A parallel plate capacitor is connected across a battery. Now, keeping the battery connected, a dielectric slab is inserted between the plates. In the process,

Options

A. no work done
B. work is done by the battery and the stored energy increases.
C. work is done by the external agent, and the stored energy decteases. lt
D. work is done by the battery as well as the external agent, but the stored energy done not charge.

Explanation

To solve the problem step by step, we will analyze the situation of a parallel plate capacitor connected to a battery, with a dielectric slab being inserted while the battery remains connected.

### Step 1: Understand the Initial Conditions
- A parallel plate capacitor with capacitance \( C \) is connected to a battery providing a voltage \( V \).
- The initial charge \( Q \) on the capacitor can be expressed as:
\[ Q = C \cdot V \]
- The initial energy \( U_{\text{initial}} \) stored in the capacitor is given by:
\[ U_{\text{initial}} = \frac{1}{2} C V^2 \]

**Hint:** Remember that the charge and energy of a capacitor are directly related to its capacitance and the voltage across it.

### Step 2: Insert the Dielectric Slab
- When a dielectric slab with dielectric constant \( K \) is inserted between the plates of the capacitor, the capacitance increases to:
\[ C' = K \cdot C \]
- Since the battery remains connected, the voltage \( V \) across the capacitor remains constant.

**Hint:** The dielectric constant \( K \) increases the capacitance of the capacitor, which is crucial for the next calculations.

### Step 3: Calculate the New Charge
- The new charge \( Q' \) on the capacitor after inserting the dielectric is:
\[ Q' = C' \cdot V = K \cdot C \cdot V \]

**Hint:** The new charge is directly proportional to the new capacitance and the constant voltage.

### Step 4: Determine the Change in Charge
- The change in charge \( \Delta Q \) that flows from the battery can be calculated as:
\[ \Delta Q = Q' - Q = (K \cdot C \cdot V) - (C \cdot V) = C \cdot V (K - 1) \]

**Hint:** The difference in charge indicates how much additional charge is supplied by the battery due to the insertion of the dielectric.

### Step 5: Work Done by the Battery
- The work done by the battery during this process can be expressed as:
\[ W = \Delta Q \cdot V = (C \cdot V (K - 1)) \cdot V = C V^2 (K - 1) \]

**Hint:** The work done by the battery is related to the additional charge supplied and the voltage across the capacitor.

### Step 6: Calculate the Final Energy Stored
- The final energy \( U_{\text{final}} \) stored in the capacitor after the dielectric is inserted is:
\[ U_{\text{final}} = \frac{1}{2} C' V^2 = \frac{1}{2} (K \cdot C) V^2 = \frac{1}{2} K C V^2 \]

**Hint:** The final energy is calculated using the new capacitance and the constant voltage.

### Step 7: Compare Initial and Final Energy
- To determine the change in energy:
\[ U_{\text{final}} - U_{\text{initial}} = \frac{1}{2} K C V^2 - \frac{1}{2} C V^2 = \frac{1}{2} C V^2 (K - 1) \]
- Since \( K > 1 \), it follows that \( U_{\text{final}} > U_{\text{initial}} \).

**Hint:** The increase in stored energy indicates that energy is being supplied to the capacitor by the battery.

### Conclusion
- The insertion of the dielectric slab while keeping the battery connected results in an increase in charge on the capacitor and an increase in the energy stored in the capacitor. The work done by the battery is also positive, indicating that energy is being transferred from the battery to the capacitor.

**Final Answer:** The work done by the battery increases the energy stored in the capacitor.