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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

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

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प्रश्न

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.)

संख्यात्मक
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उत्तर

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.

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पाठ 3: Current Electricity - Exercise [पृष्ठ १२८]

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एनसीईआरटी Physics Part I and II [English] Class 12
पाठ 3 Current Electricity
Exercise | Q 3.14 | पृष्ठ १२८

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