Respuesta :
Explanation:
It is given that,
Electric field strength, E = 20,000 N/C
Spacing between parallel-plate capacitor, d = 1 mm = 0.001 m
Initial velocity of electron, u = 0
Let v is the electronās speed when it reaches the positive plate. The force acting on the electron is :
[tex]F=qE[/tex]
Also, [tex]ma=qE[/tex]
[tex]a=\dfrac{qE}{m}[/tex]
[tex]a=\dfrac{1.6\times 10^{-19}\times 20,000}{9.1\times 10^{-31}}[/tex]
[tex]a=3.51\times 10^{15}\ m/s^2[/tex]
Using third equation of motion as :
[tex]v^2-u^2=2ad[/tex]
[tex]v=\sqrt{2ad}[/tex]
[tex]v=\sqrt{2\times 3.51\times 10^{15}\times 0.001}[/tex]
v = 2649528.2599 m/s
or
[tex]v=2.64\times 10^6\ m/s[/tex]
So, the velocity of the electron when it reaches the positive plate is [tex]2.64\times 10^6\ m/s[/tex]. Hence, this is the required solution.
Answer:
2.65 x 10ā¶ m/s
Explanation:
E = magnitude of electric field = 20,000 N/C
d = spacing between the plates of the capacitor = 1.0 mm = 0.001 m
ĪV = Potential difference between the plates = ?
Potential difference between the plates is given as
ĪV = E d
ĪV = (20000) (0.001)
ĪV = 20 Volts
q = magnitude of charge on electron = 1.6 x 10ā»Ā¹ā¹ C
m = mass of electron = 9.1 x 10ā»Ā³Ā¹ kg
v = speed of the electron as it reach the positive plate
using the conservation of energy
Kinetic energy gained = electric potential energy lost
(0.5) m vĀ² = q ĪV
(0.5) (9.1 x 10ā»Ā³Ā¹) vĀ² = (1.6 x 10ā»Ā¹ā¹) (20)
v = 2.65 x 10ā¶ m/s
