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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 electricfield, 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 tghe coontainual thunderstorms and lighting in different parts of the globe). (Radius of earth =6.37 xx 10 ^(6) m.) |
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Answer» Solution :Amount of charge on the spherical surface of Earth, `Q = 4 pi R^(2) sigma` `( because " Surface charge density " sigma = (Q)/(A) = (Q)/(4 pi R^(2)) )` Constant CURRENT is, `I = (Q)/(t) ` `therefore = (Q)/(I) ` `therefore t = (4 pi R^(2) sigma)/(I)` `therefore t = ((4) (3.14) (6.37 xx 10^(6))^(2) (10^(-9)))/(1800)` `therefore t = 283.1 ` S |
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