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 = `V/d`
where V = potential difference across the capacitor
∴ V = Ed 
= `3xx10^6xx0.1xx10^-3`
= `3 xx 10^2 = 300 V`
Thus, peak value of voltage is 300 V.
Maximum rms value of voltage `(V_{rms})` is given by,
`V_{rms} = V_0/ sqrt2`
= `300/sqrt2 = 212    V`