Two capacitors of `3 pF` and `6pF` are connected in series and a potential difference of `5000 V` is applied across the combination. They are then disconnected and reconnected in parallel. The potential between the plates is

Options

A. `2250 V`
B. `2222 V`
C. `2.25 xx 10^(6)V`
D. `1.1 xx 10^(6 V`

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Calculate the Equivalent Capacitance in Series
When capacitors are connected in series, the equivalent capacitance \( C_{eq} \) can be calculated using the formula:

\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \]

For the given capacitors:
- \( C_1 = 3 \, \text{pF} = 3 \times 10^{-12} \, \text{F} \)
- \( C_2 = 6 \, \text{pF} = 6 \times 10^{-12} \, \text{F} \)

Substituting the values:

\[ \frac{1}{C_{eq}} = \frac{1}{3 \times 10^{-12}} + \frac{1}{6 \times 10^{-12}} \]

Calculating this gives:

\[ \frac{1}{C_{eq}} = \frac{2}{6 \times 10^{-12}} + \frac{1}{6 \times 10^{-12}} = \frac{3}{6 \times 10^{-12}} = \frac{1}{2 \times 10^{-12}} \]

Thus,

\[ C_{eq} = 2 \, \text{pF} = 2 \times 10^{-12} \, \text{F} \]

### Step 2: Calculate the Charge on the Capacitors
The total charge \( Q \) stored in the series combination can be calculated using the formula:

\[ Q = C_{eq} \times V \]

Where \( V = 5000 \, \text{V} \):

\[ Q = 2 \times 10^{-12} \times 5000 = 10^{-8} \, \text{C} \]

### Step 3: Determine the Charge on Each Capacitor
In a series connection, the charge on each capacitor is the same. Therefore, the charge \( Q \) on both capacitors is:

\[ Q_1 = Q_2 = Q = 10^{-8} \, \text{C} \]

### Step 4: Disconnect and Reconnect in Parallel
When the capacitors are disconnected and then reconnected in parallel, the total charge remains the same, but the voltage across each capacitor will change.

### Step 5: Calculate the Voltage Across Each Capacitor in Parallel
In parallel, the total charge \( Q \) is the sum of the charges on each capacitor:

\[ Q = Q_1 + Q_2 \]

Where:
- \( Q_1 = C_1 \times V_1 \)
- \( Q_2 = C_2 \times V_2 \)

Since they are in parallel, \( V_1 = V_2 = V \). Thus, we have:

\[ Q = C_1 \times V + C_2 \times V = (C_1 + C_2) \times V \]

Substituting the values:

\[ 10^{-8} = (3 \times 10^{-12} + 6 \times 10^{-12}) \times V \]

This simplifies to:

\[ 10^{-8} = 9 \times 10^{-12} \times V \]

### Step 6: Solve for Voltage \( V \)
Rearranging the equation gives:

\[ V = \frac{10^{-8}}{9 \times 10^{-12}} = \frac{10^{-8}}{9 \times 10^{-12}} = \frac{10^4}{9} \approx 1111.11 \, \text{V} \]

### Final Answer
The potential difference between the plates when the capacitors are connected in parallel is approximately:

\[ V \approx 1111.11 \, \text{V} \]