A parallel-plate air-capacitor has area 20 cm 2 and plate separation 0.10 mm. Find the maximum rms voltage (in volt) of an AC source which can be safely connected to this capacitor. Max electric field till capacitor will not break down is E=3 × 106 V / m

Question

A parallel-plate air-capacitor has area 20 cm 2 and plate separation 0.10 mm. Find the maximum rms voltage (in volt) of an AC source which can be safely connected to this capacitor. Max electric field till capacitor will not break down is E=3 × 106 V / m (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Given E=3 × 106 V / m d =0.1 mm =10-4 m A=20 cm 2 V max = E.d. =3 × 106 × 10-4=300 volt V text rms =0.707 V max =0.707 × 300=210 volt

Q. A parallel-plate air-capacitor has area $20 c m^{2}$ and plate separation $0.10 mm$ . Find the maximum rms voltage (in volt) of an AC source which can be safely connected to this capacitor. Max electric field till capacitor will not break down is $E = 3 \times 1 0^{6} V / m$

Given $E = 3 \times 1 0^{6} V / m$
$d = 0.1 mm = 1 0^{- 4} m$
$A = 20 c m^{2}$
$V_{m a x} = E . d . = 3 \times 1 0^{6} \times 1 0^{- 4} = 300$ volt
$V_{rms} = 0.707 V_{m a x} = 0.707 \times 300 = 210$ volt

Q. A parallel-plate air-capacitor has area 20cm2 and plate separation 0.10mm. Find the maximum rms voltage (in volt) of an AC source which can be safely connected to this capacitor. Max electric field till capacitor will not break down is E=3×106V/m

Given E=3×106V/m d=0.1mm=10−4m A=20cm2 Vmax​=E.d.=3×106×10−4=300 volt Vrms ​=0.707Vmax​=0.707×300=210 volt

Q. A parallel-plate air-capacitor has area $20 c m^{2}$ and plate separation $0.10 mm$ . Find the maximum rms voltage (in volt) of an AC source which can be safely connected to this capacitor. Max electric field till capacitor will not break down is $E = 3 \times 1 0^{6} V / m$

Given $E = 3 \times 1 0^{6} V / m$
$d = 0.1 mm = 1 0^{- 4} m$
$A = 20 c m^{2}$
$V_{m a x} = E . d . = 3 \times 1 0^{6} \times 1 0^{- 4} = 300$ volt
$V_{rms} = 0.707 V_{m a x} = 0.707 \times 300 = 210$ volt