1. Given data

• Plate width $= 4\text{ cm} = 0.04\text{ m}$
• Plate length $= 8\text{ cm} = 0.08\text{ m}$
• Plate separation $d = 4\text{ mm} = 0.004\text{ m}$
• Battery voltage $V = 20\text{ V}$
• Dielectric constant $K = 5$
• Dielectric slab dimensions:
  • length $= 1\text{ cm} = 0.01\text{ m}$
  • width $= 4\text{ cm} = 0.04\text{ m}$
  • thickness $= 4\text{ mm} = 0.004\text{ m}$

Since the slab thickness equals the plate separation, it completely fills the gap in the region where it is inserted.

2. Interpretation of geometry

The slab covers only part of the plate area, so the arrangement is equivalent to two capacitors in parallel:

• one part with dielectric,
• one part with air.

Total plate area: $A = (0.04)(0.08) = 0.0032\ \text{m}^2$

Area covered by dielectric slab: $A_1 = (0.04)(0.01) = 0.0004\ \text{m}^2$

Remaining air-filled area: $A_2 = 0.0032 - 0.0004 = 0.0028\ \text{m}^2$

3. Capacitances of the two parts

For the dielectric-filled part: $C_1 = \frac{K\epsilon_0 A_1}{d}$ $C_1 = \frac{5\epsilon_0(0.0004)}{0.004} = 5\epsilon_0(0.1) = 0.5\epsilon_0$

For the air-filled part: $C_2 = \frac{\epsilon_0 A_2}{d}$ $C_2 = \frac{\epsilon_0(0.0028)}{0.004} = 0.7\epsilon_0$

Hence total capacitance: $C = C_1 + C_2 = 0.5\epsilon_0 + 0.7\epsilon_0 = 1.2\epsilon_0$

4. Electrostatic energy

Since the capacitor remains connected to the battery, voltage remains constant.

Energy stored is: $U = \frac{1}{2}CV^2$

Substitute values: $U = \frac{1}{2}(1.2\epsilon_0)(20)^2$ $U = 0.6\epsilon_0 \cdot 400$ $U = 240\epsilon_0\ \text{J}$

5. Final answer

The electrostatic energy of the system is: $\boxed{240\epsilon_0\ \text{J}}$

So the required integer is: $\boxed{240}$

6. Comparison with stored answer

Stored correct answer: $240$

My derived answer: $240$

They agree.