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

Step 1: Convert units and find the electric field.

Length of the plates, L = 10 cm = 0.1 m.

Horizontal velocity, \(v_x = 10^6\) m/s.

Electric field, E = 9.1 V/cm = 910 V/m.

Mass of electron, m = \(9.1 \times 10^{-31}\) kg.

Charge of electron, q = \(1.6 \times 10^{-19}\) C.

Step 2: Calculate the acceleration.

The force on the electron due to the electric field is F = qE.

The acceleration of the electron is a = F/m = qE/m.

\(a = \frac{(1.6 \times 10^{-19} \text{ C})(910 \text{ V/m})}{9.1 \times 10^{-31} \text{ kg}} = \frac{1.6 \times 910}{9.1} \times 10^{12} \text{ m/s}^2 

= 1.6 \times 10^{14} \text{ m/s}^2\)

Step 3: Calculate the time spent in the electric field.

The time taken by the electron to travel the length of the plates is:

\(t = \frac{L}{v_x} = \frac{0.1 \text{ m}}{10^6 \text{ m/s}} = 10^{-7} \text{ s}\)

Step 4: Calculate the vertical velocity.

The vertical velocity of the electron is given by:

\(v_y = at = (1.6 \times 10^{14} \text{ m/s}^2)(10^{-7} \text{ s}) = 1.6 \times 10^7 \text{ m/s} = 16 \times 10^6 \text{ m/s}\)