Two idential capacitors are joined in parallel, charged to a potential `V` and then separated and then connected in series i.e. the positive plate of one is connected to negative of the other

Options

A. The charges on the free plated connected together are destoyed.
B. The energy stored in the system increases.
C. The potential difference between the free plates is 2V.
D. The potential difference remains constant.

Explanation

To solve the problem, we will analyze the situation step by step.

### Step 1: Understand the Initial Setup
We have two identical capacitors, each with capacitance \( C \), connected in parallel and charged to a potential \( V \).

### Step 2: Calculate the Charge on Each Capacitor
When two capacitors are connected in parallel, the total charge \( Q \) on the system is given by:
\[ Q = C_{\text{eq}} \cdot V \]
where \( C_{\text{eq}} = C + C = 2C \) (since both capacitors have the same capacitance \( C \)). Thus, the total charge is:
\[ Q = 2C \cdot V \]
Each capacitor will have the same charge \( Q_1 = Q_2 = CV \).

### Step 3: Analyze the Separation
After charging, the capacitors are separated. Each capacitor retains its charge:
\[ Q_1 = CV \quad \text{and} \quad Q_2 = CV \]

### Step 4: Connect Capacitors in Series
Now, we connect the two capacitors in series, with the positive plate of one connected to the negative plate of the other. The total charge \( Q \) remains the same for the series combination.

### Step 5: Calculate the Voltage Across Each Capacitor
In a series connection, the total voltage \( V_{\text{total}} \) across the series combination is the sum of the voltages across each capacitor:
\[ V_{\text{total}} = V_1 + V_2 \]
Where \( V_1 \) and \( V_2 \) are the voltages across each capacitor. Since both capacitors have the same charge \( Q \), we can express the voltages as:
\[ V_1 = \frac{Q}{C} = \frac{CV}{C} = V \]
\[ V_2 = \frac{Q}{C} = \frac{CV}{C} = V \]
Thus, the total voltage across the series combination is:
\[ V_{\text{total}} = V + V = 2V \]

### Step 6: Energy Consideration
Next, we calculate the energy stored in the system. The energy stored in a capacitor is given by:
\[ E = \frac{1}{2} C V^2 \]
For the parallel combination, the total energy \( E_{\text{parallel}} \) is:
\[ E_{\text{parallel}} = \frac{1}{2} (2C) V^2 = CV^2 \]
For the series combination, the equivalent capacitance \( C_{\text{eq}} \) is:
\[ C_{\text{eq}} = \frac{C \cdot C}{C + C} = \frac{C}{2} \]
The energy stored in the series combination \( E_{\text{series}} \) is:
\[ E_{\text{series}} = \frac{1}{2} \left(\frac{C}{2}\right) (2V)^2 = \frac{1}{2} \left(\frac{C}{2}\right) (4V^2) = CV^2 \]

### Conclusion
We find that the energy stored in both configurations (parallel and series) is the same, \( CV^2 \).

### Final Answer
The potential difference between the plates when connected in series is \( 2V \).