InterviewSolution
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InterviewSolution
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Let us suppose that the earth had a net surface charge density of 1 electron per m^(2). What would its potential be? Also what is the field just outside the earth's surface? |
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Answer» SOLUTION :DATA supplied, `sigma=1e//m^(2) =1.6 xx 10^(-19)C//m^(2), " RADIUS of the earth , "R=6.4 xx 10^(6)m` Potential `V=(1)/(4pi epsi_(0) ) q/R=(1)/(4pi epsi_(0)). (sigma 4pi R^(2))/(R) =(sigma R)/(epsi_(0)) =(1.6 xx 10^(-19) xx 6.4 xx 10^(6))/(8.854 xx 10^(-12)) =0.1156V` Field, `E=(1)/(4pi epsi_(0)).q/R^(2)=(1)/(4pi epsi_(0)) xx (sigma 4piR^(2))/(R^(2)) =(sigma)/(epsi_(0)) =(1.6 xx 10^(-19))/(8.854 xx 10^(-12))=1.81 xx 10^(-8) N//C` |
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