An electron is made to enters 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 10^6 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 : (1) 1 × 10^6 m/s (2) 0 (3) 16 × 10^6 m/s (4) 16 × 10^4 m/s

Key ideas and theory

1. Uniform electric field: Between parallel plates the electric field $E$ is uniform. A charged particle in this field experiences a constant force $F = qE$.
2. Acceleration: For an electron (charge $q = -e$) the magnitude of acceleration is
   $a = \frac{eE}{m_e}$
   where $m_e$ is the electron mass.
3. Kinematic independence: Horizontal and vertical motions are independent.
   • Horizontal: no electric force → constant speed $v_x$.
   • Vertical: constant acceleration $a$.
4. Time of flight inside the plates:
   $t = \frac{\text{length of plates}}{v_x}$
5. Vertical velocity after time $t$ (starting from rest vertically):
   $v_y = a\,t.$

Logical chain to solve

1. Convert the field unit from V cm⁻¹ to V m⁻¹ so SI units match.
2. Compute $a$ with known $e$ and $m_e$.
3. Find $t$ using the given horizontal component of velocity and plate length.
4. Plug $a$ and $t$ into $v_y = a t$ to get the required vertical component.

Imagine this setup like a toy car in a tall hallway

1. Horizontal motion: The electron is like a toy car that is already moving straight down the hallway at a speed of $10^6\,\text{m/s}$.
2. Vertical push: Two big metal plates on the walls create an invisible "wind" (the electric field) that blows up (or down) on the car. This wind has a strength of 9.1 V per centimetre.
3. Time under the wind: The hallway is only 10 cm long, so the car feels the wind only while it is between the plates.
4. Vertical speed: While the car zips horizontally, the wind keeps pushing it up (or down). The longer it stays, the faster it goes vertically.
5. Goal of the question: Find out how fast (vertically) the car is moving the moment it leaves the hallway.

Step-by-step calculation

1. Convert field unit
   $E = 9.1\,\text{V/cm} = 9.1 \times 100 = 910\,\text{V/m}$

2. Electron acceleration
   Electron charge $e = 1.6\times10^{-19}\,\text{C}$, mass $m_e = 9.1\times10^{-31}\,\text{kg}$.
   $a = \frac{eE}{m_e} = \frac{1.6\times10^{-19}\times910}{9.1\times10^{-31}}$ $\Rightarrow a \approx \frac{1.456\times10^{-16}}{9.1\times10^{-31}} \approx 1.6\times10^{14}\,\text{m/s}^2$

3. Time inside the plates
   Plate length $L = 10\,\text{cm} = 0.1\,\text{m}$; horizontal speed $v_x = 10^6\,\text{m/s}$.
   $t = \frac{L}{v_x} = \frac{0.1}{10^6} = 1\times10^{-7}\,\text{s}$

4. Vertical component of velocity
   Starting with zero vertical speed,
   $v_y = a t = \left(1.6\times10^{14}\right)\left(1\times10^{-7}\right) = 1.6\times10^{7}\,\text{m/s}$

5. Express in the requested form
   $v_y = 16\times10^{6}\,\text{m/s}$

Answer: Option (3) $16\times10^{6}\,\text{m/s}$