Question

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 field 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 = \(1.6 \times 10^{-19}\) C):

• A. \(1 \times 10^6\) m/s
• B. 0
• C. \(16 \times 10^6\) m/s
• D. \(16 \times 10^4\) m/s

Hint:

Remember to convert all units to SI units before performing calculations. Also, the electron's motion inside the plates is similar to projectile motion, where horizontal velocity remains constant and vertical velocity changes due to acceleration.

Answer 1

The Correct Option is C

To solve for the vertical component of velocity of the electron, we can utilize the electric force acting on the electron and equate it to the mass times acceleration. The steps are as follows:

Step 1: Calculate the force on the electron

The electric force \( F \) on the electron is given by:

\( F = e \cdot E \)

Where \( e = 1.6 \times 10^{-19} \) C (charge of electron) and \( E = 9.1 \) V/cm, which is \( 910 \) V/m when converted to SI units.

\( F = 1.6 \times 10^{-19} \times 910 = 1.456 \times 10^{-16} \, \text{N} \)

Step 2: Calculate the acceleration of the electron

Using the formula \( F = m \cdot a \), where \( m = 9.1 \times 10^{-31} \) kg (mass of the electron),

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

Step 3: Determine the time electron spends between the plates

The horizontal velocity \( v_x \) is constant at \( 10^6 \) m/s. The time \( t \) to traverse the 10 cm (0.1 m) length of the plates is:

\( t = \frac{0.1}{10^6} = 10^{-7} \, \text{s} \)

Step 4: Calculate the vertical component of velocity

The vertical velocity \( v_y \) can be found using the equation:

\( v_y = a \cdot t \)

\( v_y = 1.6 \times 10^{14} \times 10^{-7} = 16 \times 10^6 \, \text{m/s} \)

Thus, the vertical component of velocity of the electron is \( 16 \times 10^6 \) m/s.

Answer 2

Given: 
Length of plates, \( L = 10\, \text{cm} = 0.1\, \text{m} \)
Horizontal velocity, \( v_x = 10^6\, \text{m/s} \)
Electric field, \( E = 9.1\, \text{V/cm} = 9.1 \times 10^2\, \text{V/m} \)
Charge of electron, \( e = 1.6 \times 10^{-19}\, \text{C} \)
Mass of electron, \( m = 9.1 \times 10^{-31}\, \text{kg} \)

Time spent between plates:
\[ t = \frac{L}{v_x} = \frac{0.1}{10^6} = 1 \times 10^{-7}\, \text{s} \]

Force on electron:
\[ F = eE = (1.6 \times 10^{-19})(9.1 \times 10^2) = 1.456 \times 10^{-16}\, \text{N} \]

Acceleration:
\[ a = \frac{F}{m} = \frac{1.456 \times 10^{-16}}{9.1 \times 10^{-31}} = 1.6 \times 10^{14}\, \text{m/s}^2 \]

Vertical velocity:
\[ v_y = a t = (1.6 \times 10^{14})(1 \times 10^{-7}) = 1.6 \times 10^7\, \text{m/s} \]

∴ The vertical component of velocity of the electron is:
\[ \boxed{v_y = 1.6 \times 10^7\, \text{m/s}} \]

Hence, correct option: Option 3 — \(16 \times 10^6\, \text{m/s}\)