Suppose n identical capacitors are joined in parallel and charged to potential V. Now, they are separated and joined in series. If the energy possessed by each capacitor is U, then on joining them in series, the energy and potential difference for the combination are

Options

A. nU, V
B. U, nV
C. nU, nV
D. Less than nU, nV

Answer

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

### Step 1: Understanding Capacitors in Parallel
When \( n \) identical capacitors, each with capacitance \( C \), are connected in parallel, the total capacitance \( C_{\text{eq}} \) is given by:
\[ C_{\text{eq}} = nC \]
The potential difference across each capacitor in parallel is the same, which is \( V \).

### Step 2: Calculating Energy in Parallel Configuration
The energy \( U \) stored in each capacitor when charged to potential \( V \) is given by the formula:
\[ U = \frac{1}{2} C V^2 \]
For \( n \) capacitors, the total energy \( U_{\text{total}} \) when they are in parallel is:
\[ U_{\text{total}} = nU = n \left( \frac{1}{2} C V^2 \right) = \frac{n}{2} C V^2 \]

### Step 3: Transitioning to Series Configuration
Now, when these \( n \) capacitors are disconnected and connected in series, the charge on each capacitor remains the same, but the potential across each capacitor adds up.

### Step 4: Calculating Equivalent Capacitance in Series
For capacitors in series, the equivalent capacitance \( C_{\text{eq, series}} \) is given by:
\[ \frac{1}{C_{\text{eq, series}}} = \frac{1}{C} + \frac{1}{C} + \ldots + \frac{1}{C} = \frac{n}{C} \]
Thus,
\[ C_{\text{eq, series}} = \frac{C}{n} \]

### Step 5: Total Potential in Series Configuration
The total potential difference \( V_{\text{total}} \) across the series combination is:
\[ V_{\text{total}} = V + V + \ldots + V = nV \]

### Step 6: Calculating Energy in Series Configuration
The energy stored in the series combination can be calculated using the equivalent capacitance and the total potential:
\[ U_{\text{series}} = \frac{1}{2} C_{\text{eq, series}} V_{\text{total}}^2 = \frac{1}{2} \left( \frac{C}{n} \right) (nV)^2 \]
Simplifying this gives:
\[ U_{\text{series}} = \frac{1}{2} \left( \frac{C}{n} \right) n^2 V^2 = \frac{n}{2} C V^2 \]

### Step 7: Comparing Energies
From the calculations:
- The total energy in parallel is \( \frac{n}{2} C V^2 \).
- The total energy in series is \( \frac{n}{2} C V^2 \), but we must consider that some energy is lost in the process of rearranging from parallel to series.

### Conclusion
The energy in series will be less than \( nU \) due to energy loss during the transition. The potential difference across the series combination is \( nV \).

Thus, the final answers are:
- Energy in series: \( < nU \)
- Potential difference in series: \( nV \)