Question

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 1

Due to negative charge appeared on the earth, an electric field is into the earth's surface due to which the positive ions of atmosphere are constantly pumped in and an equivalent current of 1800A is established across the globe.

given, surface charge density, $\sigma$ = 10^-9 C/m²

radius of the earth , r = 6.37 × 10^6 m

so, surface area of earth, A = 4πr²

= 4π(6.37 × 10^6) m²

now total negative charge on the earth, Q = $\sigma A$

= (10^-9 c/m²) × 4π(6.37 × 10^6)²

= 5.0965 × 10^5 C

using formula, Q = It

so, t = Q/I

here, I = 1800A , Q = 5.0965 × 10^5 C

= 5.0965 × 10^5 C/(5.0985 × 10^5A)

= 283.14sec

hence, time required to neutralise the earth's surface, t = 283.14 s

Answer 2

Explanation:

$\Large{\red{\underline{\underline{\sf{\orange{Solution:}}}}}}$

 ‎ ‎ ‎ ‎ ‎ ‎ ‎

$\hookrightarrow$ Surface charge density of the earth, $\sf \sigma\:=\:10^{-9}C\,m^{-2}$

$\hookrightarrow$ Current over the entire globe, $\sf I\:=\:1800\,A$

$\hookrightarrow$ Radius of the earth, $\sf r\:=\:6.37\times 10^6m$

 ‎ ‎ ‎ ‎ ‎ ‎

$\longrightarrow$ Surface area of the earth, $\sf A\:=\:4\pi r^2$

$\sf A\:=\:4\pi r^2\times (6.37\times 10^6)^2$

$\sf A\:=\:5.09\times 10^{14}m^2$

 ‎ ‎ ‎ ‎ ‎ ‎

$\longrightarrow$ Charge on the earth surface,

$\sf q\:=\: \sigma\times A$

$\sf q\:=\: 10^{-9}\times 5.09\times 10^{14}$

$\sf q\:=\:5.09\times 10^5C$

 ‎ ‎ ‎ ‎ ‎ ‎

$\longrightarrow$ Time taken to neutralize the earth's surface = t

Current, $\sf I\:=\: \dfrac{q}{t}$

$\sf t\:=\: \dfrac{q}{I}$

$\sf t\:=\: \dfrac{5.09\times 10^5}{1800}$

$\sf t\:=\:282.77s$

 ‎ ‎ ‎ ‎ ‎ ‎

$\longrightarrow$ Hence, time required to neutralize earth's surface is$\sf{\blue{\:\:282.77s}}$