Educate Sansar

Let us consider that, an electron of mass ‘m’ moves with initial horizontal velocity ‘v_x‘. 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,
F_e = eE
or, F_e = ^eV⁄_d …. (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 = ^eV⁄_d
or, a = ^eV⁄_md …. (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 + ¹⁄₂at² (Horizontal velocity remains constant.)

or, X = v_xt
or, t = ^X⁄_vx …. (iv)

Again, in vertical displacement,

Y = ut + ¹⁄₂at²

Since the initial velocity is zero,

Y = ¹⁄₂at²
Now, substituting the values of ‘a’ and ‘t’ in the above equation,
Y = ^{eVX2⁄_2mdvx2}

Here, Y = kX²
or, X² = ^2mdv_xY⁄_eV
∴ 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 = ^D⁄_Vx …. (v)

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

Now, the angle of deflection is given by,

tanθ = ^V_y/_vx

or, θ = tan^-1(^eVD/_mdvx²)

Again, the resultant velocity is given by, v_R = √(v_x² + v_y²)

= √ (v_x² + ^{e2V2D2/_m2d2vx2})

Increase in Kinetic Energy

Initial Kinetic Energy (E_i) = ¹/₂mv_x²
Final Kinetic Energy (E_f) = ¹/₂mv_R²

= ¹/₂mv_x² + ¹/₂mv_y²

Hence, the increase in kinetic energy is given by,

E = E_f – E_i
= ¹/₂mv_x² + ¹/₂mv_y² – ¹/₂mv_x²
= ¹/₂mv_y²
= ¹/₂×^me2V2D2/_m²d²vx²
= ¹/₂×^e2V2D2/_md²vx²