Motion of Electron in Electric Field

Let us consider an electron having mass ‘me’ and charge ‘e’ enters a uniform electric field E=V/d between two parallel conducing plates perpendicular to the electric field. Let ‘d’ be the distance between the plates and ‘D’ be the length of the each plates.

Motion of electron in electric field

As soon as the electron enters the electric field it experiences an electric force (eE) towards positive plate. The motion of electron can be split into two components X along horizontal and Y along vertical. Then, we  can get the following information.

  X motion    Y motion
      fx=0     fy = eE
     ax=0       ay = eE/ me
    ux = v          uy = 0
    sx = x            sy = y

Let at any time‘t’ the electron be at P(x,y) travelling horizontal distance x and vertical distance ‘y’ as shown in figure

Since velocity is constant in horizontal motion, acceleration is zero, Now:

$${y = \text{u}_xt + \frac{1}{2}at^2}$$
$${y = \text{u}_xt}$$
$${t = \frac{x}{v}}………(i)$$

In vertical motion, initial velocity (u) = 0,

$${y = \text{u}_yt + \frac{1}{2}at^2}$$

$$ y = \frac{\text{1}}{\text{2}} \frac{\text{eE}}{\text{m}_e} t^2$$

Putting the value of t from equation (i),

$$ y = \frac{\text{1}}{\text{2}} \frac{\text{eE}}{\text{m}_e} \frac{x^2}{v^2} $$
$$ \text{y} = \text{k}x^2 ……..(ii) $$ where,
$$k = \frac{\text{Ee}}{2m_ev^2}$$

The equation (ii) represent the equation of parabola.  Hence it is proved that the path of an electron in uniform electric field is parabolic in nature.

Time taken by electron to come out of electric field:

It is the time taken by a electron to travel horizontal distance.

Now, In horizontal motion:

$${y = \text{u}_xt + \frac{1}{2}at^2}$$

$$\text{D} = \text{Vt}$$

$$\text{t} = \frac{V}{D}$$

Vertical displacement (y): The vertical shift of the electron due to electric field is given by:
$${y = \text{u}_yt + \frac{1}{2}at^2}$$

$$\text{y} = 0 \times \text{t} +\frac{1}{2} \frac{\text{eE}}{\text{m}_e} t^2$$

$$\text{y} = \frac{1}{2} \frac{\text{eE}}{\text{m}_e} \frac{D^2}{V^2}$$

Final velocity

We have,

Final horizontal velocity vx = v

Final vertical velocity vy = uy + ayt

or, $${V_y = \frac{\text{eE}}{\text{m}_e} \frac{D}{V}}$$

$$ \therefore \text{Final Velocity}{(V_f)} = \sqrt{x^2 +y^2}$$

Angle at which the electron emerges out is given by:

$$\text{Direction} = \text{tan}(\beta) = \frac{v_y}{v_x} or, \beta = \text{Tan}^{-1}\frac{\text{eED}}{\text{m}_eV}$$

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