Storm clouds build up large negative charges, as described in the chapter. The c

Question

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

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

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

C) 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?

D)

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?

Explanation / Answer

Part A:
Since : V (r) = Q/4*pi*ebsoleneo * (1/r )
So,
shape must be like a
Answer: option 1

Part B:
Capacitance: C= 4*pi*ebsoleneo*a*b/(b-a)
Here: b = 10 km =10,000 m
a = radius of charge sphere in cloud = 0.5 km = 500 m
Q = 25 C
Capacitance: C= 4*pi*ebsoleneo*a*b/(b-a)
= 4*3.14*(8.854*10^-12)*10000*500 / (10000-50)
= 5.2*10^-8 F
Answer: a

Part C:
V= Q/4pi*ebsoleno* (1/a- 1/b)
Energy = 0.5*Q*V
initial energy when charge was 25 C :
E1 = 0.5*Q* Q/4pi*ebsoleno* (1/a- 1/b)
= 0.5*25*25/4pi*ebsoleno* (1/a- 1/b)

Final energy when charge is 12.5 C :
E2 = 0.5*Q* Q/4pi*ebsoleno* (1/a- 1/b)
= 0.5*12.5*12.5/4pi*ebsoleno* (1/a- 1/b)

E2/E1 = 12.5*12.5 / (25*25) = 1/4
So, energy becomes 1/4 of initial value.
so, 75% of energy is dissipated.
Answer: b

Part D:
Energy of cloud:
E = 0.5*Q* Q/4pi*ebsoleno* (1/a- 1/b)
= 0.5*25*25/(4*3.14*8.854*10^-12) * (1/500 - 1/10000)
=2.81*10^12 * (1/500 - 1/10000)
=6*10^9 J
If time taken to dissipate this energy is 1 s
Power = energy/time = 6*10^9 J / 1s = 6*10^9 W
Answer: 6.0 GW or option c

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