Calculate the magnitude of the electric field intensity in a vacuum at a distance of 25cm from a charge of 5 ×…

Analyze the question: We need the electric field intensity at a point 25 cm from a point charge of 5 × 10⁻³ C in a vacuum.

Concept: The electric field intensity due to a point charge is

E = (1/4πε₀) × Q / r²

where Q is the charge (C), r is the distance (m), and 1/4πε₀ = 9 × 10⁹ N m² C⁻².

Solution:

1. Convert the distance to metres: r = 25 cm = 0.25 m
2. Find r²: r² = (0.25)² = 0.0625 m²
3. Substitute into the formula:

E = (9 × 10⁹ × 5 × 10⁻³) / 0.0625

4. Numerator: 9 × 10⁹ × 5 × 10⁻³ = 45 × 10⁶ = 4.5 × 10⁷
5. Divide: E = 4.5 × 10⁷ / 0.0625 = 7.2 × 10⁸ N C⁻¹

Answer: E = 7.2 × 10⁸ N C⁻¹, which is option A.

Tip: Always convert cm to m before squaring. Dividing by 0.0625 is the same as multiplying by 16, so 4.5 × 10⁷ × 16 = 72 × 10⁷ = 7.2 × 10⁸. Common mistakes are forgetting to square r (which gives 1.8 × 10⁸) or leaving r in cm.