Electrostatics
Intermediate
Electric Potential
Source: High school physics (Chinese)
Problem Sets:
electrostatics 2
Problem
Two conducting spheres $A$ and $B$ with radii $R_1$ and $R_2$ respectively are placed very far apart, so each may be treated as an isolated sphere. Initially $A$ carries charge $Q$ and $B$ is uncharged. A thin long wire is then used to connect them. Ignore any charge stored on the wire.
After electrostatic equilibrium is reached, how much charge is on each sphere?
$Q_A = \dfrac{R_1}{R_1 + R_2}\, Q$; $Q_B = \dfrac{R_2}{R_1 + R_2}\, Q$ (charge splits in proportion to radius).
At equilibrium, the wire forces the two spheres to share the same potential. For an isolated sphere of radius $R$ carrying charge $q$, the surface potential is $V = kq/R$. Setting $V_A = V_B$:
$$\dfrac{k Q_A}{R_1} = \dfrac{k Q_B}{R_2} \;\Longrightarrow\; \dfrac{Q_A}{Q_B} = \dfrac{R_1}{R_2}.$$Charge conservation: $Q_A + Q_B = Q$. Solving:
$$Q_A = \dfrac{R_1}{R_1 + R_2}\, Q, \qquad Q_B = \dfrac{R_2}{R_1 + R_2}\, Q.$$