Question

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 ?

• A. 210 V
• B. 300 V
• C. 435 V
• D. 280 V

Answer 1

Answer: (A)

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
$V_{r m s}=\frac{V_{0}}{\sqrt{2}}=0.707 \times 300 = 210\, V$