`n` identical capacitors are connected in parallel to a potential difference `V`. These capacitors are then reconnected in series, their charges being left undisturbed. The potential difference obtained is

Options

A. zero
B. `(n-1)V`
C. `nV`
D. `n^2V`

Explanation

To solve the problem step by step, we will analyze the situation of `n` identical capacitors connected first in parallel and then in series.

### Step 1: Understanding Capacitors in Parallel
When `n` identical capacitors are connected in parallel to a potential difference `V`, each capacitor experiences the same voltage `V`. The charge `Q` on each capacitor can be expressed as:
\[ Q = C \cdot V \]
where \( C \) is the capacitance of each capacitor.

### Step 2: Total Charge in Parallel
Since there are `n` identical capacitors connected in parallel, the total charge \( Q_{total} \) stored in the system is the sum of the charges on each capacitor:
\[ Q_{total} = n \cdot Q = n \cdot (C \cdot V) = nCV \]

### Step 3: Reconnecting Capacitors in Series
Next, we reconnect these `n` capacitors in series. In a series connection, the charge \( Q \) on each capacitor remains the same, and the total potential difference \( V_{total} \) across the series combination is the sum of the potential differences across each capacitor.

### Step 4: Potential Difference Across Each Capacitor in Series
The potential difference across each capacitor when connected in series is given by:
\[ V_i = \frac{Q}{C} \]
Since the charge \( Q \) remains the same for each capacitor, we can substitute \( Q \) with \( nCV \) (from Step 2):
\[ V_i = \frac{nCV}{C} = nV \]

### Step 5: Total Potential Difference in Series
The total potential difference \( V_{total} \) across the series connection of `n` capacitors is:
\[ V_{total} = V_1 + V_2 + ... + V_n = nV \]

### Conclusion
Thus, when `n` identical capacitors are connected in parallel to a potential difference `V` and then reconnected in series, the total potential difference obtained is:
\[ \boxed{nV} \]

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