The plate area of a parallel-plate capacitors is `10cm^(2)` and its capacitance is `2pF`. The separation between the plates of the capacitors is close to (in mm).
(permittivity of vacuum, `epsi_(0)=(80)/(9)xx10^(-12)m^(-3)kg^(-1)s^(4)A^(2),1pF=10^(-12)F` )

Options

A. 2.2
B. 4.4
C. 8.8
D. 17.6

Explanation

To find the separation between the plates of a parallel-plate capacitor, we can use the formula for capacitance:

\[ C = \frac{\varepsilon_0 \cdot A}{d} \]

Where:
- \(C\) is the capacitance,
- \(\varepsilon_0\) is the permittivity of free space,
- \(A\) is the area of the plates,
- \(d\) is the separation between the plates.

We need to rearrange this formula to solve for \(d\):

\[ d = \frac{\varepsilon_0 \cdot A}{C} \]

### Step 1: Convert the area from cm² to m²
The area \(A\) is given as \(10 \, \text{cm}^2\). 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 capacitance from pF to F
The capacitance \(C\) is given as \(2 \, \text{pF}\). To convert this to farads:

\[ C = 2 \, \text{pF} = 2 \times 10^{-12} \, \text{F} \]

### Step 3: Substitute the values into the formula
Now, we can substitute the values into the rearranged formula for \(d\):

\[ d = \frac{\varepsilon_0 \cdot A}{C} = \frac{\left(\frac{80}{9} \times 10^{-12} \, \text{m}^{-3} \text{kg}^{-1} \text{s}^4 \text{A}^2\right) \cdot (1 \times 10^{-3} \, \text{m}^2)}{2 \times 10^{-12} \, \text{F}} \]

### Step 4: Calculate the numerator
Calculating the numerator:

\[ \varepsilon_0 \cdot A = \left(\frac{80}{9} \times 10^{-12}\right) \cdot (1 \times 10^{-3}) = \frac{80 \times 10^{-15}}{9} \, \text{m}^{-1} \text{kg}^{-1} \text{s}^4 \text{A}^2 \]

### Step 5: Calculate \(d\)
Now substituting this into the equation for \(d\):

\[ d = \frac{\frac{80 \times 10^{-15}}{9}}{2 \times 10^{-12}} = \frac{80 \times 10^{-15}}{18 \times 10^{-12}} = \frac{80}{18} \times 10^{-3} \, \text{m} \]

### Step 6: Simplify the fraction
Calculating the fraction:

\[ \frac{80}{18} = \frac{40}{9} \approx 4.44 \]

Thus,

\[ d \approx 4.44 \times 10^{-3} \, \text{m} = 4.44 \, \text{mm} \]

### Final Answer
The separation between the plates of the capacitor is approximately \(4.44 \, \text{mm}\).

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