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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)
A{'text': '1200 V', 'label': 'A'}
B{'text': '900 V', 'label': 'B'}
C{'text': '600 V', 'label': 'C'}
D{'text': '300 V', 'label': 'D'}
Answer & Solution
Correct answer: C. {'text': '600 V', 'label': 'C'}
1. V = kq / r.
2. V = (9 × 10⁹)(20 × 10⁻⁹) / 0.30.
3. V = 180 / 0.30 = 600 V.
_Source: NCERT Class 12 Physics Part I, Ch 2 §2.3, mirrors Example 2.1 style_
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