In the circuit shown, the capacitance $C_{0}=10 \; \mu \text{F}$ and inductance $\text{L}_{0} = 1 \; \text{mH}$ and the diode is ideal. The capacitor is initially charged to $10 \; \text{V}$ and the current in the inductor is initially zero. If the switch is closed at $t=0 \; \text{s},$ the voltage $V_{c}(t)$ (in volts) across the capacitor at $t=0.5 \; \text{s}$ is (round off to one decimal place)
