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Step-by-Step Solution
Step 1: Identify the Initial Energy Stored in the Charged Capacitor
Initially, a capacitor of capacitance $C$ is fully charged to voltage $V_0$. The energy $U_i$ stored in this capacitor is given by:
$U_i = \tfrac{1}{2}\,C\,V_0^2.$
Step 2: Understand the Final Configuration
After disconnecting the voltage source, the charged capacitor is connected in parallel with an uncharged capacitor of capacitance $C/2$. Because they are connected in parallel, the total charge $Q = C\,V_0$ initially present on the first capacitor will now be shared between the two capacitors.
Step 3: Determine the Final Voltage Across Both Capacitors
When two capacitors are connected in parallel, they share the same final voltage $V_f$. By conservation of charge:
$Q_{\text{total}} = C\,V_0 = (C + \tfrac{C}{2})\,V_f = \tfrac{3C}{2}\,V_f.$
Hence,
$V_f = \dfrac{C\,V_0}{\tfrac{3C}{2}} = \dfrac{2}{3}\,V_0.$
Step 4: Calculate the Final Energy
The combined capacitance of the two capacitors in parallel is $C + \tfrac{C}{2} = \tfrac{3C}{2}$. The final energy $U_f$ stored in the two capacitors together is:
$U_f = \tfrac{1}{2}\,\bigl(C + \tfrac{C}{2}\bigr)\,V_f^2
= \tfrac{1}{2} \times \tfrac{3C}{2} \times \Bigl(\dfrac{2}{3}V_0\Bigr)^2.$
Simplifying,
$\Bigl(\dfrac{2}{3}V_0\Bigr)^2 = \dfrac{4}{9}V_0^2,$
so
$U_f = \tfrac{1}{2} \times \tfrac{3C}{2} \times \dfrac{4}{9}\,V_0^2
= \tfrac{1}{2} \times \dfrac{12C}{18}\,V_0^2
= \tfrac{1}{2} \times \dfrac{2C}{3}\,V_0^2
= \dfrac{1}{3}\,C\,V_0^2.$
Step 5: Find the Energy Loss
The energy lost (often appearing as heat or radiation) is the difference between the initial energy and the final energy:
$\text{Energy Lost} = U_i - U_f
= \tfrac{1}{2}\,C\,V_0^2 - \dfrac{1}{3}\,C\,V_0^2
= \bigl(\tfrac{3}{6} - \tfrac{2}{6}\bigr)\,C\,V_0^2
= \tfrac{1}{6}\,C\,V_0^2.$
This matches the given correct answer: $ \tfrac{1}{6}\,C\,V_0^2.$
