The capacity of a parallel plate condenser consisting of two plates each `10 cm` square and are seperated by a distance of `2 mm` is (Take air as the medium between the plates).

Options

A. `8.85 xx 10^(-13) F`
B. `4.42 xx 10^(-12) F`
C. `44.25 xx 10^(-12) F`
D. `88.5 xx 10^(-13) F`

Explanation

To find the capacity of a parallel plate condenser, we can use the formula:

\[ C = \frac{A \epsilon_0}{d} \]

where:
- \(C\) is the capacitance,
- \(A\) is the area of one of the plates,
- \(\epsilon_0\) is the permittivity of free space (air in this case),
- \(d\) is the separation between the plates.

### Step 1: Convert the area from cm² to m²
The area of each plate is given as \(10 \, \text{cm}^2\). We need to convert this to square meters.

\[ A = 10 \, \text{cm}^2 = 10 \times 10^{-4} \, \text{m}^2 = 1 \times 10^{-3} \, \text{m}^2 \]

### Step 2: Convert the distance from mm to m
The distance between the plates is given as \(2 \, \text{mm}\). We need to convert this to meters.

\[ d = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \]

### Step 3: Use the value of \(\epsilon_0\)
The permittivity of free space is given as:

\[ \epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \]

### Step 4: Substitute the values into the capacitance formula
Now we can substitute the values of \(A\), \(\epsilon_0\), and \(d\) into the capacitance formula.

\[ C = \frac{A \epsilon_0}{d} = \frac{(1 \times 10^{-3} \, \text{m}^2)(8.85 \times 10^{-12} \, \text{F/m})}{2 \times 10^{-3} \, \text{m}} \]

### Step 5: Calculate the capacitance
Calculating the above expression:

\[ C = \frac{(1 \times 10^{-3})(8.85 \times 10^{-12})}{2 \times 10^{-3}} = \frac{8.85 \times 10^{-15}}{2 \times 10^{-3}} = 4.425 \times 10^{-12} \, \text{F} \]

Thus, the capacitance of the parallel plate condenser is:

\[ C \approx 4.425 \, \text{pF} \quad (\text{picoFarads}) \]

### Summary of the Solution
The capacity of the parallel plate condenser is approximately \(4.425 \, \text{pF}\).

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