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

Answer

To solve the problem step by step, we will follow the logical flow as described in the video transcript.

### Step 1: Calculate the initial charge on the capacitor
The charge \( Q \) on the capacitor can be calculated using the formula:
\[ Q = C \times V \]
Where:
- \( C = 60 \, \text{pF} = 60 \times 10^{-12} \, \text{F} \)
- \( V = 20 \, \text{V} \)

Substituting the values:
\[ Q = 60 \times 10^{-12} \, \text{F} \times 20 \, \text{V} = 1200 \times 10^{-12} \, \text{C} = 1.2 \times 10^{-9} \, \text{C} \]

### Step 2: Determine the final charge on each capacitor
When the charged capacitor is connected to an identical uncharged capacitor, the total charge is shared equally between the two capacitors. Therefore, the charge on each capacitor after connection will be:
\[ Q_f = \frac{Q}{2} = \frac{1.2 \times 10^{-9} \, \text{C}}{2} = 0.6 \times 10^{-9} \, \text{C} \]

### Step 3: Calculate the initial potential energy stored in the charged capacitor
The initial potential energy \( U_i \) stored in the charged capacitor can be calculated using the formula:
\[ U_i = \frac{Q^2}{2C} \]
Substituting the values:
\[ U_i = \frac{(1.2 \times 10^{-9})^2}{2 \times (60 \times 10^{-12})} \]
Calculating \( (1.2 \times 10^{-9})^2 \):
\[ (1.2 \times 10^{-9})^2 = 1.44 \times 10^{-18} \]
Now substituting this value:
\[ U_i = \frac{1.44 \times 10^{-18}}{2 \times 60 \times 10^{-12}} = \frac{1.44 \times 10^{-18}}{120 \times 10^{-12}} = \frac{1.44}{120} \times 10^{-6} = 0.012 \times 10^{-6} = 1.2 \times 10^{-5} \, \text{J} \]

### Step 4: Calculate the final potential energy stored in the two capacitors
The final potential energy \( U_f \) when both capacitors have equal charge \( Q_f \):
\[ U_f = 2 \times \frac{(Q_f)^2}{2C} = \frac{(0.6 \times 10^{-9})^2}{60 \times 10^{-12}} \]
Calculating \( (0.6 \times 10^{-9})^2 \):
\[ (0.6 \times 10^{-9})^2 = 0.36 \times 10^{-18} \]
Now substituting this value:
\[ U_f = \frac{0.36 \times 10^{-18}}{60 \times 10^{-12}} = \frac{0.36}{60} \times 10^{-6} = 0.006 \times 10^{-6} = 6 \times 10^{-9} \, \text{J} \]

### Step 5: Calculate the heat loss
The heat loss \( Q_{loss} \) is the difference between the initial and final potential energies:
\[ Q_{loss} = U_i - U_f \]
Substituting the values:
\[ Q_{loss} = 1.2 \times 10^{-5} - 6 \times 10^{-9} \]
Converting \( 1.2 \times 10^{-5} \) to nanojoules:
\[ 1.2 \times 10^{-5} \, \text{J} = 12 \, \text{nJ} \]
Thus:
\[ Q_{loss} = 12 \, \text{nJ} - 6 \, \text{nJ} = 6 \, \text{nJ} \]

### Final Answer
The heat loss is \( \boxed{6 \, \text{nJ}} \).