Dry air will break down if the electric field exceeds about 3.0×106Vm Part A What amount of charge can be placed on a capacitor if the area of each plate is 8.6cm2 ? Express your answer using two significant figures. I want the answer in nC

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Dry air will break down if the electric field exceeds about 3.0×106Vm   Part A   What amount of charge can be placed on a capacitor if the area of each plate is 8.6cm2 ?   Express your answer using two significant figures.   I want the answer in nC

Bukvald5z · Accepted Answer

electric field E=QmaxA⋅e0 A=area of cross section of plate=6.9⋅10−4m2 e0=8.85⋅10−12NC2m2 Qmax=E⋅A⋅e0 Qmax=3⋅106⋅6.9⋅10−4⋅8.85⋅10−12 Qmax=18nC

RizerMix · Accepted Answer

The charge of the capacitor is, Q=CV (ξ0A)/d(Ed) =ξ0EA The amount of charge can be placed on a capacitor is, Q=ξ0EA =(8.85×10(−12) (Nm2)/C2)(3×106 V/m)(8.6×10(−4) m2) =22.83×10(−9) C =22.83nC

alenahelenash · Accepted Answer

C=ξ∗A/d&amp;nbsp;Where&amp;nbsp;ξ&amp;nbsp;is the permittivity of free-space, A is the plate area, and d is the distance separating the plates. The maximum voltage can be calculated:&amp;nbsp;V(max)=3∗106∗d&amp;nbsp;Q = C*VQ=(ξ∗A/d)∗3∗106∗d&amp;nbsp;The d cancels:&amp;nbsp;Q=ξ∗A∗3∗106=8.854∗10(−12)∗(68/104)∗3∗106=181∗10(−9)&amp;nbsp;C

Dry air will break down if the electric field exceeds

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