carry less charge but have greater density.

Question 4 - Solution

Unequal conducting spheres need not share charge equally

(a) Method and calculation

$$Q_B=\frac{R_B}{R_A}Q_A=3Q_A,\quad Q_A+Q_B=16\,\mathrm{nC}.$$$$Q_A=4.0\,\mathrm{nC},\quad Q_B=12.0\,\mathrm{nC}.$$

(b) Interpretation and further calculation

$$\frac{\sigma_A}{\sigma_B}=\frac{Q_A}{Q_B}\left(\frac{R_B}{R_A}\right)^2=\frac13(3)^2=3.$$$$Q\propto R,\qquad\sigma\propto R^{-1}\quad\text{at the stated equilibrium}.$$Area grows as $R^2$, faster than the equilibrium charge grows with radius.

Physical check

The supplied ratio assumes widely separated spheres; nearby conductors alter each other charge distributions.