The dielectric strength of air is `3.0 xx 10^6 V/m`. A parallel-plate air-capacitor has area `20 cm^2` and plate separation `0.10 mm`.Find the maximum rms voltage of an AC source which can be safely connected to this capacitor.

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Understand the given values
- Dielectric strength of air, \( E = 3.0 \times 10^6 \, \text{V/m} \)
- 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, \( D = 0.10 \, \text{mm} = 0.10 \times 10^{-3} \, \text{m} = 1.0 \times 10^{-4} \, \text{m} \)

### Step 2: Calculate the maximum voltage (V₀) using the dielectric strength
The maximum voltage that can be applied across the capacitor plates before breakdown occurs can be calculated using the formula:
\[ V_0 = E \times D \]
Substituting the values:
\[ V_0 = (3.0 \times 10^6 \, \text{V/m}) \times (1.0 \times 10^{-4} \, \text{m}) = 300 \, \text{V} \]

### Step 3: Calculate the maximum RMS voltage (V_rms)
The maximum RMS voltage can be calculated from the maximum voltage using the relationship:
\[ V_{\text{rms}} = \frac{V_0}{\sqrt{2}} \]
Substituting the value of \( V_0 \):
\[ V_{\text{rms}} = \frac{300 \, \text{V}}{\sqrt{2}} \approx \frac{300 \, \text{V}}{1.414} \approx 212.2 \, \text{V} \]

### Final Answer
The maximum RMS voltage that can be safely connected to the capacitor is approximately \( 212.2 \, \text{V} \).

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