An electron is released from rest in a weak electric field given by vector E = - 1.30 10-10 N/C ĵ. After the electron has traveled a vertical distance of 1.6 µm, what is its speed? (Do not neglect the gravitational force on the electron.)

v = 6.45 10⁻³ m / s

Explanation:

Electric force is

F = q E

Where q is the charge and E is the electric field

Let's use Newton's second law to find acceleration

F - W = m a

a = F / m - g

a = q / m E g

Let's calculate

a = - 1.6 10⁻¹⁹ / 9.1 10⁻³¹ (-1.30 10⁻¹⁰) - 9.8

a = 0.228 10² - 9.8

a = 13.0 m / s²

Now we can use kinematics, knowing that the resting parts electrons

v² = v₀² + 2 a y

v = √ (0 + 2 13.0 1.6 10⁻⁶)

v = 6.45 10⁻³ m / s