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2019年6月10日星期一

earth-cloud capacitor

試想地面與地面上方800米處的雲層是平行極板電容。 如果雲層的面積為1.0平方公里= 1.0×106平方米,那麼電容是多少? 如果電場強度大於3.0×106 N / C會導致閃電,那麼雨雲可以承受的最大電荷是多少?

Consider the Earth and a cloud layer 800 meters above the Earth to be the plates of a parallel-plate capacitor. If the cloud layer has an area of 1.0 km² = 1.0×106 m², what is the capacitance? If an electric field strength greater than 3.0×106 N/C causes the air to break down and conduct charge (lightning), what is the maximum charge the cloud can hold?

The definition of capacitance is given by this equation: capacitance C, measured in farads, equals charge Q, measured in coulombs, divided by voltage V, measured in volts.

C = Q/ΔV

V = P / I (unit: J*s-1A-1)
(Unit: F = A*s / J*s-1A-1 = A2s2J-1)

Force, F’ = ma
(Unit: kg*m*s-2)
J = N*1 meter
(Unit: kg*m2*s-2)

a = (vf - vi) / (tf - ti)
vf: final velocity
vi: initial velocity
tf: final time
ti: initial time

(Unit: F = A2s2J-1 = A2s2kg-1m-2s2 = A2s4kg-1m-2)

Farad (F) is the unit of capacitance. The farad measures how much electric charge is accumulated on the capacitor. 1 farad is the capacitance of a capacitor that has charge of 1 coulomb when applied voltage drop of 1 volt.

Coulomb (C, A*s) is the SI unit of electric charge, equal to the quantity of electricity conveyed (傳送) in one second by a current of one ampere.

Voltage can also be stated as electric potential along a wire when an electric current of one ampere dissipates (消散) one watt (W) of power (W = J/s).

Ampere (A) represents a flow of one coulomb of electricity per second. A flow of one ampere is produced in a resistance of one ohm by a potential difference of one volt.

Joule (J) is equal to the energy transferred to an object when a force of one newton acts on that object in the direction of its motion through a distance of one meter.

Newton (N), the absolute unit of force in the International System of Units (SI units). It is defined as that force necessary to provide a mass of one kilogram with an acceleration of one meter per second per second.

The capacitance depends on the area of the plates, A, the size of the gap, d, and the insulator between the plates. We will assume that the insulator is air here:

Ɛ0 = 1/(4πk)
where k is the Coulomb constant.
(Unit: C²/Nm²)

C = ƐA / d
where Ɛis a constant called the permittivity of free space.
(Unit: C²N-1m-2*m2m-1 = C2N-1m-1)

Electric field strength, E = V/d (Unit: N/C = V/m)

Q = CΔV = CEd
(Unit: C2N-1m-1NC-1m = C)

N/C = kg*m*s-2*A-1s-1 = kg*m*s-3A-1
Newton Per Coulomb (N/C) has a dimension of MLT-3I-1 where M is mass, L is length, T is time, and I is electric current.


V/m = J*s-1A-1m-1 = kg*m2*s-2*s-1A-1m-1 = kg*m*s-3A-1

Source:
hep.physics.wayne.edu
www.britannica.com
rapidtables.com
school-for-champions.com
study.com
efunda.com
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