ong>series:

$ \frac{1}{C_{\text{eq1}}} = \frac{1}{C_1} + \frac{1}{C_1} + \dots \,(10 \text{ terms}) = \frac{10}{C_1} $

$ C_{\text{eq1}} = \frac{C_1}{10}. $

Energy stored when connected across $4 \, \text{V}$:

$ U_1 = \tfrac{1}{2} C_{\text{eq1}} V^2 = \tfrac{1}{2} \cdot \frac{C_1}{10} \cdot (4)^2 = \tfrac{1}{2} \cdot \frac{C_1}{10} \cdot 16 = \frac{8}{10} C_1 = 0.8\, C_1. $

Case 2: Parallel combination of $8$ capacitors, each of capacitance $C_2$.

Equivalent capacitance of capacitors in parallel:

$ C_{\text{eq2}} = 8 C_2. $

Energy stored when connected across $V$:

$ U_2 = \tfrac{1}{2}\,C_{\text{eq2}} V^2 = \tfrac{1}{2}\,(8C_2)\, V^2 = 4 C_2 V^2. $

Condition: Energy equal

$ U_1 = U_2 $

$ 0.8 C_1 = 4 C_2 V^2 $

$ C_2 = \frac{0.8}{4 V^2} C_1 = \frac{0.2}{V^2}\, C_1 = \frac{C_1}{5 V^2}. $

Check the problem statement

The second system ("parallel combination… charged by a source of potential difference $V$") uses the same total energy as the first.

So the relation indeed gives:

$ C_2 = \frac{C_1}{5 V^2}. $

Now, look at the given options:

A. $\frac{C_1}{5}$
B. $\frac{8}{5} C_1$
C. $\frac{64}{5} C_1$
D. $\frac{C_1}{40}$

This suggests in the problem, $V$ may actually equal 1 V (i.e., the wording "charged by source potential difference 'V'" intends $V = 1$ volt).

If $V = 1$:

$ C_2 = \frac{C_1}{5}. $

✅ Final Answer:

$ \boxed{\tfrac{C_1}{5}} $

So the correct option is A.

Capacitor

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