Step 1: Understanding the Concept:
When capacitors are connected in parallel to a voltage source \( V \), each capacitor charges to the same potential \( V \).
If they are disconnected, they hold their respective charges.
When these charged capacitors are then connected in series (typically with aiding polarities), their individual potential differences add up.
The total energy of the system is the sum of the energies stored in each individual capacitor, regardless of how they are connected as long as charge is conserved.
Step 2: Key Formula or Approach:
Energy stored in a single capacitor is \( U = \frac{1}{2}CV^2 \).
Total voltage for series connection of \( n \) capacitors is \( V_{total} = V_1 + V_2 + \dots + V_n \).
Step 3: Detailed Explanation:
Let the capacitance of each identical capacitor be \( C \).
Initially, the \( n \) capacitors are in parallel and charged to a potential \( V \).
The charge on each capacitor is \( q = CV \).
The energy stored in one capacitor is \( U_1 = \frac{1}{2}CV^2 \).
The total initial energy of the combination is the sum of the energies of all \( n \) capacitors:
\[ U_{initial} = n \times U_1 = n \left(\frac{1}{2}CV^2\right) \]
Now, the capacitors are separated. Each capacitor still retains its charge \( q = CV \) and potential difference \( V \).
When they are joined in series, the total potential difference across the combination is the sum of the individual potential differences:
\[ V_{final} = V + V + \dots \text{ (n times)} = nV \]
The equivalent capacitance of the series combination is \( C_{eq} = \frac{C}{n} \).
The total final energy of the series combination can be calculated using the equivalent capacitance and total voltage:
\[ U_{final} = \frac{1}{2} C_{eq} V_{final}^2 \]
\[ U_{final} = \frac{1}{2} \left(\frac{C}{n}\right) (nV)^2 \]
\[ U_{final} = \frac{1}{2} \left(\frac{C}{n}\right) (n^2 V^2) \]
\[ U_{final} = \frac{1}{2} n C V^2 \]
Comparing the initial and final states, we observe that the total energy remains the same (\( U_{final} = U_{initial} \)).
The potential difference across the combination has become \( nV \).
Step 4: Final Answer:
The potential difference becomes nV and energy remains the same.