31. Maximum Current between Spheres
Step 1: System Parameters
The two conducting spheres act as capacitors. Since they are far apart, mutual capacitance is negligible.
$$ C_1 = 4\pi\epsilon_0(3r), \quad C_2 = 4\pi\epsilon_0(r) $$
Initially, $Q_1 = Q$ and $Q_2 = 0$.
Step 2: Condition for Maximum Current
The system is an LC oscillator. Charge flows from the higher potential sphere to the lower one. Maximum current occurs when the potentials equalize (voltage across inductor is zero, so $di/dt = 0$, meaning $i$ is max).
$$ V_1 = V_2 \implies \frac{q_1}{C_1} = \frac{q_2}{C_2} $$
Let $q$ be the charge transferred. $q_1 = Q-q$, $q_2 = q$.
$$ \frac{Q-q}{3r} = \frac{q}{r} \implies Q-q = 3q \implies q = \frac{Q}{4} $$
Step 3: Energy Conservation
Loss in potential energy equals gain in magnetic (kinetic) energy.
$$ U_i = \frac{Q^2}{2C_1} $$
$$ U_f = \frac{(Q-Q/4)^2}{2C_1} + \frac{(Q/4)^2}{2C_2} + \frac{1}{2} L I_{max}^2 $$
Substitute $C_1 = 3C_2$ (let $C_2 = C = 4\pi\epsilon_0 r$):
$$ \frac{Q^2}{6C} = \frac{(3Q/4)^2}{6C} + \frac{(Q/4)^2}{2C} + \frac{1}{2} L I_{max}^2 $$
$$ \frac{Q^2}{6C} \left( 1 – \frac{9}{16} \right) – \frac{Q^2}{32C} = \frac{1}{2} L I^2 $$
$$ \frac{Q^2}{C} \left( \frac{7}{96} – \frac{3}{96} \right) = \frac{1}{2} L I^2 \implies \frac{4 Q^2}{96 C} = \frac{1}{2} L I^2 $$
$$ I_{max} = \frac{Q}{\sqrt{12 LC}} = \frac{Q}{\sqrt{48 \pi \epsilon_0 r L}} $$