Question

A charged particle of mass 5 g and charge 20 μC is thrown with a velocity of 16 m/s\(^-1\) in a direction opposite to the direction of a uniform electric field of \( 2 \times 10^5 \) N/C. The distance travelled by the particle before coming to rest is:

• A. \( 24 \text{ cm} \)
• B. \( 12 \text{ cm} \)
• C. \( 16 \text{ cm} \)
• D. \( 20 \text{ cm} \)

Hint:

- The motion of a charged particle in a uniform electric field is governed by Newton’s second law.
- The kinematic equation \( v^2 = u^2 - 2as \) helps determine the stopping distance.
- Acceleration is positive if force and motion are in the same direction, negative otherwise.

Answer 1

The Correct Option is C

Step 1: Identifying the Forces Acting on the Particle
- The charged particle experiences a force \( F = qE \) due to the electric field: \[ F = (20 \times 10^{-6} \text{C}) \times (2 \times 10^5 \text{ N/C}) \] \[ F = 4 \text{ N} \] Step 2: Calculating Acceleration using Newton’s Second Law
- Using \( F = ma \), the acceleration is: \[ a = \frac{F}{m} = \frac{4}{5 \times 10^{-3}} \] \[ a = 800 \text{ m/s}^2 \] Step 3: Applying Kinematic Equation
- Using the equation \( v^2 = u^2 - 2as \) (since motion is opposite to acceleration): \[ 0 = (16)^2 - 2(800)s \] \[ 256 = 1600s \] \[ s = \frac{256}{1600} = 0.16 \text{ m} = 16 \text{ cm} \]