In fair weather over flat ground there is a downward electric field of 150 N/C.
ID: 1317211 • Letter: I
Question
In fair weather over flat ground there is a downward electric field of 150 N/C. Assume that the earth is a conducting sphere with a charge on its surface. if the electric field just outside is 150 N/C pointing radially inward, calculate the
a) total (excess) charge on the earth's surface
b) the charge per unit area.
At an altitude of 250 meters above the earth's surface the fields only 120 N/C. Assume that the charge is uniformly distributed, ie, the volume charge density is constant. Calculate
c) the total (excess) charge in the air and
d) the charge density of the atmosphere (charge per unit volume).
Solution:
E surface = electric field at earth's surface = 150 N/C pointing downward. This means charge on surface is negative.
E atm = electric field at altitude of 250 m = 120 N/C, also downward.
r = radius of earth = 6.37*10^6 m
h = altitude = 250 m
q surface = charge on surface
q atm = charge in air
q total = q surface + q atm
Explanation / Answer
a)
Earth radius, R = 6370 km = 6.38*10^6 m
Let Q is the charge on the surface of the earth.
so, E = k*Q/R^2
==> Q = E*R^2/k
= 150*(6.37*10^6)^2/(9*10^9)
= 6.763*10^5 C
Here Q is negaitve beacuse Eectric filed is raially inward.
Q = -6.763*10^5 C <<<<-----------Answer
b) charge per unit area, sugma = Q/A
= Q/(4*pi*R^2)
= -6.763*10^5/(4*pi*(6.37*10^6)^2)
= -1.326*10^-9 C/m^2 or -1.326 nC/m^2
c) at, r = R + h
= 6.37*10^6 + 250
= 6.37012*10^6
let q is the excess charge in the air.
E = k*(Q+q)/r^2
==> Q+q = E*r^2/k
= -120*(6.37012*10^6)^2/(9*10^9)
= -5.41*10^5 C
q = -5.41*10^5 - Q
= -5.41*10^5 - (-6.763*10^5)
= +1.353*10^5 C <<<<-----------Answer
d)
volume charge density of atmosphere, rho = q/V
V = 4/3*pi*((R+h)^3 - R^3))
= (4/3)*pi*((6.37012)^3 - 6.37^3))*10^18
= 6.116*10^16 m^3
rho = 1.353*10^5/(6.116*10^16)
= 2.212*10^-12 C/m^3 or 2.212 pc/m^3 <<<<-----------Answer