Storm clouds build up large negative charges as described in

Storm clouds build up large negative charges, as described in the chapter. The charges dwell in charge centers, regions of concentrated charge. Suppose a cloud has -25 C in a 1.0-km-diameter spherical charge center located 10 km above the ground, as sketched in (Figure 1) . The negative charge center attracts a similar amount of positive charge that is spread on the ground below the cloud. The charge center and the ground function as a charged capacitor, with a potential difference of approximately 3.1×108 V . The large electric field between these two \"electrodes\" may ionize the air, leading to a conducting path between the cloud and the ground. Charges will flow along this conducting path, causing a discharge of the capacitora lightning strike.

Part B

Which of the curves sketched below best approximates the shape of an equipotential drawn halfway between the charge center and the ground?

Part C

What is the approximate capacitance of the charge center + ground system?

Part D

If 12.5 C of charge is transferred from the cloud to the ground in a lightning strike, what fraction of the stored energy is dissipated?

If 12.5  of charge is transferred from the cloud to the ground in a lightning strike, what fraction of the stored energy is dissipated?

Part E

If the cloud transfers all of its charge to the ground via several rapid lightning flashes lasting a total of 1 s, what is the average power?

If the cloud transfers all of its charge to the ground via several rapid lightning flashes lasting a total of 1 , what is the average power?

https://session.masteringphysics.com/problemAsset/1871972/8/Knight3e.ch21.p87a.jpg

Solution

a)The cloud is a spherical charge on the ground the earth surface can be approximated to be flat over the area of 10 km

Equipotential lines will be curved upward as in the 1^st figure, they are at equal distance from the charge center. and at equal potential.

capacitance C = Q/V

given Q= -25C and v = 3.1e+8 V

C = 25/3.1e-8 = 8.1e-8 F

energy stored = QV/2 = = Q²/2C

as the charge drains out potential also drops but the capacitance is constant

when Q has reduced from 25 to 12.5 energy drops by ¼

stored energy is dissipated by 75%

total energy transferred = 625/2*8.1e-8

= 3.86e+9 J

it is transferred in 1s

Power = 3.86e+9 watts

     3.9 GW