(a) Find the magnitude of the electric field in the capacitor. (b) Find the spee

Question

(a) Find the magnitude of the electric field in the capacitor.

(b) Find the speed of the electron when it exits the capacitor.

.The figure below shows an electron entering a parallel-plate capacitor with a speed of v = 5.30 .The figure below shows an electron entering a par (a) Find the magnitude of the electric field in the capacitor. (b) Find the speed of the electron when it exits the capacitor. .The figure below shows an electron entering a par 106 m/s. The electric field of the capacitor has deflected the electron downward by a distance of d = 0.634 cm at the point where the electron exits the capacitor.

Explanation / Answer

PART-A

The time the electron is between the plates = L/vx = 0.0225/5.30x10^6 = 4.245x10^-9s

Now the acceleration of the electron is

F/m = E*q/m

y = 1/2*a*t^2

Y= 1/2*E*q/m*t^2

E = 2*y*m/(q*t^2)

E= 2*0.634x10^-2*9.11x10^-31/(1.60x10^-19*(4.24528x10^-9)^2

E =4005.944 N/C

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PART-B

vx doesn't change so vx = 5.3x10^6

vy = a*t

Vy= Eq/m*t

Vy= 4005.944*1.60x10^-19 *(4.24528x10^-9)/ 9.11x10^-31

Vy= 2.9868x10^6m/s

speed = sqrt((5.3x10^6)^2 + (3.24x10^6)^2)

speed= 6.083x10^6m/s

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