A positively charged particle initially at rest on the ground accelerates upward to 160 m/s in 2.90 s . The particle has a charge-to-mass ratio of 0.100 C/kg and the electric field in this region is constant and uniform.
What are the magnitude and direction of the electric field?Express your answer to two significant digits and include the appropriate units. Enter positive value if the electric field is upward and negative value if the electric field is downward

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moya1316 moya1316 · Answer 1 · 2021-07-13T03:25:54+02:00

Answer:

E = 5.5 10² N /C

Explanation:

For this exercise let's use Newton's second law

F = m a

the force is electric

F = q E

we substitute

q E = m a

E = [tex]\frac{m}{q} \ a[/tex] (1)

the acceleration can be found with kinematics,

v = v₀ + a t

as the particle starts from rest v₀ = 0

a = v / t

a = 160 / 2.90

a = 55.17 m / s²

the relationship

q / m = 0.100 C / kg

m / q = 10 kg / C

we use equation 1

E = 10 55.17

E = 551.7 N / C

E = 5.517 10² N / C

ask for the result with two significant figures

E = 5.5 10² N /C