The dielectric strength of air is 3-0 - 108 V-m- A parallel plate air capacitor has area 20 cm2 and plate separation -2 mm- Find the maximum rms voltage of an AC source which can be safely connected to this capacitor-

Dielectric strength of air = 3 × 10^8 v/m Area , A = 20cm^2 V= E.d = 3 × 10^8 ×√(2)× 10^ 3 = 3√(2)× 10^5 V_rms = V√(2) = 3√(2)× 10^5√(2) = 3 ...

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Standard XII

Physics

Capacitor in an AC Circuit

The dielectri...

Question

The dielectric strength of air is $3.0 \times 10^{8}$ V/m. A parallel plate air capacitor has area $20 c m^{2}$ and plate separation $\sqrt{2} m m .$ Find the maximum rms voltage of an AC source which can be safely connected to this capacitor.

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Solution

Dielectric strength of air $= 3 \times 10^{8} v / m$
Area , $A = 20 c m^{2}$
$V = E . d = 3 \times 10^{8} \times \sqrt{2} \times 10^{- 3} = 3 \sqrt{2} \times 10^{5}$
$V_{r m s} = \frac{V}{\sqrt{2}} = \frac{3 \sqrt{2} \times 10^{5}}{\sqrt{2}} = 3 \times 10^{4} V$

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The dielectric strength of air is 3.0×108 V/m. A parallel plate air capacitor has area 20cm2 and plate separation √2mm. Find the maximum rms voltage of an AC source which can be safely connected to this capacitor.

Dielectric strength of air =3×108v/mArea ,A=20cm2V=E.d=3×108×√2×10−3=3√2×105Vrms=V√2=3√2×105√2=3×104V

The dielectric strength of air is $3.0 \times 10^{8}$ V/m. A parallel plate air capacitor has area $20 c m^{2}$ and plate separation $\sqrt{2} m m .$ Find the maximum rms voltage of an AC source which can be safely connected to this capacitor.

Dielectric strength of air $= 3 \times 10^{8} v / m$
Area , $A = 20 c m^{2}$
$V = E . d = 3 \times 10^{8} \times \sqrt{2} \times 10^{- 3} = 3 \sqrt{2} \times 10^{5}$
$V_{r m s} = \frac{V}{\sqrt{2}} = \frac{3 \sqrt{2} \times 10^{5}}{\sqrt{2}} = 3 \times 10^{4} V$