The area of the positive plate is `A_(1)` and the area of the negative plate is `A_(2)(A_(2) lt A_(1))`. They are parallel to each other and are separated by a distance `d`. The capacity of a condenser with air dielectric is.

Options

A. `(epsilon_(0)A_(1))/(d)`
B. `(epsilon_(0)A_(2))/(d)`
C. `(epsilon_(0)A_(1)A_(2))/(d)`
D. `(epsilon_(0)A_(1))/(A_(2)d)`

Explanation

To solve the problem of finding the capacitance of a capacitor with two parallel plates where the area of the positive plate is \( A_1 \) and the area of the negative plate is \( A_2 \) (with \( A_2 < A_1 \)), separated by a distance \( d \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Configuration**:
- We have two parallel plates, one with area \( A_1 \) (positive plate) and the other with area \( A_2 \) (negative plate).
- The plates are separated by a distance \( d \).

2. **Identify the Effective Area**:
- The capacitance of a capacitor depends on the effective area that can store charge. Since \( A_2 < A_1 \), the effective area for capacitance will be determined by the smaller plate, which is \( A_2 \).

3. **Use the Capacitance Formula**:
- The formula for the capacitance \( C \) of a parallel plate capacitor is given by:
\[ C = \frac{k \cdot A}{d} \]
- Here, \( k \) is the dielectric constant. For air, \( k = 1 \).

4. **Substitute the Values**:
- Since the effective area \( A \) is \( A_2 \) and \( k = 1 \), we can substitute these values into the formula:
\[ C = \frac{1 \cdot A_2}{d} = \frac{A_2}{d} \]

5. **Final Expression**:
- Therefore, the capacitance of the capacitor is:
\[ C = \frac{A_2 \epsilon_0}{d} \]
- Where \( \epsilon_0 \) is the permittivity of free space, which is a constant.

### Final Answer:
The capacitance of the capacitor is:
\[ C = \frac{A_2 \epsilon_0}{d} \]