The magnitude of the force experienced by a charge of 1.6 x 10$^{-8}$C in a uniform electric field of intensity 5 x 10$^5$NC$^{-1}$  is

Analyze the question: We're given a charge q = 1.6 × 10⁻⁸ C placed in a uniform electric field of intensity E = 5 × 10⁵ N/C. We need the force on the charge.

Concept: Electric field intensity is the force per unit charge:

E = F/q, so F = qE

Solution:

F = (1.6 × 10⁻⁸ C) × (5 × 10⁵ N/C)

F = (1.6 × 5) × 10⁻⁸⁺⁵

F = 8.0 × 10⁻³ N

Check against the options:

• A (3.2 × 10⁻¹⁴ N): wrong, this is far too small.
• B (8.0 × 10⁻¹³ N): wrong, the powers of 10 were not combined correctly.
• C (8.0 × 10⁻³ N): matches our result.
• D (3.1 × 10¹³ N): wrong, this comes from dividing (5 × 10⁵ ÷ 1.6 × 10⁻⁸ ≈ 3.1 × 10¹³) instead of multiplying.

Tip: Remember the triangle F = qE. Force is multiplied by charge and field. When multiplying powers of ten, add the exponents: −8 + 5 = −3. Option D is a trap for students who divide instead of multiply, so always check which operation the formula needs.