Show work : a particle with mass 2.00 μg and a charge of – 200 nc has a velocity of 3000 m/s in the x-direction. there is a constant electric field of 1000 n/c in the x-direction too. how far will the charge move before reversing direction?
if the charge was stationary, what is the magnitude and direction of the electric field that would keep the charge suspended in air?

 x =4.5 10⁴ m

Explanation:

To find the distance that the particle moves we must use the equations of motion in one dimension and to find the acceleration of the particle we will use Newton's second law

     m = 2.00 mg (1 g / 1000 ug) (1 Kg / 1000g) = 2.00 10-6 Kg

     q = -200 nc (1C / 10 9 nC) = -200 10-9 C

Let's calculate the acceleration

     F = ma

     F = q E

     a = qE / m

     a = -200 10⁻⁹ 1000 / 2.00 10⁻⁶

     a = 1 10² m / s²

Let's use kinematics to find the distance traveled before stopping, where it has zero speed (Vf = 0)

     Vf² = Vo² -2 a x

     0 = Vo² - 2 a x

     x = Vo² / 2a

     x = 3000²/ 2100

     x =4.5 10⁴ m

This is the distance the particule stop, after this distance in the field accelerates in the opposite direction of the initial

Second part

In this case Newton's second law is applied on the y axis

      F -W = 0

      F = w = mg

      E q = mg

      E = mg / q

      E = 2.00 10⁻⁶ 9.8 / 200 10⁻⁹

      E = 9.8 10⁵ C

The direction of the field is such that the force on the particle is up, as the particle has a negative charge, the field must be directed downwards F = qE = (-q) E

grav forces affect all objects magnetic forces affect magnetic objects and electric currents

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