CURRENT BYU 50

System Analysis:

Consider the conducting ball of radius $r$ and the capacitor $C$. The ball initially holds a charge $Q$. When the key is closed, charge flows from the ball to the capacitor plate until they reach a common electrostatic potential. The other plate of the capacitor is earthed.

1. Charge Redistribution:
Let $q$ be the charge transferred from the ball to the capacitor.

• Final charge on the ball: $Q – q$
• Final charge on the capacitor: $q$

At steady state, the potential of the ball equals the potential of the capacitor plate connected to it (since no current flows through $R_1$). $$ V_{\text{ball}} = V_{\text{capacitor}} $$ $$ \frac{1}{4\pi\epsilon_0} \frac{Q-q}{r} = \frac{q}{C} $$ Solving for $q$: $$ C(Q-q) = 4\pi\epsilon_0 r q $$ $$ CQ = q(C + 4\pi\epsilon_0 r) \implies q = \frac{CQ}{C + 4\pi\epsilon_0 r} $$

2. Energy Analysis:
The total heat dissipated ($H$) is the loss in the stored electrostatic potential energy of the system.
Initial Energy ($U_i$): Only the ball has charge. $$ U_i = \frac{1}{2} \frac{Q^2}{C_{\text{ball}}} = \frac{Q^2}{8\pi\epsilon_0 r} $$ Final Energy ($U_f$): The ball and capacitor are effectively in parallel. The equivalent capacitance is $C_{eq} = C_{\text{ball}} + C = 4\pi\epsilon_0 r + C$. Since charge is conserved: $$ U_f = \frac{1}{2} \frac{Q^2}{C_{eq}} = \frac{Q^2}{2(4\pi\epsilon_0 r + C)} $$ Total Heat Loss ($H_{\text{total}}$): $$ H_{\text{total}} = U_i – U_f = \frac{Q^2}{2} \left[ \frac{1}{4\pi\epsilon_0 r} – \frac{1}{4\pi\epsilon_0 r + C} \right] $$ $$ H_{\text{total}} = \frac{Q^2}{2} \left[ \frac{4\pi\epsilon_0 r + C – 4\pi\epsilon_0 r}{4\pi\epsilon_0 r(4\pi\epsilon_0 r + C)} \right] = \frac{Q^2 C}{8\pi\epsilon_0 r(C + 4\pi\epsilon_0 r)} $$

3. Heat Dissipation in Resistors:
During the charging process, the current $I(t)$ flows from the ball, through $R_1$, to the capacitor, and an equal displacement current effectively causes current to flow from the ground through $R_2$. Thus, the resistors $R_1$ and $R_2$ are in series with respect to the current path.
Since the same current flows through both, the heat dissipated is proportional to their resistance values ($H \propto R$). $$ H_{R1} = \frac{R_1}{R_1 + R_2} H_{\text{total}} \quad \text{and} \quad H_{R2} = \frac{R_2}{R_1 + R_2} H_{\text{total}} $$

Substituting $H_{\text{total}}$: $$ H_{R1} = \frac{R_1 C Q^2}{8\pi\epsilon_0 r (R_1 + R_2)(C + 4\pi\epsilon_0 r)} $$ $$ H_{R2} = \frac{R_2 C Q^2}{8\pi\epsilon_0 r (R_1 + R_2)(C + 4\pi\epsilon_0 r)} $$