Question

The dielectric strength of air is 3.0 × 10⁶ V/m. A parallel-plate air-capacitor has area 20 cm² and plate separation 0.10 mm. Find the maximum rms voltage of an AC source that can be safely connected to this capacitor.

Answer 1

Given:
Area of parallel-plate air-capacitor, A = 20 cm²
Separation between the plates, d = 0.1 mm
Dielectric strength of air, E= 3 × 10⁶ V/m
E = $\frac{V}{d}$,
where V = potential difference across the capacitor
$\therefore$ V = Ed
= 3 × 10⁶ × 0.1 × 10⁻³
= 3 × 10² = 300 V
Thus, peak value of voltage is 300 V.
Maximum rms value of voltage $\left(V_{\mathrm{rms}}\right)$ is given by,
$$\begin{gathered} V_{\mathrm{rms}}=\frac{V_0}{\sqrt{2}} \\ =\frac{300}{\sqrt{2}}=212\mathrm{V} \end{gathered}$$