A parallel plate capacitor of capacitance `200muF` is connected to a battery of 200 V. A dielectric slab of dielectric 2 is now inserted into the space between plates of capacitor while the battery remain connected .The change in the electrostatic energy in the capacitor will be __________J.

Answer

To solve the problem, we need to find the change in the electrostatic energy of a parallel plate capacitor when a dielectric slab is inserted while the battery remains connected.

### Step-by-Step Solution:

1. **Calculate the Initial Energy of the Capacitor:**
The formula for the energy stored in a capacitor is given by:
\[ U = \frac{1}{2} C V^2 \]
where:
- \( C = 200 \, \mu F = 200 \times 10^{-6} \, F \)
- \( V = 200 \, V \)

Substituting the values:
\[ U_i = \frac{1}{2} \times (200 \times 10^{-6}) \times (200)^2 \]
\[ U_i = \frac{1}{2} \times 200 \times 10^{-6} \times 40000 \]
\[ U_i = 100 \times 10^{-6} \times 40000 \]
\[ U_i = 4 \, J \]

2. **Calculate the New Capacitance with the Dielectric:**
When a dielectric material is inserted, the capacitance increases by a factor equal to the dielectric constant (\( k \)):
\[ C' = kC \]
Given \( k = 2 \):
\[ C' = 2 \times 200 \, \mu F = 400 \, \mu F = 400 \times 10^{-6} \, F \]

3. **Calculate the New Energy of the Capacitor:**
The voltage across the capacitor remains the same (200 V) since the battery is connected. Thus, the new energy \( U_f \) is:
\[ U_f = \frac{1}{2} C' V^2 \]
Substituting the new capacitance:
\[ U_f = \frac{1}{2} \times (400 \times 10^{-6}) \times (200)^2 \]
\[ U_f = \frac{1}{2} \times 400 \times 10^{-6} \times 40000 \]
\[ U_f = 200 \times 10^{-6} \times 40000 \]
\[ U_f = 8 \, J \]

4. **Calculate the Change in Energy:**
The change in energy \( \Delta U \) is given by:
\[ \Delta U = U_f - U_i \]
Substituting the values we calculated:
\[ \Delta U = 8 \, J - 4 \, J = 4 \, J \]

### Final Answer:
The change in the electrostatic energy in the capacitor is **4 J**.