# 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 - Physics

Numerical

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.)

#### Solution

Surface charge density of the earth, σ = 10−9 C m−2

Current over the entire globe, I = 1800 A

Radius of the earth, r = 6.37 × 106 m

Surface area of the earth,

A = 4πr2

= 4π × (6.37 × 106)2

= 5.09 × 1014 m2

Charge on the earth surface,

q = σ × A

= 10−9 × 5.09 × 1014

= 5.09 × 10C

Time taken to neutralize the earth’s surface = t

Current, I = "q"/"t"

t = "q"/"I"

= (5.09 xx 10^5)/1800

= 282.77 s

Therefore, the time taken to neutralize the earth’s surface is 282.77 s.

Concept: Cells, Emf, Internal Resistance
Is there an error in this question or solution?

#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 3 Current Electricity
Exercise | Q 3.14 | Page 128
NCERT Class 12 Physics Textbook
Chapter 3 Current Electricity
Exercise | Q 14 | Page 128

Share