Given:
• Distance between plates = 1.396 mm
,
• Diameter = 2 cm
,
• Electric field = 549,707.592 V/m
,
• Speed = 19,424,119.514 m/s
Let's find the electron's speed as it left the negative plate.
Apply the formula:

Where:
vf is the final speed = 19,424,119.514 m/s
q = 1.60 x 10⁻¹⁹
V = E x d = 549,707.592 V/m x 1.396 mm
m is the mass of electron = 9.11 x 10⁻³¹ kg
Thus, we have:

Solving further, we have:

Therefore, the speed as it left the negative plate is 10379815.66 m/s.
ANSWER:
10,379,815.66 m/s