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Step-by-Step Solution
Step 1: Determine the initial charge on the first capacitor
The first capacitor has a capacitance of $10\,\mu\text{F}$ and is charged to a potential difference of $50\,\text{V}$. The charge $Q_{\text{initial}}$ stored on this capacitor is given by:
$Q_{\text{initial}} = C \times V = 10\,\mu\text{F} \times 50\,\text{V} = 500\,\mu\text{C}.$
Step 2: Find the final charge on the first capacitor after connecting in parallel
When the two capacitors are connected in parallel and allowed to share charge, the final potential difference across both becomes $20\,\text{V}$. For the first capacitor (of $10\,\mu\text{F}$) at $20\,\text{V}$, the final charge $Q_{\text{final, first}}$ is:
$Q_{\text{final, first}} = 10\,\mu\text{F} \times 20\,\text{V} = 200\,\mu\text{C}.$
Step 3: Apply charge conservation
Before connection, the total charge was $500\,\mu\text{C}$ on the first capacitor. After sharing, the first capacitor has $200\,\mu\text{C}$. Hence, the remaining charge $Q_{\text{unknown}}$ must reside on the second (unknown) capacitor:
$Q_{\text{unknown}} = Q_{\text{initial}} - Q_{\text{final, first}} = 500\,\mu\text{C} - 200\,\mu\text{C} = 300\,\mu\text{C}.$
Step 4: Calculate the unknown capacitor’s value
Since both capacitors share the same final potential of $20\,\text{V}$, the capacitance of the second capacitor, $C_{\text{unknown}}$, is found using:
$C_{\text{unknown}} = \frac{Q_{\text{unknown}}}{V} = \frac{300\,\mu\text{C}}{20\,\text{V}} = 15\,\mu\text{F}.$
Final Answer
The capacitance of the second capacitor is $15\,\mu\text{F}$.
