Case 1: 10 identical capacitors, each of capacitance \(C_1\), connected in series
For capacitors connected in series,
\[ \frac{1}{C_{\text{eq1}}} = \frac{1}{C_1} + \frac{1}{C_1} + \cdots \text{ (10 terms)} = \frac{10}{C_1} \]
Hence,
\[ C_{\text{eq1}}=\frac{C_1}{10} \]
The combination is connected across a potential difference of \(4\,\text{V}\).
Energy stored in a capacitor is:
\[ U=\frac{1}{2}CV^2 \]
Therefore,
\[ U_1 = \frac{1}{2}\left(\frac{C_1}{10}\right)(4)^2 \]
\[ U_1 = \frac{1}{2}\times\frac{C_1}{10}\times16 \]
\[ U_1 = \frac{16C_1}{20} = \frac{4C_1}{5} \]
\[ U_1=0.8C_1 \]
Case 2: 8 identical capacitors, each of capacitance \(C_2\), connected in parallel
For capacitors connected in parallel,
\[ C_{\text{eq2}} = C_2+C_2+\cdots \text{ (8 terms)} = 8C_2 \]
If this combination is connected across a potential difference \(V\), then the energy stored is:
\[ U_2 = \frac{1}{2}C_{\text{eq2}}V^2 \]
\[ U_2 = \frac{1}{2}(8C_2)V^2 \]
\[ U_2 = 4C_2V^2 \]
According to the question, both combinations store the same energy.
So,
\[ U_1=U_2 \]
\[ \frac{4C_1}{5}=4C_2V^2 \]
Dividing both sides by \(4\),
\[ \frac{C_1}{5}=C_2V^2 \]
Therefore,
\[ C_2=\frac{C_1}{5V^2} \]
If \(V=1\),
\[ C_2=\frac{C_1}{5} \]
Final Answer:
\[ \boxed{\frac{C_1}{5}} \]