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

- The electric force on a charged particle in a uniform electric field is: \[ F = qE \]
- Given:
- \( q = 20 \times 10^{-6} \, C \)
- \( E = 2 \times 10^5 \, N/C \)
- Force: \[ F = (20 \times 10^{-6})(2 \times 10^5) \] \[ F = 4 \, N \]
- Using Newton’s second law: \[ a = \frac{F}{m} \]
- Given mass:
- \( m = 5 \times 10^{-3} \, kg \)
- Acceleration: \[ a = \frac{4}{5 \times 10^{-3}} = 800 \, m/s^2 \]
- Now using kinematic equation (retardation case): \[ v^2 = u^2 - 2as \]
- Given:
- \( u = 16 \, m/s \), \( v = 0 \)
- Substitute: \[ 0 = 16^2 - 2(800)s \]
- Solve: \[ 256 = 1600s \] \[ s = \frac{256}{1600} = 0.16 \, m \]
- Convert to cm: \[ s = 16 \, cm \]
- Hence, the stopping distance is: 16 cm