1. Initial energy stored in capacitor $C$

A capacitor of capacitance $C$ is charged to voltage $V_0$.

So the initial charge on it is: $Q_0 = CV_0$

Initial energy stored is: $U_i = \frac{1}{2}CV_0^2$

2. After connecting to another uncharged capacitor $\frac{C}{2}$ in parallel

Since the battery is disconnected, total charge is conserved.

The two capacitors are connected in parallel, so final common voltage = $V_f$.

Total capacitance: $C_{\text{eq}} = C + \frac{C}{2} = \frac{3C}{2}$

Total charge remains: $Q_0 = CV_0$

Hence, $V_f = \frac{Q_0}{C_{\text{eq}}} = \frac{CV_0}{\frac{3C}{2}} = \frac{2V_0}{3}$

3. Final energy of the system

$U_f = \frac{1}{2}C_{\text{eq}}V_f^2$

Substitute values: $U_f = \frac{1}{2}\cdot \frac{3C}{2} \cdot \left(\frac{2V_0}{3}\right)^2$

$U_f = \frac{1}{2}\cdot \frac{3C}{2} \cdot \frac{4V_0^2}{9}$

$U_f = \frac{C V_0^2}{3}$

4. Energy loss

$\Delta U = U_i - U_f$

$\Delta U = \frac{1}{2}CV_0^2 - \frac{1}{3}CV_0^2$

$\Delta U = \left(\frac{3-2}{6}\right)CV_0^2$

$\Delta U = \frac{1}{6}CV_0^2$

5. Option check

• A: $\frac{1}{2}CV_0^2$ ❌
• B: $\frac{1}{4}CV_0^2$ ❌
• C: $\frac{1}{3}CV_0^2$ ❌
• D: $\frac{1}{6}CV_0^2$ ✅

Therefore, the correct answer is Option D.