Answer:
Explanation:
Force on charge in electric field = q E , q is charge and E is electric field
= 1.6 x 10⁻¹⁹ x 12000
= 19200 x 10⁻¹⁹ N
acceleration = force / mass
= 19200 x 10⁻¹⁹ / 9.1 x 10⁻³¹
a = 2109.9 x 10¹² m /s²
initial velocity of electron u = 2 x 10⁷ m /s
v² = u² + 2as ( s is distance between plate , u is initial velocity )
v² = ( 2 x 10⁷)² + 2 x 2109.9 x 10¹² x 40 x 10⁻³
= 4 x 10¹⁴ + 1.68 x 10¹⁴
= 5.68 x 10¹⁴
v = 2.38 x 10⁷ m /s
The final velocity of the electron is required.
The final velocity of the electron is 15231546.21 m/s.
E = Electric field = 12000 N/C
s = Distance between plates = 40 mm
u = Initial velocity = [tex]2\times 10^7\ \text{m/s}[/tex]
q = Charge of electron = [tex]-1.6\times 10^{-19}\ \text{C}[/tex]
m = Mass of electron = [tex]9.11\times 10^{-31}\ \text{kg}[/tex]
v = Final velocity
a = Acceleration
The force balance of the system is given by
[tex]ma=qE\\\Rightarrow a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{-1.6\times 10^{-19}\times 12000}{9.11\times 10^{-31}}\\\Rightarrow a=-2.11\times 10^{15}\ \text{m/s}^2[/tex]
From the kinematic equations we have
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times (-2.1\times 10^{15})\times 40\times 10^{-3}+(2\times 10^7)^2}\\\Rightarrow v=15231546.21\ \text{m/s}[/tex]
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