A capacitor of capacitance $$C$$ and potential $$V$$ has energy $$E$$. It is connected to another capacitor of capacitance $$2C$$ and potential $$2V$$. Then the loss of energy is $$\frac{x}{3}E$$, where $$x$$ is ______.

The energy stored in a capacitor is given by: $$E = \frac{1}{2}CV^2$$.

Initially, for the first capacitor, $$E_1 = \frac{1}{2}CV^2 = E$$ (given), so $$CV^2 = 2E$$. For the second capacitor, $$E_2 = \frac{1}{2}(2C)(2V)^2 = \frac{1}{2} \times 2C \times 4V^2 = 4CV^2 = 8E$$. Hence the total initial energy is $$E_i = E + 8E = 9E$$.

After connecting the capacitors, charge redistributes while the total charge remains conserved: $$Q_{total} = CV + 2C(2V) = CV + 4CV = 5CV$$. The total capacitance becomes $$C_{total} = C + 2C = 3C$$, so the common potential is $$V_f = \frac{Q_{total}}{C_{total}} = \frac{5CV}{3C} = \frac{5V}{3}$$.

The final energy stored is $$E_f = \frac{1}{2}(3C)\left(\frac{5V}{3}\right)^2 = \frac{1}{2} \times 3C \times \frac{25V^2}{9} = \frac{25CV^2}{6} = \frac{25 \times 2E}{6} = \frac{25E}{3}$$.

The energy loss is given by $$\Delta E = E_i - E_f = 9E - \frac{25E}{3} = \frac{27E - 25E}{3} = \frac{2E}{3} = \frac{x}{3}E$$, which yields $$x = 2$$.