The question reads: Electric Field of the Earth. The earth has anet electric cha

Question

The question reads: Electric Field of the Earth. The earth has anet electric charge that causes a field at points near its surfaceequal to 150 N/C and directed in toward the center of the earth.. A- (a) What magnitude and sign of charge would a 67 kg human have to acquire to overcome his or herweight by the force exerted by the earth's electricfield?
The answer to A is -4.38 C (b) What would be the force of repulsion between two peopleeach with the charge calculated in part (a) and separated by adistance of 43 m? This is where im stuck help would be appreciated please showformulas***
The question reads: Electric Field of the Earth. The earth has anet electric charge that causes a field at points near its surfaceequal to 150 N/C and directed in toward the center of the earth.. A- (a) What magnitude and sign of charge would a 67 kg human have to acquire to overcome his or herweight by the force exerted by the earth's electricfield?
The answer to A is -4.38 C (b) What would be the force of repulsion between two peopleeach with the charge calculated in part (a) and separated by adistance of 43 m? This is where im stuck help would be appreciated please showformulas***

Explanation / Answer

(a). Electric field E = 150 N / C we know E = KQ / R ^ 2 where Q = charge           K =Coulomb's constant = 8.99 * 10 ^ 9 N m ^ 2/ C ^ 2            R= Radius of earth = 6.38 * 10 ^ 6 m from above charge Q = E R ^ 2 / K                                  = 679.16 * 10 ^3 C (a). Mass of human m = 67 kg Eq = mg from this Magnitude of the charge q = mg / E                                                        = 4.3777 C                                                        ~ 4.38 C Electric field is along the center of the earth. So, force must be act away from the center of the earth. Force is opposite to the electric field for negativecharge So, charge q = -4.38 C (b). separation r= 43 m the force of repulsion between two people each with thecharge F ' = Kq ^ 2/ r ^ 2                                                                                                       = ( 8.99*10^9)(4.38)^2 / 43^2                                                                                                       = 0.0931 * 10^ 9 N

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