Answer:
2903446.22825 m/s
Explanation:
m = Mass of electron = [tex]9.11\times 10^{-31}\ kg[/tex]
q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]
E = Electric field = [tex]1.6\times 10^{4}\ N/C[/tex]
t = Time taken
u = Initial velocity = 0
v = Final velocity
s = Displacement = 1.5 mm
Acceleration is given by
[tex]a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times 1.6\times 10^{4}}{9.11\times 10^{-31}}\\\Rightarrow a=2.81\times 10^{15}\ m/s^2[/tex]
Displacement is given by
[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0+\dfrac{1}{2}at^2\\\Rightarrow t=\sqrt{\dfrac{2s}{a}}\\\Rightarrow t=\sqrt{\dfrac{2\times 1.5\times 10^{-3}}{2.81\times 10^{15}}}\\\Rightarrow t=1.0332548855\times 10^{-9}\ s[/tex]
Velocity is given by
[tex]v=u+at\\\Rightarrow v=0+2.81\times 10^{15}\times 1.0332548855\times 10^{-9}\\\Rightarrow v=2903446.22825\ m/s[/tex]
The velocity of the electron is 2903446.22825 m/s
The speed of the electron when it reaches the positive plate is 1.11 × 10⁵ m/s
The electric field between the plates is E = 1.6 × 10⁴ N/C.
the distance between the plates is r = 1.5mm = 1.5 × 10⁻³ m
So, the potential difference is:
V = E×r
V = 1.6 × 10⁴ × 1.5 × 10⁻³ Volts
V = 24V
The force acting on the electron is given by:
ma = eV
where m is the mass of the electron = 9.31×10⁻³¹ kg
a is the acceleration
e is the charge on electron = 1.6×10⁻¹⁹ C
a = eV/m
a = (1.6×10⁻¹⁹)×24 / (9.31×10⁻³¹)
a = 4.12×10¹² m/s²
The initial speed of the electron is u = 0
Let the final speed be v and the distance traveled is r = 1.5 × 10⁻³ m
applying the third equation of motion we get:
v² = u² + 2ar
v² = 0 + 2×4.12×10¹²×1.5 × 10⁻³
v = 1.11 × 10⁵ m/s
Learn more about potential difference:
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