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

Maximum RMS Voltage is 210 V

Explanation:

Given,  

Dielectric Strength of Air V = $3 \times 10^6\ Vm^-^1$

Area A = 20 $cm^2$

Plate Separation d = 0.1 mm

E = $\frac {V}{d}$

or V = $E\times d$

= $3 \times 10^-^6 \times 10^-^4$

= 300 V

$v_r_m_s = \frac {V_p}{\sqrt 2}$

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

= $300 \times 0.707$

= 210 V

Hence, The Maximum rms voltage of an AC source which can be connected is 210 V.

Answer 2

Answer:

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