If capacitor with capacitance C,2C are initially at potential v, 2v resp. Now they are connected in parallel combination such that end of charge of `(capacitor)_1` & -ve charge of `(capacitor)_2` are on same side. Find energy loss

Answer

To solve the problem of energy loss when two capacitors are connected in parallel, we will follow these steps:

### Step 1: Calculate Initial Energy Stored in Each Capacitor
1. **Capacitor 1 (C)**:
- Capacitance = C
- Initial potential = V
- Energy stored \( E_1 = \frac{1}{2} C V^2 \)

2. **Capacitor 2 (2C)**:
- Capacitance = 2C
- Initial potential = 2V
- Energy stored \( E_2 = \frac{1}{2} (2C) (2V)^2 = \frac{1}{2} (2C) (4V^2) = 4C V^2 \)

3. **Total Initial Energy**:
\[ E_{initial} = E_1 + E_2 = \frac{1}{2} C V^2 + 4C V^2 = \frac{1}{2} C V^2 + \frac{8C V^2}{2} = \frac{9C V^2}{2} \]

### Step 2: Determine Charge on Each Capacitor
- **Charge on Capacitor 1**:
\[ Q_1 = C \cdot V = CV \]

- **Charge on Capacitor 2**:
\[ Q_2 = 2C \cdot (2V) = 4CV \]

### Step 3: Analyze the Connection of Capacitors
When connected in parallel, the positive plate of Capacitor 1 (C) is connected to the negative plate of Capacitor 2 (2C). The effective charge on the positive plate of Capacitor 1 will be \( CV - x \) and on the negative plate of Capacitor 2 will be \( -4CV + x \), where \( x \) is the charge that flows from Capacitor 1 to Capacitor 2.

### Step 4: Set Up the Voltage Equation
Since the capacitors are connected in parallel, they will have the same final voltage \( V_0 \):
\[ V_0 = \frac{CV - x}{C} = \frac{-4CV + x}{2C} \]

### Step 5: Solve for Final Voltage
Equating the two expressions for voltage:
\[ \frac{CV - x}{C} = \frac{-4CV + x}{2C} \]
Cross-multiplying gives:
\[ 2(CV - x) = -4CV + x \]
Expanding and rearranging:
\[ 2CV - 2x = -4CV + x \]
\[ 2CV + 4CV = 3x \]
\[ 6CV = 3x \implies x = 2CV \]

### Step 6: Calculate Final Charges
- **Final charge on Capacitor 1**:
\[ Q_{1, final} = CV - x = CV - 2CV = -CV \]

- **Final charge on Capacitor 2**:
\[ Q_{2, final} = -4CV + x = -4CV + 2CV = -2CV \]

### Step 7: Calculate Final Energy
1. **Final Energy in Capacitor 1**:
\[ E_{1, final} = \frac{1}{2} C \left(\frac{-CV}{C}\right)^2 = \frac{1}{2} C \left(V^2\right) = \frac{1}{2} CV^2 \]

2. **Final Energy in Capacitor 2**:
\[ E_{2, final} = \frac{1}{2} (2C) \left(\frac{-2CV}{2C}\right)^2 = \frac{1}{2} (2C) \left(2V\right)^2 = 4CV^2 \]

3. **Total Final Energy**:
\[ E_{final} = E_{1, final} + E_{2, final} = \frac{1}{2} CV^2 + 4CV^2 = \frac{1}{2} CV^2 + \frac{8}{2} CV^2 = \frac{9}{2} CV^2 \]

### Step 8: Calculate Energy Loss
\[ \text{Energy Loss} = E_{initial} - E_{final} = \frac{9}{2} CV^2 - \frac{3}{2} CV^2 = \frac{6}{2} CV^2 = 3CV^2 \]

### Final Answer
The energy loss when the capacitors are connected in parallel is:
\[ \text{Energy Loss} = 3CV^2 \]