A parallel plate capacitor with a dielectric slab with dielectric constant `k= 3` filling the space between the plates is charged to potential `V` and isolated. Then the dielectric slab is drawn out and another dielectric slab of equal thickness but dielectric constant `k'= 2` is introduced between the plates. The ratio of the energy stored in the capacitor later to that initially is:

Options

A. `2:3`
B. `3:2`
C. `4:9`
D. `9:4`

Answer

To find the ratio of the energy stored in the capacitor after the dielectric slab is changed to that initially, we will follow these steps:

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- The initial dielectric constant \( k = 3 \).
- The capacitor is charged to a potential \( V \) and is isolated.

2. **Calculate Initial Capacitance**:
- The capacitance \( C_1 \) of the capacitor with the dielectric is given by:
\[ C_1 = \frac{k \cdot A \cdot \epsilon_0}{d} = \frac{3 \cdot A \cdot \epsilon_0}{d} \]

3. **Calculate Initial Charge**:
- The charge \( Q \) on the capacitor is:
\[ Q = C_1 \cdot V = \frac{3 \cdot A \cdot \epsilon_0}{d} \cdot V \]

4. **Calculate Initial Energy**:
- The energy \( E_1 \) stored in the capacitor is given by:
\[ E_1 = \frac{Q^2}{2C_1} = \frac{Q^2}{2 \cdot \frac{3 \cdot A \cdot \epsilon_0}{d}} = \frac{Q^2 \cdot d}{6 \cdot A \cdot \epsilon_0} \]

5. **Change Dielectric**:
- The dielectric slab is removed and replaced with a new slab of dielectric constant \( k' = 2 \).

6. **Calculate New Capacitance**:
- The new capacitance \( C_2 \) is given by:
\[ C_2 = \frac{k' \cdot A \cdot \epsilon_0}{d} = \frac{2 \cdot A \cdot \epsilon_0}{d} \]

7. **Charge Remains Constant**:
- Since the capacitor is isolated, the charge \( Q \) remains the same.

8. **Calculate New Energy**:
- The new energy \( E_2 \) stored in the capacitor is:
\[ E_2 = \frac{Q^2}{2C_2} = \frac{Q^2}{2 \cdot \frac{2 \cdot A \cdot \epsilon_0}{d}} = \frac{Q^2 \cdot d}{4 \cdot A \cdot \epsilon_0} \]

9. **Calculate the Ratio of Energies**:
- The ratio of the energy stored in the capacitor later to that initially is:
\[ \frac{E_2}{E_1} = \frac{\frac{Q^2 \cdot d}{4 \cdot A \cdot \epsilon_0}}{\frac{Q^2 \cdot d}{6 \cdot A \cdot \epsilon_0}} = \frac{6}{4} = \frac{3}{2} \]

### Final Answer:
The ratio of the energy stored in the capacitor later to that initially is \( \frac{3}{2} \).