Q. The dielectric strength of air is $3 \times 10^{6} Vm ^{-1}$. A parallel plate capacitor has area $20\, cm ^{2}$ and plate separation $0.1 \,mm$. Find the maximum rms voltage of an AC source which can be connected ?

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A

210 V

B

300 V

C

435 V

D

280 V

Solution:

Dielectric strength of air $=3 \times 10^{6} V / m$.
Then pd required to conduct electricity through $0.1 \,mm$ air is $300\, V$.
If $300\, V$ is peak value, then rms value is
$V_{r m s}=\frac{V_{0}}{\sqrt{2}}=0.707 \times 300 = 210\, V$

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