Calculating Voltage from Electric Field between Parallel Plates

[mmoadi]

Homework Statement

Electrons, which evaporated from the filament accelerated in homogeneous electric field that was created by using two parallel metal plates, connected to the voltage
V = 1 kV. The resulting stream of electrons is deviated from the initial direction for a
φ = 15 ° in electrical field between two parallel flat surfaces. The length of the plate, in the direction of the initial movement of electrons, is s = 10 cm and the distance between the plates is d = 2 cm. What is the voltage V' between the two plates?

Homework Equations

E= V/d
ΔV= V’ – V
ΔV= -E*d*cos φ

The Attempt at a Solution

E= V/d
E= 1000 v/ 0.02 m
E= 50000 V/m

ΔV= -E*d*cos φ
V’ – V= -E*d*cos φ
V’ – 1000 V= -50000 V/m*0.02m*cos15º
V’= -965.9 V – 1000 V
V’= 34.1 V

Are my calculations correct?

Thank you for helping!

 

[nasu]

You have two sets of plates. The first one is used to accelerate the electrons. The 1kV is applied between the plates in this first set.
The second set is used to deviate the electron beam. You need to calculate the volatge between these two.
The first voltage (1KV) is given so that you can calculate the initial speed of the electrons before they enter the second set of plates.

 

[mmoadi]

<sub>RECALCULATION:

F= m*a
E= U/d
W= ΔKE

- First I solved for vector velocity in the w-direction:

W= F*d= q*U
W= KE’ – KE0 ≈ 0
q*U= m*v(x)²/ 2
v(x)²= 2q*U/ m
v(x)= sqrt[2q*U/ m]

- Now I solved for acceleration in the y-direction:

a(y)= F(y)/ m
F(y)= E(y)*q
F(y)= (U’/d)*q
a(y)= (U’*q)/ (d*m)

- Now velocity in the y-direction:

v(y)= a(y)*t
s= v(x)*t → t= s/ v(x)
v(y)= (U’*q/ d*m)/ s/v(x)
v(y)= U’*q*s/ d*m*v(x)

tan φ= v(y)/ v(x)
tan φ= (U’*q*s/ d*m*v(x))/ v(x)
tan φ= (U’*q*s)/ (d*m*v(x)²)
tan φ= (U’*q*s*m)/ (d*m*2q*U)
tan φ= (U’*s)/ (2d*U)
U’= (tan φ*d*2U)/ s
U’= 107.2 V

Are my calculations correct?
Thank you for helping!</sub>

 

[nasu]

It looks OK to me.

 

[mmoadi]

_{Thank you for helping!}