Two capacitors of `10 pF` and `20 pF` are connected to 200 V and 100 V sources, respectively. If they are connected by the wire, then what is the common potential of the capacitors?

Options

A. 133.3 V
B. 150 V
C. 300 V
D. 400 V

Answer

To solve the problem of finding the common potential of two capacitors connected together after being charged by different voltage sources, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Capacitor 1 (C1) = 10 pF
- Capacitor 2 (C2) = 20 pF
- Voltage across C1 (V1) = 200 V
- Voltage across C2 (V2) = 100 V

2. **Calculate the Initial Charges**:
- The charge on capacitor 1 (Q1) is calculated using the formula:
\[ Q1 = C1 \times V1 = 10 \, \text{pF} \times 200 \, \text{V} = 2000 \, \text{pC} \]
- The charge on capacitor 2 (Q2) is calculated using the formula:
\[ Q2 = C2 \times V2 = 20 \, \text{pF} \times 100 \, \text{V} = 2000 \, \text{pC} \]

3. **Total Initial Charge**:
- The total initial charge (Q_initial) when the capacitors are connected to their respective voltage sources is:
\[ Q_{\text{initial}} = Q1 + Q2 = 2000 \, \text{pC} + 2000 \, \text{pC} = 4000 \, \text{pC} \]

4. **Connecting the Capacitors**:
- Once the batteries are removed and the capacitors are connected together, they will share the total charge. Let V be the common potential across both capacitors when they are connected.

5. **Using Charge Conservation**:
- The total charge after connecting the capacitors (Q_final) is given by:
\[ Q_{\text{final}} = Q1' + Q2' = C1 \times V + C2 \times V \]
- This can be expressed as:
\[ Q_{\text{final}} = (C1 + C2) \times V \]

6. **Setting Up the Equation**:
- Since charge is conserved, we can set the initial charge equal to the final charge:
\[ 4000 \, \text{pC} = (C1 + C2) \times V \]
- Substituting the values of C1 and C2:
\[ 4000 \, \text{pC} = (10 \, \text{pF} + 20 \, \text{pF}) \times V \]
- This simplifies to:
\[ 4000 \, \text{pC} = 30 \, \text{pF} \times V \]

7. **Solving for the Common Potential (V)**:
- Rearranging the equation to solve for V gives:
\[ V = \frac{4000 \, \text{pC}}{30 \, \text{pF}} = \frac{4000}{30} \, \text{V} \approx 133.33 \, \text{V} \]

### Final Answer:
The common potential of the capacitors when connected together is approximately **133.33 V**.

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