Suppose that the electric field in the Earth's atmosphere is E = 131 N/C, pointing downward. Determine the electric charge (in C) in the Earth. (Assume the electric field is measured at the surface of the Earth. Include the sign of the value in your answer.) Hint

the electric charge in the Earth is -5.91 × 10⁵ C

Explanation:

Given the data in the question;

Electric field E = 131 N/C

we know that; radius of the earth r = 6,371 km = 6371000 m

and Coulomb's constant k = 8.99 × 10⁹ Nm²/c²

Now, using the following formula to calculate the charge;

E = k × Q/r²

we make Q the subject of the formula

Q = Er² / k

so we substitute

Q = [ 131 N/C × ( 6371000 m )² ] / 8.99 × 10⁹ Nm²/c²

Q = [ 5.317242971 × 10¹⁵ ] / [ 8.99 × 10⁹ ]

Q = 5.91 × 10⁵ C

Since the electric field pointing downward

Q = -5.91 × 10⁵ C

Therefore,  the electric charge in the Earth is -5.91 × 10⁵ C

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer

c) power lifting, will be your answer.

,

deku ❤