Question

An electron is accelerated horizontally from rest by a potential difference of 2100 V . It then passes between two horizontal plates 6.5 cm long and 1.3 cm apart that have a potential difference of 200 V

At what angle θ will the electron be traveling after it passes between the plates?

Express your answer to two significant figures and include the appropriate units.

Answer 1

Method:1

V = 200 V for deflection
V = 2100 V for acceleration

W = eV = 1.6*10^-19*2100 = 3.36*10^-16 J

delta KE = KEf - KEi = KEf = (1/2)*m*Vf^2
W=deltaKE

1/2*m*Vf^2 = 3.36*10^-16 J
solve for Vf
Vf^2 = 2*3.36*10^-16/m = 6.72*10^-16/m
Vf = root(6.72*10^-16/9.11*10^-31)
Vf = 2.7*10^7 m/s

constant speed through the plates so takes
the plate length 6.5 cm
Vf = x/t solve for t
t = x/Vf = 0.065/2.7*10^7
t = 2.41*10^-9 seconds
this is the time it spends in the plates so that will be used to find the deflection

When it passes through the plates..
F = Ee = (V/d)*e = eV/d
d = 1.3 cm = 0.013 m
a = F/m = eV/(m*d)

Vy = a*t
= eV/(m*d)*(2.41*10^-9)
= (1.6*10^-19*200/(9.11*10^-31*0.013))*( 2.41*10^-9)
= 6.5* 10^6 m/s

The angle is arctan(Vy/Vx) = arctan((6.5* 10^6 m/s)/( 2.7*10^7 m/s)) = 13.54 degs..

ΔE = 0

-½ m Vo² + V q = 0

-½ (9.11 * 10^(-31)) Vo² + 2100 * 1.6 * 10^(-19) = 0

Vo = 2.7* 10^7 m/s



Method :2
a = qE/m

a = qV/(dm)

a = (1.6 * 10^(-19))(200)/(1.3 * 10^(-2) * 9.11 * 10^(-31))

a = 2.7* 10^15 m/s²


t = x/Vo

t = 0.065/(2.7 * 10^7)

Vy = a t

Vy = 2.7 * 10^15 *(0.065/(2.7 * 10^7))

Vy = 6.5 * 10^6 m/s


tan θ = Vy/vo

θ =tan^-1(6.5* 10^6)/(2.7 * 10^7) =13.54 deg