Question

One can think of the bottom of a thundercloud as a capacitor plate and the ground beneath it as the other plate. If the bottom of the cloud is located a height of 0.5km above the ground, has a width of 2km and a length of 3km, approximated as a rectangle, carries a negative charge of 160 C, while the ground has the same amount of positive charge, (a) what is the capacitance of the cloud-ground system? (b) What is the potential difference between the cloud bottom and the ground? (c) If lightning strikes when the electric field strength between the cloud bottom and the ground exceeds approximately 2.5MV/m, are the conditions for lightning to occur in this case fulfilled? (d) What is the electric potential energy stored in this “capacitor”? (e) If we could tap into this energy, for how long will we be able to run a 100W bulb off it? How about an average household with a need of 2000W? Make sure to express these results in a unit of time that is appropriate in order to grasp the numbers!

Answer 1

Given

capacitor with

length 3 km , width 2 km , Q = 160 C , d = 0.5 km

we know that the capacitanceof the parallel plate capacitor is

a)

C = epsilon not A/d

C = 8.854*10^-12 *(3000*2000)/(500)

C = 1.06248*10^-7 F

C = 0.106248*10^-6 F

b) Q =C*V

V = Q/C

V = 160/(0.106248*10^-6) V

V = 1505910699.49552 V

c) we have the relation V = E*d

E = V/d

E = 1505910699.49552/500 V/m

E= 3011821.399 V/m = 3.011821399 MV/m

yes the condition fulfilled

d) energy stored in the capacitor is  

U = 0.5*C*V^2

U = 0.5*0.106248*10^-6*(1505910699.49552)^2 J

U = 120472855959.64162 J

e) U = P*t

t = U/p = 120472855959.64162/100 s = 1204728559.6 s

t = 334646.82 hr

f) t = U/p = 120472855959.64162/2000 s = 60236427.98 s

t = 16732.34111 hr