Question

An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.43 m in the first 3.24 μs after it is released. What is the magnitude of the electric field

Answer 1

Here ,

let the magnitude of electric field is E

let the acceleration is a

Using second equation of motion

d = 0.50 a * t^2

4.43 = 0.50 * a * (3.24 *10^-6)^2

a = 8.44 *10^11 m/s^2

Using second law of motion

m *a = e * E

9.11 *10^-31 * 8.44 *10^11 = E * 1.602 *10^-19

E = 4.8 N/C

the electric field magnitude is 4.8 N/C