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The potential at a distance 0.30 m from a point charge of +20 nC is (k = 9 × 10⁹ N m² C⁻²)

A1200 V
B900 V
C600 V
D300 V
Answer & Solution
Correct answer: C. 600 V
1. V = kq / r. 2. V = (9 × 10⁹)(20 × 10⁻⁹) / 0.30. 3. V = 180 / 0.30 = 600 V. _Source: syllabus Physics Part I, Ch 2 §2.3, mirrors Example 2.1 style_
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