1. Initial data

• Initial capacitance: $C_1 = 200\,\mu F = 200 \times 10^{-6} F = 2 \times 10^{-4} F$
• Battery voltage remains connected: $V = 200\,V$
• Dielectric constant: $K = 2$

2. Initial electrostatic energy

For a capacitor connected to a battery, energy stored is $U = \frac{1}{2}CV^2$

So initially, $U_1 = \frac{1}{2}(2 \times 10^{-4})(200)^2$

Now, $(200)^2 = 40000$

Thus, $U_1 = \frac{1}{2}(2 \times 10^{-4})(40000)$ $U_1 = \frac{1}{2}(8) = 4\,J$

3. Capacitance after inserting dielectric

Since the dielectric completely fills the space, new capacitance becomes $C_2 = KC_1 = 2 \times 200\,\mu F = 400\,\mu F$

So, $C_2 = 4 \times 10^{-4} F$

4. Final electrostatic energy

Battery remains connected, so voltage stays constant at $200\,V$.

Hence, $U_2 = \frac{1}{2}C_2V^2 = \frac{1}{2}(4 \times 10^{-4})(200)^2$ $U_2 = \frac{1}{2}(4 \times 10^{-4})(40000)$ $U_2 = \frac{1}{2}(16) = 8\,J$

5. Change in electrostatic energy

$\Delta U = U_2 - U_1 = 8 - 4 = 4\,J$

Therefore, the change in electrostatic energy is $\boxed{4}$