Potential of a Charged Isolated Conductor — Question 2

Question 2

Connected distant spheres share voltage in proportion to their radii

Two conducting spheres of radii R and 2R are very far apart. Initially the smaller carries Q>0 and the larger is neutral. A thin wire connects them until electrostatic equilibrium. Neglect wire charge and mutual influence.

1. Find the final charges and common potential relative to infinity.

2. Find the two surface field magnitudes and their ratio.

Question 2 - Solution

Connected distant spheres share voltage in proportion to their radii

(a) Method and calculation

$$Q_1+Q_2=Q,\quad\frac{kQ_1}{R}=\frac{kQ_2}{2R}\Rightarrow Q_2=2Q_1.$$$$Q_1=Q/3,\quad Q_2=2Q/3,\quad V_f=kQ/(3R).$$

(b) Interpretation and further calculation

$$E_1=\frac{kQ}{3R^2},\quad E_2=\frac{k(2Q/3)}{(2R)^2}=\frac{kQ}{6R^2}.$$$$E_1/E_2=2.$$

Physical check

Equal voltage does not mean equal charge or equal field; the distant-sphere approximation neglects the other sphere contribution at each surface.