A parallel plate capacitor (without dielectric) is charged by a battery and kept connected to the battery. A dielectric salb of dielectric constant `'k'` is inserted between the plates fully occupying the space between the plates. The energy density of electric field between the plates will:

Options

A. increase `k^(2)` times
B. decrease `k^(2)` times
C. increase k times
D. decrease k times

Answer

To solve the problem, we need to analyze the effect of inserting a dielectric slab into a parallel plate capacitor that is already connected to a battery.

### Step-by-Step Solution:

1. **Understanding the Initial Setup**:
- A parallel plate capacitor is charged by a battery, which means it has an initial electric field \( E \) and energy density \( u \) without the dielectric.
- The energy density \( u \) of the electric field in a capacitor is given by the formula:
\[ u = \frac{1}{2} \epsilon E^2 \]
where \( \epsilon \) is the permittivity of the medium between the plates.

2. **Inserting the Dielectric**:
- When a dielectric slab with dielectric constant \( k \) is inserted between the plates, the electric field \( E' \) in the capacitor changes.
- The new electric field \( E' \) is given by:
\[ E' = \frac{E}{k} \]
This is because the dielectric reduces the electric field by a factor of \( k \).

3. **Calculating the New Energy Density**:
- The new energy density \( u' \) after inserting the dielectric can be calculated using the new electric field \( E' \):
\[ u' = \frac{1}{2} \epsilon E'^2 \]
- Substituting \( E' \) into the equation gives:
\[ u' = \frac{1}{2} \epsilon \left(\frac{E}{k}\right)^2 = \frac{1}{2} \epsilon \frac{E^2}{k^2} \]

4. **Relating New Energy Density to Original**:
- We can express \( u' \) in terms of the original energy density \( u \):
\[ u' = \frac{1}{k^2} \cdot \frac{1}{2} \epsilon E^2 = \frac{u}{k^2} \]
- This shows that the new energy density \( u' \) is decreased by a factor of \( k^2 \).

5. **Conclusion**:
- Therefore, the energy density of the electric field between the plates decreases by \( k^2 \) times when the dielectric slab is inserted.

### Final Answer:
The energy density of the electric field between the plates will **decrease by \( k^2 \) times**.