An electron is made to enter symmetrically between two parallel and equally but oppositely charged metal plates, each of 10 cm length. The electron emerges out of the field region with a horizontal component of velocity 10⁶ m/s. If the magnitude of the electric between the plates is 9.1 V/cm, then the vertical component of velocity of electron is 
$($mass of electron $=9.1\times {10}^{-31}$kg and charge of electron $\left.=1.6\times {10}^{-19}\mathrm{C}\right)$

To calculate the vertical component of the velocity of the electron when it exits the electric field, we use the following steps:

Step 1: Force acting on the electron

The force $F$ on the electron due to the electric field $E$ is given by:

$$F = qE$$

where:

• $q = 1.6 \times 10^{-19} \, \text{C}$ (charge of the electron),
• $E = 9.1 \, \text{V/cm} = 9.1 \times 10^2 \, \text{V/m}$.

$$F = (1.6 \times 10^{-19}) \cdot (9.1 \times 10^2) = 1.456 \times 10^{-16} \, \text{N}.$$

Step 2: Acceleration of the electron

The vertical acceleration $a$ of the electron is given by Newton's second law:

$$a = \frac{F}{m}$$

where:

• $m = 9.1 \times 10^{-31} \, \text{kg}$ (mass of the electron).

$$a = \frac{1.456 \times 10^{-16}}{9.1 \times 10^{-31}} = 1.6 \times 10^{14} \, \text{m/s}^2.$$

Step 3: Time spent in the electric field

The time $t$ the electron spends in the field is determined by the horizontal motion. Since the electron moves with a horizontal velocity $v_x = 10^6 \, \text{m/s}$ and the length of the plates is $L = 10 \, \text{cm} = 0.1 \, \text{m}$, the time is:

$$t = \frac{L}{v_x}$$

 $$t = \frac{0.1}{10^6} = 10^{-7} \, \text{s}.$$ 

Step 4: Vertical velocity

The vertical velocity $v_y$ acquired by the electron due to acceleration in the electric field is:

$$v_y = a \cdot t$$

Substitute $a = 1.6 \times 10^{14} \, \text{m/s}^2$ and $t = 10^{-7} \, \text{s}$:

$v_y=(1.6\times {10}^{14})\cdot \left({10}^{-7}\right)=16\times {10}^6\text{m/s}\mathrm{.}$