Let us consider that, an electron of mass ‘m’ moves with initial horizontal velocity ‘vx‘. While the electron enters the electric field between the parallel plates, it is attracted towards the upper positive plate.

The electrostatic force with which electron is attracted upwards is given by,
Fe = eE
or, Fe = eVd …. (i)

Let ‘a’ be the acceleration produced in an electron while it moves upwards.

Then, from Newton’s second law of motion,

F = ma …. (ii)

From equations (i) and (ii), we get,

ma = eVd
or, a = eVmd …. (iii)

Let at any time ‘t’, electron is at a point ‘P’ and X and Y be its horizontal and vertical displacement.

In horizontal displacement,

X = ut + 12at2 (Horizontal velocity remains constant.)

or, X = vxt
or, t = Xvx …. (iv)

Again, in vertical displacement,

Y = ut + 12at2

Since the initial velocity is zero,

Y = 12at2
Now, substituting the values of ‘a’ and ‘t’ in the above equation,
Y = eVX22mdvx2

Here, Y = kX2
or, X2 = 2mdvxYeV
∴ X2 = bY

This is the equation of parabola.

Hence, the path followed by electron in electric field is parabolic. Let ‘D’ be the length of the plate and ‘T’ be the time taken to travel the distance equal to the length of the plate.

T = DVx …. (v)

The electron moves in a straight line after leaving the plates. Let ‘θ’ be the angle of deflection with respect to vx after leaving the plate by an electron.
The vertical component of velocity vy is given by,
vy = u + aT
Since u = 0,
vy = eVDmdvx

Now, the angle of deflection is given by,

tanθ = Vy/vx

or, θ = tan-1(eVD/mdvx2)

Again, the resultant velocity is given by, vR = √(vx2 + vy2)

= √ (vx2 + e2V2D2/m2d2vx2)

## Increase in Kinetic Energy

Initial Kinetic Energy (Ei) = 1/2mvx2
Final Kinetic Energy (Ef) = 1/2mvR2

= 1/2mvx2 + 1/2mvy2

Hence, the increase in kinetic energy is given by,

E = Ef – Ei
= 1/2mvx2 + 1/2mvy21/2mvx2
= 1/2mvy2
= 1/2×me2V2D2/m2d2vx2
= 1/2×e2V2D2/md2vx2