Question

You are working as an intern for a meteorological laboratory. You are out in the field taking measurements with a device that measures electric fields. You measure the electric field in the air immediately above the Earth's surface to be 147 N/C directed downward. (Assume the radius of the Earth is 6.37 x 106 m.) (a) Determine the surface charge density (in C/m2) on the ground. -1.30e-9 C/m2 (b) Imagine the surface charge density is uniform over the planet. Determine the charge (in C) of the whole surface of the Earth. 6.63e5 X Use the given electric field of the Earth directly to find the charge, or find the charge from the charge density in part (a). C (C) Determine the Earth's electric potential (in V) due to the charge found in (b). 9.37e8 The potential of the Earth should be negative if the charge on the Earth is negative. Is the electric field due to the Earth uniform over its entire radius? V (d) Determine the difference in potential (in V) between the head and the feet of a person 1.59 m tall. (Ignore any charges in the atmosphere.) 242.55 X Jim of + Orto Life field

Answer 1

Let a) The electric field on the surface of earth as E. E= 147 N/c=0 Eo here, t= surface charge density Eo electric permittivity of free space. (47) (0-254131612) 1-301X 169 cima on the surface of earth is a' Q Q 1-301 X10 T: EEO - q 11 b) Change TE 2 A GARZ =) Q = (1301x109)/40R2)=(-301x16 9) (4*) (6-37*30492 ar Orto4433 6.6338 losc. the charge is negative. and

c) 9 VE electric potential on the surface of enthis, kQ 9X10%(5-633*18) R (6-37 X106 +(9 V 9.372 X 10 volt potential is ugative. potential cliff. is :- DV=V₂ -V, EkQ ka (Rth) 5 (9.372 X10 Q R b) (9x10976.633x108) 6.372)+(59) DV =41678. 19 volt - o