A `600 pF` capacitor is charged by a `200 V` supply. It is then disconnected from the supply and is connected to another uncharged `600 pF` capacitor. What is the common potential in `V` and energy lost in `J` after reconnection?

Options

A. `100, 6xx10^(-6)`
B. `200, 6xx10^(-5)`
C. `200, 5xx10^(-6)`
D. `100, 6xx10^(-5)`

Explanation

To solve the problem step by step, we will calculate the common potential and the energy lost after connecting the two capacitors.

### Step 1: Calculate the initial charge on the charged capacitor
The initial charge \( Q_1 \) on the first capacitor (C1) can be calculated using the formula:
\[ Q_1 = C_1 \times V_1 \]
Where:
- \( C_1 = 600 \, \text{pF} = 600 \times 10^{-12} \, \text{F} \)
- \( V_1 = 200 \, \text{V} \)

Substituting the values:
\[ Q_1 = 600 \times 10^{-12} \, \text{F} \times 200 \, \text{V} = 120000 \times 10^{-12} \, \text{C} = 120 \times 10^{-9} \, \text{C} = 120 \, \text{nC} \]

### Step 2: Determine the total charge after reconnection
The second capacitor (C2) is initially uncharged, so its charge \( Q_2 = 0 \).
When the two capacitors are connected, the total charge \( Q \) is conserved:
\[ Q = Q_1 + Q_2 = 120 \, \text{nC} + 0 = 120 \, \text{nC} \]

### Step 3: Calculate the common potential after reconnection
After reconnection, the two capacitors will share the total charge, and the common potential \( V \) can be calculated using:
\[ Q = (C_1 + C_2) \times V \]
Where:
- \( C_2 = 600 \, \text{pF} = 600 \times 10^{-12} \, \text{F} \)

Substituting the values:
\[ 120 \times 10^{-9} \, \text{C} = (600 \times 10^{-12} \, \text{F} + 600 \times 10^{-12} \, \text{F}) \times V \]
\[ 120 \times 10^{-9} = (1200 \times 10^{-12}) \times V \]
\[ V = \frac{120 \times 10^{-9}}{1200 \times 10^{-12}} = 100 \, \text{V} \]

### Step 4: Calculate the initial and final energy
The initial energy \( U_i \) stored in the charged capacitor is given by:
\[ U_i = \frac{1}{2} C_1 V_1^2 \]
Substituting the values:
\[ U_i = \frac{1}{2} \times 600 \times 10^{-12} \times (200)^2 \]
\[ U_i = \frac{1}{2} \times 600 \times 10^{-12} \times 40000 = 12 \times 10^{-6} \, \text{J} = 12 \, \mu\text{J} \]

The final energy \( U_f \) stored in both capacitors after reconnection is:
\[ U_f = \frac{1}{2} (C_1 + C_2) V^2 \]
Substituting the values:
\[ U_f = \frac{1}{2} \times (600 \times 10^{-12} + 600 \times 10^{-12}) \times (100)^2 \]
\[ U_f = \frac{1}{2} \times 1200 \times 10^{-12} \times 10000 = 6 \times 10^{-6} \, \text{J} = 6 \, \mu\text{J} \]

### Step 5: Calculate the energy lost
The energy lost \( \Delta U \) is given by:
\[ \Delta U = U_i - U_f \]
Substituting the values:
\[ \Delta U = 12 \times 10^{-6} - 6 \times 10^{-6} = 6 \times 10^{-6} \, \text{J} = 6 \, \mu\text{J} \]

### Final Answers
- Common potential \( V = 100 \, \text{V} \)
- Energy lost \( \Delta U = 6 \, \mu\text{J} \)