ELECTROSTATICS BYU 33

1. System Analysis

We have two spheres connected by a wire, so they share the same potential $V$.

• Sphere 1: Radius $r_1$, isolated. Capacitance $C_1 = 4\pi\epsilon_0 r_1$.
• Sphere 2: Radius $r_2$, surrounded by a grounded shell at distance $d$ ($d \ll r_2$). This forms a spherical capacitor. Capacitance $C_2 \approx \frac{4\pi\epsilon_0 r_2^2}{d}$.

Total charge given to the conducting system is $Q$.

2. Charge Distribution

Charges distribute according to capacitance ($q = CV$).

$$ q_1 = C_1 V \quad \text{and} \quad q_2 = C_2 V $$ $$ \frac{q_1}{q_2} = \frac{C_1}{C_2} = \frac{4\pi\epsilon_0 r_1}{4\pi\epsilon_0 r_2^2 / d} = \frac{r_1 d}{r_2^2} $$

Using conservation of charge $q_1 + q_2 = Q$:

$$ q_2 \left( \frac{r_1 d}{r_2^2} + 1 \right) = Q $$ $$ q_2 = Q \left( \frac{r_2^2}{r_1 d + r_2^2} \right) $$

Consequently for $q_1$:

$$ q_1 = Q \left( \frac{r_1 d}{r_1 d + r_2^2} \right) $$