After charging a capacitor the battery is removed. Now by placing a dielectric slab between the plates

Options

A. the potential difference between the plates and the energy stored will decreases but the charge on plates will remains same
B. the charge on the plates will decreases and the potential difference between the plates will increases
C. the potential difference between the plates will increases and energy stored will decreases but the charge on the plates will remains same
D. the potential difference, energy stored and the charge will remain unchanged

Explanation

To solve the problem step by step, let's analyze the situation of a capacitor after it has been charged and the battery has been removed, followed by the introduction of a dielectric slab between the plates.

### Step 1: Initial Setup
- Initially, we have a capacitor with capacitance \( C_0 \) connected to a battery providing a voltage \( V \).
- The charge \( Q \) on the capacitor can be expressed as:
\[ Q = C_0 V \]

### Step 2: Removing the Battery
- Once the battery is removed, the charge \( Q \) on the capacitor remains constant because there is no path for charge to flow away from the capacitor.
- Therefore, we have:
\[ Q = \text{constant} \]

### Step 3: Introducing the Dielectric
- When a dielectric slab with dielectric constant \( k \) is inserted between the plates of the capacitor, the capacitance increases to:
\[ C' = k C_0 \]

### Step 4: Calculating the New Voltage
- The relationship between charge, capacitance, and voltage is given by:
\[ V' = \frac{Q}{C'} \]
- Substituting for \( C' \):
\[ V' = \frac{Q}{k C_0} \]
- Since \( Q = C_0 V \), we can substitute this into the equation:
\[ V' = \frac{C_0 V}{k C_0} = \frac{V}{k} \]
- This shows that the new voltage \( V' \) is less than the original voltage \( V \), indicating that the potential difference decreases when the dielectric is introduced.

### Step 5: Calculating the Energy Stored
- The energy stored in a capacitor is given by:
\[ U = \frac{1}{2} C V^2 \]
- The new energy \( U' \) after inserting the dielectric becomes:
\[ U' = \frac{1}{2} C' (V')^2 = \frac{1}{2} (k C_0) \left(\frac{V}{k}\right)^2 \]
- Simplifying this:
\[ U' = \frac{1}{2} k C_0 \frac{V^2}{k^2} = \frac{1}{2} \frac{C_0 V^2}{k} \]
- This indicates that the energy stored in the capacitor decreases by a factor of \( k \).

### Conclusion
From the analysis:
- The charge \( Q \) remains constant.
- The potential difference \( V' \) decreases to \( \frac{V}{k} \).
- The energy stored \( U' \) decreases to \( \frac{U}{k} \).

Thus, the correct option is:
- The potential difference between the plates and the energy stored will decrease, but the charge will remain the same.