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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.)

Respuesta :

Answer:

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

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