In: Physics
We can treat the earth’s surface as one plate of a capacitor, and the upper part of the atmosphere as the other plate. This is a first order approximation: A parallel plate capacitor, with plates that have an area equal to the area of the surface of the earth, and the height of the atmosphere as the distance between plates:
a) Find the capacitance of this system
b) If the potential difference between ground and upper atmosphere is 100,000 V, calculate how much energy is stored in Earth’s atmosphere via this “capacitor” effect.
c)What is the electric field at a point just above the ground in the Earth-as-capacitor scenario?
Part A.We know, capacitance C of the parallel plate capacitor is given by the formula
C = AE/D ...eqn 1
Where A is the area of plates, E is the absolute permittivity of the free space having value
= 8.854*10-12 C2 N-1 m^(-2)
Where ^ denotes power.
Now, Radius of Earth, R = 6.38*106 m
Hence, Area of surface of earth, A = 4*pie*r2
i.e A = 4*3.14*(6.38*106)2
= 4*3.14*6.38*6.38*1012
= 511.24*1012 m2
Also, the upper atmosphere starts from a distance 85 km , hence seperation between the plates,
D = 85 km = 85000 m
Now, using values of D, E and A in eqn 1, we get
C = 511.24*1012* 8.854*10-12/ 85000
= 4526.58328/85000
C = 0.0532 F ..... Required capacitance.
Part b.Energy E stored in capacitor is given by the formula :
E = CV2/2
Since V = 100000 V = 105 V and C = 0.0532 F
Hence E = 0.0532*(105)2 / 2
= 0.0532*1010/2
E = 0.0266*1010 J.....required energy stored in the capacitor.
Part c. The inner core of earth' surface is positively charged. However, the surface of earth is negatively charges in order to maintain neutrality i.e net charge = 0.
Also, air above the surface of earth, i.e the atmosphere is positively charged.
Since the electric field lines always point from positive to negatively charged plate, hence the direction of Electric field at a point just above the surface of earth will be towards the surface of earth, i.e in downward direction.