A capacitor of capacitance `2.0 mu F` is charged to a potential difference of 12 V. It is then connected to an uncharged capacitor of capacitance `4.0 mu F`. Find (a) the charge flow through connecting wire upto study strate on each of the two capacitors after the connection in `mu C` (b) The total electrostatic energy stored in both capacitors in `mu J` (c) the heat produced during the charge transfer from one capacitor to the other in `mu J`.

Answer

To solve the problem step by step, we will break it down into three parts as per the question.

### Given Data:
- Capacitance of capacitor 1, \( C_1 = 2.0 \, \mu F \)
- Potential difference across capacitor 1, \( V_1 = 12 \, V \)
- Capacitance of capacitor 2, \( C_2 = 4.0 \, \mu F \)
- Initial charge on capacitor 1, \( Q_1 = C_1 \times V_1 \)

### Step 1: Calculate the initial charge on capacitor 1
Using the formula for charge on a capacitor:
\[ Q_1 = C_1 \times V_1 = 2.0 \, \mu F \times 12 \, V = 24 \, \mu C \]

### Step 2: Determine the final potential after connecting the capacitors
When the charged capacitor (capacitor 1) is connected to the uncharged capacitor (capacitor 2), charge will redistribute until both capacitors reach the same final potential \( V \).

Using the conservation of charge:
\[ Q_{\text{initial}} = Q_{\text{final}} \]
\[ Q_1 + Q_2 = Q_1' + Q_2' \]
Where:
- \( Q_2 = 0 \) (initial charge on capacitor 2)
- \( Q_1' = C_1 \times V \)
- \( Q_2' = C_2 \times V \)

Substituting the values:
\[ 24 \, \mu C + 0 = C_1 V + C_2 V \]
\[ 24 = (C_1 + C_2) V \]
\[ 24 = (2.0 + 4.0) V \]
\[ 24 = 6.0 V \implies V = 4 \, V \]

### Step 3: Calculate the charge on each capacitor after connection
Now we can find the final charges on both capacitors:
\[ Q_1' = C_1 \times V = 2.0 \, \mu F \times 4 \, V = 8 \, \mu C \]
\[ Q_2' = C_2 \times V = 4.0 \, \mu F \times 4 \, V = 16 \, \mu C \]

### Step 4: Calculate the charge flow through the connecting wire
The charge that flows through the wire is the difference between the initial charge on capacitor 1 and the final charge on capacitor 1:
\[ q = Q_1 - Q_1' = 24 \, \mu C - 8 \, \mu C = 16 \, \mu C \]

### Step 5: Calculate the total electrostatic energy stored in both capacitors
The energy stored in a capacitor is given by:
\[ U = \frac{1}{2} C V^2 \]
Calculating the energy for both capacitors:
\[ U_1 = \frac{1}{2} C_1 V^2 = \frac{1}{2} \times 2.0 \, \mu F \times (4 \, V)^2 = \frac{1}{2} \times 2.0 \times 16 = 16 \, \mu J \]
\[ U_2 = \frac{1}{2} C_2 V^2 = \frac{1}{2} \times 4.0 \, \mu F \times (4 \, V)^2 = \frac{1}{2} \times 4.0 \times 16 = 32 \, \mu J \]
Total energy stored:
\[ U_{\text{total}} = U_1 + U_2 = 16 \, \mu J + 32 \, \mu J = 48 \, \mu J \]

### Step 6: Calculate the heat produced during the charge transfer
The heat produced can be calculated as the difference between the initial energy stored in capacitor 1 and the final total energy stored in both capacitors:
\[ U_{\text{initial}} = \frac{1}{2} C_1 V_1^2 = \frac{1}{2} \times 2.0 \, \mu F \times (12 \, V)^2 = \frac{1}{2} \times 2.0 \times 144 = 144 \, \mu J \]
The heat produced:
\[ Q_{\text{heat}} = U_{\text{initial}} - U_{\text{total}} = 144 \, \mu J - 48 \, \mu J = 96 \, \mu J \]

### Final Answers:
(a) Charge flow through connecting wire: \( 16 \, \mu C \)
(b) Total electrostatic energy stored: \( 48 \, \mu J \)
(c) Heat produced during charge transfer: \( 96 \, \mu J \)