Problem Analysis

We have a system of two concentric conducting spherical shells with radii $r$ (inner) and $R$ (outer). The outer shell is given a charge $Q$. The inner shell is connected via a switch S.

• Position 1: Inner shell connected to Earth (Potential = 0).
• Position 2: Inner shell connected to the positive terminal of a battery with voltage $V$ (Potential = $V$). The negative terminal is grounded.

We need to find the heat released after the switch is thrown from position 1 to position 2.

Key Principle: The heat released ($H$) in an electrical process is the difference between the work done by the battery ($W_{batt}$) and the change in the stored electrostatic potential energy ($\Delta U$) of the system. $$ H = W_{batt} – \Delta U $$ where $\Delta U = U_{final} – U_{initial}$.

Step-by-Step Calculation

Step 1: Analyze Initial State (Switch at 1)

The inner shell (radius $r$) is grounded ($V_{inner} = 0$). The outer shell (radius $R$) has charge $Q$. Let charge on inner shell be $q_1$. The potential at the surface of the inner shell is the sum of potentials due to $q_1$ and $Q$: $$ V_{inner} = \frac{1}{4\pi\epsilon_0} \left( \frac{q_1}{r} + \frac{Q}{R} \right) = 0 $$ $$ \frac{q_1}{r} = -\frac{Q}{R} \implies q_1 = -\frac{r}{R}Q $$ Let’s calculate the initial energy $U_i$. The energy of a system of charged shells can be calculated using $U = \frac{1}{2} \sum q V$. Here, for the inner shell $V=0$. For the outer shell, the potential is: $$ V_{outer, i} = \frac{1}{4\pi\epsilon_0} \left( \frac{q_1}{R} + \frac{Q}{R} \right) = \frac{1}{4\pi\epsilon_0 R} (q_1 + Q) $$ $$ U_i = \frac{1}{2} q_1 (0) + \frac{1}{2} Q V_{outer, i} = \frac{1}{2} Q \frac{1}{4\pi\epsilon_0 R} \left( -\frac{r}{R}Q + Q \right) $$ $$ U_i = \frac{Q^2}{8\pi\epsilon_0 R} \left( 1 – \frac{r}{R} \right) $$

Step 2: Analyze Final State (Switch at 2)

The inner shell is now at potential $V$. The outer shell still has charge $Q$. Let the new charge on the inner shell be $q_2$. The potential condition for the inner shell is: $$ V_{inner} = \frac{1}{4\pi\epsilon_0} \left( \frac{q_2}{r} + \frac{Q}{R} \right) = V $$ Solving for $q_2$: $$ \frac{q_2}{r} = 4\pi\epsilon_0 V – \frac{Q}{R} $$ $$ q_2 = 4\pi\epsilon_0 r V – \frac{r}{R}Q $$ Notice that $q_2 = q_1 + 4\pi\epsilon_0 r V$. Let’s calculate the final energy $U_f$. $$ U_f = \frac{1}{2} q_2 V + \frac{1}{2} Q V_{outer, f} $$ Potential of outer shell: $$ V_{outer, f} = \frac{1}{4\pi\epsilon_0 R} (q_2 + Q) $$ $$ U_f = \frac{1}{2} q_2 V + \frac{Q}{8\pi\epsilon_0 R} (q_2 + Q) $$

Step 3: Calculate Change in Energy and Work Done

Change in Energy ($\Delta U$): $$ \Delta U = U_f – U_i $$ Instead of expanding everything, let’s notice that the potential of the outer shell due to $Q$ itself ($\frac{Q}{4\pi\epsilon_0 R}$) is constant. The cross terms matter. Alternatively, use the energy density or self/interaction energy formulation. $U = \frac{q_{inner}^2}{8\pi\epsilon_0 r} + \frac{Q^2}{8\pi\epsilon_0 R} + \frac{q_{inner} Q}{4\pi\epsilon_0 R}$. $U_i = \frac{q_1^2}{8\pi\epsilon_0 r} + \frac{Q^2}{8\pi\epsilon_0 R} + \frac{q_1 Q}{4\pi\epsilon_0 R}$ $U_f = \frac{q_2^2}{8\pi\epsilon_0 r} + \frac{Q^2}{8\pi\epsilon_0 R} + \frac{q_2 Q}{4\pi\epsilon_0 R}$ $\Delta U = \frac{1}{8\pi\epsilon_0 r} (q_2^2 – q_1^2) + \frac{Q}{4\pi\epsilon_0 R} (q_2 – q_1)$ Let $\Delta q = q_2 – q_1$. We found earlier $\Delta q = 4\pi\epsilon_0 r V$. Also $q_2 + q_1 = 4\pi\epsilon_0 r V – \frac{2r}{R}Q$. $\Delta U = \frac{1}{8\pi\epsilon_0 r} (q_2 – q_1)(q_2 + q_1) + \frac{Q}{4\pi\epsilon_0 R} \Delta q$ $\Delta U = \frac{\Delta q}{8\pi\epsilon_0 r} (q_2 + q_1) + \frac{Q \Delta q}{4\pi\epsilon_0 R}$ Substitute $\Delta q = 4\pi\epsilon_0 r V$: $\Delta U = \frac{V}{2} (q_2 + q_1) + \frac{r V Q}{R}$ Substitute $q_2 + q_1 = 4\pi\epsilon_0 r V – \frac{2r}{R}Q$: $\Delta U = \frac{V}{2} \left( 4\pi\epsilon_0 r V – \frac{2r}{R}Q \right) + \frac{r V Q}{R}$ $\Delta U = 2\pi\epsilon_0 r V^2 – \frac{r V Q}{R} + \frac{r V Q}{R} = 2\pi\epsilon_0 r V^2$. Work Done by Battery ($W_{batt}$): Charge $\Delta q$ flows from the battery to the inner shell at potential $V$. $$ W_{batt} = (\Delta q) V = (4\pi\epsilon_0 r V) V = 4\pi\epsilon_0 r V^2 $$

Step 4: Calculate Heat Released

$$ H = W_{batt} – \Delta U $$ $$ H = 4\pi\epsilon_0 r V^2 – 2\pi\epsilon_0 r V^2 $$ $$ H = 2\pi\epsilon_0 r V^2 $$