A capacitor C is fully charged with voltage $$V_0$$. After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance $$\frac{C}{2}$$. The energy loss in the process after the charge is distributed between the two capacitors is:

We know that the electrostatic energy stored in a capacitor having capacitance $$C$$ and potential difference $$V$$ is given by the well-known formula

$$U=\frac{1}{2}CV^{2}\,.$$

Initially only one capacitor of capacitance $$C$$ is present and it is fully charged to a voltage $$V_0$$. Applying the above formula, its initial energy is

$$U_{\text{initial}}=\frac{1}{2}C{V_0}^{2}.$$

The corresponding initial charge on this capacitor is obtained from $$Q=CV$$. Hence,

$$Q_{\text{initial}}=C\,V_0.$$

Now the battery is removed and this charged capacitor is connected in parallel with another capacitor whose capacitance is $$\dfrac{C}{2}$$ and which is completely uncharged. Because the circuit is isolated, the total charge remains conserved while the two capacitors share the charge until they reach a common final voltage, say $$V_f$$.

In the parallel combination, the equivalent capacitance becomes

$$C_{\text{eq}}=C+\frac{C}{2}=\frac{3C}{2}.$$

Using charge conservation we write

$$Q_{\text{initial}}=Q_{\text{final}}.$$

But $$Q_{\text{final}}=C_{\text{eq}}\,V_f,$$ so

$$C\,V_0=\frac{3C}{2}\,V_f.$$

Solving this for the final voltage gives

$$V_f=\frac{2}{3}\,V_0.$$

Next we compute the total energy stored after equilibrium is reached. Using the same energy formula for the equivalent capacitance at the final common voltage, we have

$$U_{\text{final}}=\frac{1}{2}\,C_{\text{eq}}\,V_f^{2}=\frac{1}{2}\left(\frac{3C}{2}\right)\left(\frac{2}{3}V_0\right)^{2}.$$

Simplifying step by step:

$$U_{\text{final}}=\frac{1}{2}\cdot\frac{3C}{2}\cdot\frac{4}{9}\,V_0^{2} =\frac{3C}{4}\cdot\frac{4}{9}\,V_0^{2} =\frac{3\times4}{4\times9}\,C\,V_0^{2} =\frac{1}{3}C\,V_0^{2}.$$

The energy lost in the process is the difference between the initial and final energies:

$$\Delta U=U_{\text{initial}}-U_{\text{final}} =\frac{1}{2}C\,V_0^{2}-\frac{1}{3}C\,V_0^{2} =\left(\frac{3}{6}-\frac{2}{6}\right)C\,V_0^{2} =\frac{1}{6}C\,V_0^{2}.$$

Hence, the correct answer is Option D.