Question

An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.65 m in the first 1.90 µs after it is released.

(a) What are the magnitude and direction of the electric field? magnitude N/C direction

(b) Are we justified in ignoring the effects of gravity?

Yes No

Answer 1

An electron is released from rest in a uniform electric field.

Initial velocity,

v₀ = 0

Acceleration,

a = 2d /t^2
F/m=2d/t^2
qE/m=2d/t^2

hence,

E =2md / qt^2


E =2*9.1*10^-31 *4.65/1.6022*10^-19*[1.9*10^-6]^2

(a)

E = 19.65 N/C
Direction of the electric field E is downward since eletron is moving upward.

(b)

a = 2*d/t^2 = 2*4.65/(1.9*10^-6)^2 >>>> g

Hence, Yes, it alright to ignore the gravity.