Question

If a particle of mass 10 mg and charge 2 µC at rest is subjected to a uniform electric field of potential difference 160 V, then the velocity acquired by the particle is

• A. 9 ms⁻¹
• B. 4 ms⁻¹
• C. 6 ms⁻¹
• D. 8 ms⁻¹

Hint:

Always convert all given quantities to their base SI units (mass to kg, charge to C, etc.) before substituting them into formulas. This prevents unit-related errors in the final calculation.

Answer 1

The Correct Option is D

Step 1: Energy Principle: Work done by the electric field equals the change in kinetic energy. \( W = qV = \frac{1}{2} m v^2 \) Where: \( q = 2 \, \mu\text{C} = 2 \times 10^{-6} \, \text{C} \) \( V = 160 \, \text{V} \) \( m = 10 \, \text{mg} = 10 \times 10^{-3} \, \text{g} = 10 \times 10^{-6} \, \text{kg} = 10^{-5} \, \text{kg} \)
Step 2: Calculation: \[ \frac{1}{2} (10^{-5}) v^2 = (2 \times 10^{-6}) (160) \] \[ v^2 = \frac{2 \times 2 \times 10^{-6} \times 160}{10^{-5}} \] \[ v^2 = \frac{4 \times 160 \times 10^{-6}}{10^{-5}} \] \[ v^2 = \frac{640 \times 10^{-6}}{10^{-5}} = 640 \times 10^{-1} = 64 \] \[ v = \sqrt{64} = 8 \, \text{ms}^{-1} \]