Question

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

Answer 1

Given E = $3\times {10}^5$ V/m ,

A = 20 $c{m}^2$, d= 0.1 mm

Potential difference across the capacitor

= Ed

= $3\times {10}^{-5}\times 0.1\times {10}^{-3}$

= $3\times {10}^2$ = 300 V

Maximum rms voltage

= $\frac{V}{\sqrt{2}}=\frac{300}{\sqrt{2}}$ =212 V