Question

Suppose that the charge distribution of a thundercloud can be approximated by two point charges, a negative charge -40 C at a heigh of 5 km (above ground) and a positive charge of 40 C at a height of 11 km. To find the electric field strength at the ground, we must take into account that the ground is a conductor and the the charge of the thundercloud induces charges on the ground. It can be shown that the effect of the induced charges can be simulated by fictitious

Answer 1

E = kq/r²

At x=2 , E = k ( 40/29 + 40/29 + 40/123 + 40/123) and putting k = 9*10^9

E = 30.68*10^9 N/C

At x = 4 , E = k ( 40/41 + 40/41 + 40/137 + 40/137) and putting k = 9*10^9

E = 22.8*10^9 N/C

At x = 6 , E = k ( 40/61 + 40/61 + 40/167 + 40/167) and putting k = 9*10^9

E = 16.11*10^9 N/C

At x = 8 , E = k ( 40/89 + 40/89 + 40/185 + 40/185) and putting k = 9*10^9

E = 11.98*10^9 N/C

At x = 10 , E = k ( 40/125 + 40/125 + 40/221 + 40/221) and putting k = 9*10^9

E = 9.01*10^9 N/C

B. a plot of field strength vs. distance will be declinin curve .