A capacitor of 60 pF charged to 20 volt. Now battery is removed and then this capacitor is connected to another identical uncharged capacitor. Find heat loss in nJ

Explanation

To solve the problem step-by-step, we will follow the outlined approach in the video transcript.

### Step 1: Calculate the initial charge on the capacitor
The charge \( Q \) stored in a capacitor is given by the formula:
\[ Q = C \cdot V \]
Where:
- \( C = 60 \, \text{pF} = 60 \times 10^{-12} \, \text{F} \)
- \( V = 20 \, \text{V} \)

Substituting the values:
\[ Q = 60 \times 10^{-12} \, \text{F} \cdot 20 \, \text{V} = 1200 \times 10^{-12} \, \text{C} = 1.2 \, \text{nC} \]

### Step 2: Calculate the initial energy stored in the capacitor
The energy \( U \) stored in a capacitor is given by:
\[ U = \frac{1}{2} C V^2 \]
Substituting the values:
\[ U = \frac{1}{2} \cdot 60 \times 10^{-12} \, \text{F} \cdot (20 \, \text{V})^2 \]
Calculating:
\[ U = \frac{1}{2} \cdot 60 \times 10^{-12} \cdot 400 = 12000 \times 10^{-12} \, \text{J} = 12 \, \text{nJ} \]

### Step 3: Determine the final voltage after connecting to an identical uncharged capacitor
When the charged capacitor is connected to an identical uncharged capacitor, the total charge remains the same, but the total capacitance doubles. The new voltage \( V_{\text{common}} \) can be calculated as:
\[ V_{\text{common}} = \frac{Q_{\text{total}}}{C_{\text{total}}} \]
Where:
- \( Q_{\text{total}} = 1.2 \, \text{nC} \)
- \( C_{\text{total}} = 2C = 2 \times 60 \times 10^{-12} \, \text{F} = 120 \times 10^{-12} \, \text{F} \)

Substituting the values:
\[ V_{\text{common}} = \frac{1.2 \times 10^{-9} \, \text{C}}{120 \times 10^{-12} \, \text{F}} = 10 \, \text{V} \]

### Step 4: Calculate the final energy stored in the two capacitors
The final energy \( U_f \) in the system can be calculated using the new voltage:
\[ U_f = \frac{1}{2} C_{\text{total}} V_{\text{common}}^2 \]
Substituting the values:
\[ U_f = \frac{1}{2} \cdot 120 \times 10^{-12} \, \text{F} \cdot (10 \, \text{V})^2 \]
Calculating:
\[ U_f = \frac{1}{2} \cdot 120 \times 10^{-12} \cdot 100 = 6000 \times 10^{-12} \, \text{J} = 6 \, \text{nJ} \]

### Step 5: Calculate the heat loss
The heat loss \( Q_{\text{loss}} \) is the difference between the initial energy and the final energy:
\[ Q_{\text{loss}} = U_i - U_f \]
Substituting the values:
\[ Q_{\text{loss}} = 12 \, \text{nJ} - 6 \, \text{nJ} = 6 \, \text{nJ} \]

### Final Answer
The heat loss is **6 nJ**.

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