On a parallel plate capacitor following operations can be performed .
P - connect the capacitor to a battery of emf V
Q - disconnect the battery
R - reconnect the battery with polarity reversed
S - insert a dielectric slab in the capacitor

Options

A. In PQR (perform P, then Q, Then R) , the stored electric energy remains unchanged and no thermal energy is developed
B. The charge appearing on the capacitor is greater after the action PSQ then after the action PQS
C. The electric energy stored in the capacitor is greater after the action SPQ then after the action PQS
D. The electric field in the capacitor after the action PS is the same as that after SP

Answer

To solve the problem regarding the operations performed on a parallel plate capacitor, we will analyze each operation step by step and determine the effects on charge, energy, and electric field.

### Step-by-Step Solution:

1. **Operation P - Connect the capacitor to a battery of emf V:**
- When the capacitor is connected to a battery of emf V, it gets charged. The charge (Q) on the capacitor can be calculated using the formula:
\[ Q_P = C \cdot V \]
- The energy (U) stored in the capacitor at this stage is given by:
\[ U_P = \frac{1}{2} C V^2 \]

2. **Operation Q - Disconnect the battery:**
- After disconnecting the battery, the charge on the capacitor remains constant because there is no path for the charge to flow away. Therefore:
\[ Q_Q = Q_P = C \cdot V \]
- The energy stored in the capacitor remains the same as it was just before disconnecting:
\[ U_Q = U_P = \frac{1}{2} C V^2 \]

3. **Operation R - Reconnect the battery with polarity reversed:**
- When the battery is reconnected with reversed polarity, the capacitor will discharge and then recharge to the new voltage. The new charge on the capacitor will be:
\[ Q_R = C \cdot (-V) = -C \cdot V \]
- The energy stored in the capacitor after reconnecting to the battery is:
\[ U_R = \frac{1}{2} C V^2 \]
- Note that the energy is the same as before, but the charge is now negative due to the reversed polarity.

4. **Operation S - Insert a dielectric slab in the capacitor:**
- When a dielectric slab is inserted, the capacitance increases by a factor of K (dielectric constant):
\[ C' = K \cdot C \]
- If the battery is connected, the voltage remains constant, and the new charge will be:
\[ Q_S = C' \cdot V = K \cdot C \cdot V \]
- The energy stored in the capacitor with the dielectric is:
\[ U_S = \frac{1}{2} C' V^2 = \frac{1}{2} (K \cdot C) V^2 = K \cdot \frac{1}{2} C V^2 \]

### Summary of Results:
- Charge after each operation:
- After P: \( Q_P = C \cdot V \)
- After Q: \( Q_Q = C \cdot V \)
- After R: \( Q_R = -C \cdot V \)
- After S (if battery connected): \( Q_S = K \cdot C \cdot V \)

- Energy after each operation:
- After P: \( U_P = \frac{1}{2} C V^2 \)
- After Q: \( U_Q = \frac{1}{2} C V^2 \)
- After R: \( U_R = \frac{1}{2} C V^2 \)
- After S (if battery connected): \( U_S = K \cdot \frac{1}{2} C V^2 \)