Question

An electron is fired at a speed  v₀ = 5.3 ✕ 10⁶ m/s and at an angle  θ₀ = −45° between two parallel conducting plates that are  D = 2.5 mm apart, as in the figure below. The voltage difference between the plates is ΔV = 105 V. (a) Determine how close, d, the electron will get to the bottom plate.  mm (b) Determine where the electron will strike the top plate. mm

Path of the electron 0

Answer 1

Given that,

vo = 5.3*10^6 m/s

$\theta$ = 45 deg

d = 2.5 mm

$\Delta$ V = 105 V

(a)

vertical component of velocity,

vy = vo*sin45

vy = 5.3*10^6*sin45 = 3.76*10^6 m/s

horizontal component, vx = vo*sin45

vx = 3.76*10^6 m/s

Electric field E = $\Delta$ V / d

E = 105 / 2.5*10^(-3) = 42*10^3 N/C

acceleration of electron,

a = qE / m

a = 1.6*10^(-19)*42*10^3 / 9.1*10^(-31) = 7.38*10^(15) m/s^2

From kinematic equation,

v^2 - u^2 = 2*a*s

s = (3.76*10^6)^2 / 2*7.38*10^15

s = 0.962 mm

d = D/2 - s

d = (2.5 / 2) - 0.962

d = 0.287 mm

(b)

Let, Time taken by electron to reach the top plate = t

From kinematic equation,

-D/2 = vy*t - (1/2)at^2

-1.25*10^(-3) = 3.76*10^6*t - (1/2)*7.38*10^15*t^2

(3.69*10^15)t^2 - (3.76*10^6)t - 1.25*10^(-3) = 0

by solving the quadratic equation,

t = 1.124*10^(-9) s

Distance where electron will strike the top plate,

x = vx*t

x = 3.76*10^6*1.124*10^(-9)

x = 4.22 mm