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:
[tex]v_i=\sqrt{v_f^2-\frac{2qV}{m}}[/tex]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:
[tex]v_i=\sqrt{(19424119.514)^2-\frac{2(1.6\times10^{-19})(549707.592)(1.396\times10^{-3})}{9.11\times10^{-31}}}[/tex]Solving further, we have:
[tex]\begin{gathered} v_i=\sqrt{1.07740573×10^{14}} \\ \\ v_i=10379815.66\text{ m/s} \end{gathered}[/tex]Therefore, the speed as it left the negative plate is 10379815.66 m/s.
ANSWER:
10,379,815.66 m/s
