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

The vertical velocity component of an electron can be determined by applying the electric force to the electron and equating it to its mass multiplied by acceleration. The process is outlined below:

Step 1: Calculate the force on the electron

The electric force \( F \) acting on the electron is calculated using:

\( F = e \cdot E \)

Here, \( e = 1.6 \times 10^{-19} \) C represents the charge of the electron, and \( E = 9.1 \) V/cm, which converts to \( 910 \) V/m in 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 equation \( F = m \cdot a \), with \( m = 9.1 \times 10^{-31} \) kg as the electron's mass:

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

Step 3: Determine the time the electron spends between the plates

The electron's horizontal velocity \( v_x \) is constant at \( 10^6 \) m/s. The time \( t \) taken to travel 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 \( v_y \) is calculated using the formula:

\( v_y = a \cdot t \)

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

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