A capacitor of capacitance `C` is charged to a potential difference `V` from a cell and then disconncted from it. A charge `+Q` is now given to its positive plate. The potential difference across the capacitor is now

Options

A. `V`
B. `V+Q/C`
C. `V+Q/(2C)`
D. `V-Q/C` if `QltCV`

Explanation

To solve the problem step by step, let's break down the process of determining the potential difference across the capacitor after a charge \( +Q \) is added to its positive plate.

### Step-by-Step Solution:

1. **Initial Conditions**:
- A capacitor of capacitance \( C \) is charged to a potential difference \( V \).
- The initial charge \( Q_{\text{initial}} \) on the capacitor can be calculated using the formula:
\[ Q_{\text{initial}} = C \cdot V \]

2. **Disconnecting the Capacitor**:
- After charging, the capacitor is disconnected from the cell. The charge on the plates remains constant at \( Q_{\text{initial}} \).

3. **Adding Charge \( +Q \)**:
- A charge \( +Q \) is added to the positive plate of the capacitor.
- The new charge on the positive plate becomes:
\[ Q_{\text{positive}} = Q_{\text{initial}} + Q \]
- The charge on the negative plate remains unchanged, which is \( Q_{\text{initial}} \).

4. **Calculating the New Charge**:
- The total charge on the positive plate after adding \( +Q \):
\[ Q_{\text{positive}} = C \cdot V + Q \]
- The charge on the negative plate is still:
\[ Q_{\text{negative}} = C \cdot V \]

5. **Using Charge Conservation**:
- For the capacitor, the charge on the plates can be expressed as:
\[ Q_{\text{positive}} = Q_{\text{negative}} + Q_{\text{added}} \]
- Here, \( Q_{\text{added}} = +Q \).

6. **Finding the New Potential Difference**:
- The potential difference \( V' \) across the capacitor can be calculated using the new charge on the positive plate:
\[ V' = \frac{Q_{\text{positive}}}{C} = \frac{C \cdot V + Q}{C} \]
- Simplifying this expression gives:
\[ V' = V + \frac{Q}{C} \]

7. **Final Result**:
- Thus, the potential difference across the capacitor after adding the charge \( +Q \) is:
\[ V' = V + \frac{Q}{C} \]

### Final Answer:
The potential difference across the capacitor is \( V + \frac{Q}{C} \).

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