Answered

Two large conducting parallel plates A and B are separated by 2.4 m. A uniform field of 1500 V/m, in the positive x-direction, is produced by charges on the plates. The center plane at x = 0.00 m is an equipotential surface on which V = 0. An electron is projected from x = 0.0 m. with an initial kinetic energy K = 300 eV, in the positive x-direction.
a. Determine the initial velocity of the electron at x = 0.0 m.
b. What is the potential difference between the plates?
c. At a certain point the electron stops momentarily and then reverses direction. What is the electric potential at this point?
d. What is the kinetic energy of the electron when it reaches plate A?

Answer :

oodaramola

Answer:

a. 1.027 x 10^7 m/s b. 3600 V c. 0 V and d. 1.08 MeV

Explanation:

a. KE =1/2 (MV^2) where the M is mass of electron

b. E = V/d

c. V= 0 V (momentarily the pd changes to zero)

d KE= 300*3600 v = 1.08 MeV

jayilych4real

The initial velocity of the electron at x = 0.0 m is 0m/s

Calculation and Parameters:

Given that there is:
A distance of 2.4m between A and B,

  • Electric field: 1500v/m
  • Distance: 2.4m
  • Kinetic Energy: 3000v
  • Initial Velocity: ?

a. Initial velocity, u= v-at

at v = 0 in a uniform field

Initial velocity= 0m/s.

b. To find the potential difference, V

V= Ed

Where:

E= Electric field

d= distance between plates

V= 1500 x 2.4

= 3600V.

c.V= kq/r

r= 2.4/2

r = 1.2m

q= [tex]1.602 * 10^-^1^9[/tex]

k= [tex]9.0 * 10^9Nm^2c^-^2[/tex]

Hence,

V= [tex]9.0 * 10^9 * 1.602 * 10^-^1^9/1.2[/tex]

V= [tex]1.20 * 10^-^1^9J/C[/tex]

d. Potential differemce:= 3600V

q= [tex]1.6 * 10^-^1^9C[/tex]

The kinetic energy, KE = q△V
= [tex]1.6 * 10^-^1^9 * 3600[/tex]

= [tex]5.76 * 10^-^1^6J[/tex]

Read more about velocity here:

https://brainly.com/question/6504879

Other Questions