Answer:
Step-by-step explanation:
Given that
E=120N/C
The radius of earth is R=6.4×10^6m
The electric field on the surface of the charged sphere depends upon the charge on the sphere and the radius of the sphere. In the case the sphere is the earth with radius R
Then,
Electric field is given as
E=kQ/R²
k=9×10^9Nm²/C²
Then, make the charge Q subject of the formula
Q=ER²/k
Q=120 × (6.4×10^6)² / (9×10^9)
Q=4.92×10^15 / 9×10^9
Q=5.461 × 10^6 C
That is a lot of charge