When a dielectric slab is inserted between the plates of a capacitor which is initially charged and disconnected from the source then what can be concluded on the amount of heat dissipated during insertion ? What would be the answer if source remain connected to the capacitor while insertion of the dielectric slab?

Answer

To solve the problem, we will analyze two scenarios: one where the capacitor is disconnected from the source before the dielectric slab is inserted, and another where the capacitor remains connected to the source during the insertion of the dielectric slab.

### Step 1: Case 1 - Capacitor Disconnected from Source

1. **Initial Condition**: The capacitor is charged with charge \( Q_0 \) and disconnected from the battery. The initial energy stored in the capacitor can be expressed as:
\[ U_i = \frac{Q_0^2}{2C} \]
where \( C \) is the capacitance of the capacitor.

2. **Inserting Dielectric**: When a dielectric slab with dielectric constant \( K \) is inserted, the capacitance changes to:
\[ C' = K \cdot C \]

3. **Final Energy**: The final energy stored in the capacitor after inserting the dielectric slab is:
\[ U_f = \frac{Q_0^2}{2C'} \]
Substituting \( C' \):
\[ U_f = \frac{Q_0^2}{2K \cdot C} \]

4. **Heat Dissipated**: The heat dissipated during the insertion of the dielectric slab is equal to the change in stored energy:
\[ H = U_i - U_f \]
Substituting the expressions for \( U_i \) and \( U_f \):
\[ H = \frac{Q_0^2}{2C} - \frac{Q_0^2}{2K \cdot C} \]
Factoring out \( \frac{Q_0^2}{2C} \):
\[ H = \frac{Q_0^2}{2C} \left(1 - \frac{1}{K}\right) \]

### Step 2: Case 2 - Capacitor Connected to Source

1. **Initial Condition**: The capacitor is connected to the battery, maintaining a constant voltage \( V \). The initial energy stored in the capacitor is:
\[ U_i = \frac{1}{2} C V^2 \]

2. **Inserting Dielectric**: After inserting the dielectric slab, the capacitance becomes \( C' = K \cdot C \).

3. **Final Energy**: The final energy stored in the capacitor is:
\[ U_f = \frac{1}{2} C' V^2 = \frac{1}{2} (K \cdot C) V^2 \]

4. **Heat Dissipated**: The heat dissipated during the insertion of the dielectric slab is given by:
\[ H = U_f - U_i \]
Substituting the expressions for \( U_f \) and \( U_i \):
\[ H = \frac{1}{2} (K \cdot C) V^2 - \frac{1}{2} C V^2 \]
Factoring out \( \frac{1}{2} C V^2 \):
\[ H = \frac{1}{2} C V^2 (K - 1) \]

### Summary of Results

- **Heat Dissipated when Disconnected**:
\[ H = \frac{Q_0^2}{2C} \left(1 - \frac{1}{K}\right) \]

- **Heat Dissipated when Connected**:
\[ H = \frac{1}{2} C V^2 (K - 1) \]