WAEC 2013 Physics · Question 26
A charge of 2.0 x 10 C experiences a force of 80N in a uniform electric field. Calculate the magnitude of the electric field intensity
Options
- A.8.0 x 106 NC-1
- B.4.0 x 106 NC-1
- C.4.0 x 104 NC-1
- D.2.0 x 104 NC-1
Answer: B. 4.0 x 106 NC-1
Explanation written with AI and checked against the answer key
Analyze the question: We're given a charge q = 2.0 × 10⁻⁵ C and the force on it, F = 80 N. We need the electric field intensity E. Concept: Electric field intensity is the force per unit charge: E = F / q Solution: E = 80 / (2.0 × 10⁻⁵) E = (80 / 2.0) × 10⁵ E = 40 × 10⁵ E = 4.0 × 10⁶ N/C Check against the options: 4.0 × 10⁶ N C⁻¹ matches option B. Answer: B Tip: When dividing by a power of ten with a negative exponent, move it to the numerator with a positive exponent (1/10⁻⁵ = 10⁵). Do the plain numbers first (80 ÷ 2 = 40), then fix the power of ten (40 × 10⁵ = 4.0 × 10⁶). Also remember the unit: N/C is the same as V/m.AI can make mistakes. Report anything that looks wrong.
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