Solution:

Let's analyze the problem step by step and derive why the potential difference is 10 V.

1. Capacitors in Parallel (Initial Condition)

When the 10 capacitors, each with capacitance $C$, are connected in parallel:

• The equivalent capacitance is: $$C_{\text{parallel}} = 10C$$
• The total charge stored when connected to a battery of potential $V$ is: $$Q = C_{\text{parallel}} \cdot V = (10C) \cdot V = 10CV$$

2. Capacitors in Series (Reconfigured System)

After disconnecting the battery, the capacitors are joined in series:

• The equivalent capacitance for 10 capacitors in series is: $$C_{\text{series}} = \frac{C}{10}$$
• The charge $Q$ stored on the series combination remains the same (as charge is conserved): $$Q_{\text{series}} = Q = 10CV$$

3. Potential Across the Series Combination

The potential difference across a capacitor or combination of capacitors is related to the charge $Q$ and capacitance $C$:

$$V_{\text{series}} = \frac{Q}{C_{\text{series}}}$$

Substitute $Q = 10CV$ and $C_{\text{series}} = \frac{C}{10}$:

$$V_{\text{series}} = \frac{10CV}{\frac{C}{10}} = V$$

Thus, the potential across the series combination is $V = 10V$,.

Final Answer:

The potential of the series combination is:

$$\boxed{10V}$$