Given data

• First capacitor: $C_1 = 600\,\text{pF} = 600 \times 10^{-12}\,\text{F}$
• Initial voltage on it: $V = 200\,\text{V}$
• Second capacitor: $C_2 = 600\,\text{pF}$, initially uncharged

Initial energy stored in the first capacitor

The electrostatic energy stored in a capacitor is $U = \frac{1}{2}CV^2$ So initially, $U_i = \frac{1}{2}(600 \times 10^{-12})(200)^2$

Now, $(200)^2 = 40000$ Hence, $U_i = \frac{1}{2} \times 600 \times 10^{-12} \times 40000$ $U_i = 300 \times 10^{-12} \times 40000$ $U_i = 12 \times 10^{-6}\,\text{J} = 12\,\mu\text{J}$

After connecting to another identical uncharged capacitor

Since the second capacitor is identical and initially uncharged, charge redistributes equally.

Therefore, the final voltage becomes $V_f = \frac{V}{2} = 100\,\text{V}$

Total capacitance of the two capacitors in parallel: $C_{\text{eq}} = C_1 + C_2 = 1200\,\text{pF} = 1200 \times 10^{-12}\,\text{F}$

Final energy of the system

$U_f = \frac{1}{2}C_{\text{eq}}V_f^2$ $U_f = \frac{1}{2}(1200 \times 10^{-12})(100)^2$

Since $(100)^2 = 10000$, $U_f = \frac{1}{2} \times 1200 \times 10^{-12} \times 10000$ $U_f = 600 \times 10^{-12} \times 10000$ $U_f = 6 \times 10^{-6}\,\text{J} = 6\,\mu\text{J}$

Energy lost

$\Delta U = U_i - U_f$ $\Delta U = 12 - 6 = 6\,\mu\text{J}$

Final answer

Electrostatic energy lost is $\boxed{6\,\mu\text{J}}$