Login/Sign Up
The earth’s surface has a negative surface charge density of 10–9 C m–2. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entireglobe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth’s surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 × 106 m.)
The earth’s surface has a negative surface charge density of 10–9 C m–2. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire
globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth’s surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 × 106 m.)
CBSE | Class 12 | Current Electricity | NCERT | Physics
The earth’s surface has a negative surface charge density of 10–9 C m–2. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire
globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth’s surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 × 106 m.)

Answer –

According to the question statement,

Surface charge density of the earth is σ = 10−9 cm−2
Potential difference between the surface and the top of the atmosphere is V= 400 kV
Current over the entire globe is given by I=1800 A
Radius of the earth is r=6.37×106 m
Surface area of the earth is given by the expression –

A = 4πr2

A = 4 x 3.14 x (6.37×106 )2

A = 5.09×1014 m2

Charge on the earth surface can be determined by making use of the surface charge density of earth in the following manner –

Q = σA

Q = 10−9  x  5.09 × 1014

Q = 5.09×105 C

Time taken to neutralize the earth’s surface is given by

 t = q/I              

t = 5.09 × 105 / 1800

Time = 283 s

i

More Material