The dielectric strength of air is `3.0 xx 10^(8)V//m`. A parallel plate air capacitor has area `20 cm^(2)` and plate separation `sqrt2mm`. Find the maximum rms voltage of an AC source which can be safely connected to this capacitor.

Answer

To find the maximum RMS voltage of an AC source that can be safely connected to a parallel plate air capacitor, we can follow these steps:

### Step 1: Understand the given values
- Dielectric strength of air, \( E = 3.0 \times 10^8 \, \text{V/m} \)
- Area of the capacitor plates, \( A = 20 \, \text{cm}^2 = 20 \times 10^{-4} \, \text{m}^2 = 2.0 \times 10^{-3} \, \text{m}^2 \)
- Plate separation, \( d = \sqrt{2} \, \text{mm} = \sqrt{2} \times 10^{-3} \, \text{m} \)

### Step 2: Calculate the maximum voltage (V) using the formula
The maximum voltage \( V \) that can be applied across the capacitor is given by the product of the dielectric strength \( E \) and the separation distance \( d \):
\[ V = E \times d \]
Substituting the values:
\[ V = (3.0 \times 10^8 \, \text{V/m}) \times (\sqrt{2} \times 10^{-3} \, \text{m}) \]

### Step 3: Calculate \( \sqrt{2} \)
The value of \( \sqrt{2} \) is approximately \( 1.414 \). Therefore:
\[ V = 3.0 \times 10^8 \times 1.414 \times 10^{-3} \]

### Step 4: Simplify the expression
Calculating the multiplication:
\[ V = 3.0 \times 1.414 \times 10^5 \, \text{V} \]
\[ V = 4.242 \times 10^5 \, \text{V} \]

### Step 5: Calculate the maximum RMS voltage
The RMS voltage \( V_{\text{rms}} \) is related to the maximum voltage \( V \) by the formula:
\[ V_{\text{rms}} = \frac{V}{\sqrt{2}} \]
Substituting the value of \( V \):
\[ V_{\text{rms}} = \frac{4.242 \times 10^5}{\sqrt{2}} \]
Since \( \sqrt{2} \approx 1.414 \):
\[ V_{\text{rms}} = \frac{4.242 \times 10^5}{1.414} \approx 3.0 \times 10^5 \, \text{V} \]

### Final Answer
The maximum RMS voltage of the AC source that can be safely connected to this capacitor is:
\[ \boxed{3.0 \times 10^5 \, \text{V}} \]