A parallel plate capacitor is filled with a dielectric of dielectric constant (relative permittivity) 5 between its plates and is charged to acquire an energy E. Then it is isolated (disconnected from the battery) and the dielectric is replaced by another dielectric of dielectric constant (relative permittivity) 2. The new energy stored in the capacitor

Options

A. A) `E`
B. B) `2.5 E`
C. C) `5E `
D. D) `10E`

Answer

To solve the problem, we need to analyze the situation of a parallel plate capacitor filled with different dielectrics and how the energy stored in the capacitor changes when the dielectric is replaced.

### Step-by-Step Solution:

1. **Understand the Initial Conditions:**
- Let \( C_0 \) be the capacitance of the capacitor when it is filled with air (no dielectric).
- When a dielectric with dielectric constant \( k_1 = 5 \) is inserted, the new capacitance \( C_1 \) becomes:
\[ C_1 = k_1 C_0 = 5 C_0 \]

2. **Calculate the Initial Energy Stored:**
- The energy \( E \) stored in the capacitor when filled with the dielectric of constant \( k_1 \) is given by:
\[ E = \frac{Q^2}{2C_1} \]
- Here, \( Q \) is the charge on the capacitor, which remains constant since the capacitor is isolated.

3. **Relate Energy to Capacitance:**
- The energy can also be expressed in terms of the initial capacitance \( C_0 \):
\[ E = \frac{Q^2}{2(5C_0)} = \frac{Q^2}{10C_0} \]

4. **Introduce the New Dielectric:**
- Now, the dielectric is replaced with another dielectric of constant \( k_2 = 2 \). The new capacitance \( C_2 \) becomes:
\[ C_2 = k_2 C_0 = 2 C_0 \]

5. **Calculate the New Energy Stored:**
- The new energy \( E' \) stored in the capacitor with the new dielectric is given by:
\[ E' = \frac{Q^2}{2C_2} = \frac{Q^2}{2(2C_0)} = \frac{Q^2}{4C_0} \]

6. **Relate the New Energy to the Initial Energy:**
- Now we can relate \( E' \) to \( E \):
\[ E' = \frac{Q^2}{4C_0} = \frac{Q^2}{10C_0} \cdot \frac{10}{4} = E \cdot \frac{10}{4} = 2.5E \]

7. **Final Result:**
- Therefore, the new energy stored in the capacitor after replacing the dielectric is:
\[ E' = 2.5E \]

### Conclusion:
The new energy stored in the capacitor is \( 2.5E \), which corresponds to option B.