every time I work this problem I get different answers. can some show me how to

Question

every time I work this problem I get different answers. can some show me how to solve?

Problem: In fair weather over flat ground there is a downward electric field of 150 N/C. N/C pointing radially inward, calculate the Assume that the earth is a conducting sphere with a charge on its surface. If the electric field just outside is 150 a) b) Total (excess) charge on the earth's surface 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) d) The total (excess) charge in the air and The charge density of the atmosphere (charge per unit volume). 150 N/C pointing downward. This means charge on surface is Euface electric field at earth's surface negative Eatm-electric field at altitude of 250 m r = radius of earth = 6.37x106 m h = altitude = 250 m E.-8.85410-12 cyNm2 Qsurface charge on surface qatm = charge in air 120 N/C, also downward. otal-Qsurface + Qatrm

Explanation / Answer

radius of earth = r = 6.400*10^6 m

E = kQ/r^2

150 = 9*10^9*Q/(6.4*10^6)^2

Q =- 6.826667*10^5 C ..............answer(a)

charge per unit area = Q/(4*pi*r^2) = 6.83*10^(5) / (4*pi*(6.4*10^6)^2) = -1.33*10^(-9) C/m^2.......answer(b)

(c):

electric field 250 m above ground = Electric field due to earth + electric field due to charge in atmosphere

Electric field due to earth = kQ/r^2 = 9*10^9 * 6.826667*10^5 / (6400250^2) = 149.99 N/C (downwards)

Electric field due to charge in atmosphere = kQ/r^2 = 9*10^9*Q(atm)/(6400250^2) N/C (upwards)

electric field 250 m above ground = Electric field due to earth + electric field due to charge in atmosphere

120 = 149.99 - 9*10^9*Q(atm)/(6400250^2

Q(atm) = 1.3650*10^5 C

volume charge density of atm = Q(atm)/[(4/3)*pi*(6400250^3-6400000^3)] = + 1.0607*10^(-12) C/m^3

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