A 600 pF capacitor is charged by 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. Electrostatic energy lost in the process is _______ $$\mu$$J.

We have two identical capacitors $$C_1 = C_2 = 600$$ pF, with $$C_1$$ initially charged to $$V = 200$$ V.

The initial energy stored in $$C_1$$ is:

$$U_i = \frac{1}{2}C_1V^2 = \frac{1}{2} \times 600 \times 10^{-12} \times (200)^2 = 12 \times 10^{-6} \text{ J} = 12 \text{ } \mu\text{J}$$

When $$C_1$$ is connected to the uncharged $$C_2$$, charge is conserved. The common voltage becomes:

$$V' = \frac{C_1 V}{C_1 + C_2} = \frac{600 \times 200}{600 + 600} = \frac{120000}{1200} = 100 \text{ V}$$

Now, the final energy of the combination is:

$$U_f = \frac{1}{2}(C_1 + C_2)V'^2 = \frac{1}{2} \times 1200 \times 10^{-12} \times (100)^2 = 6 \times 10^{-6} \text{ J} = 6 \text{ } \mu\text{J}$$

So the energy lost is:

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

Hence, the answer is $$6$$ $$\mu$$J.