Two point charges of +2 μC each are placed 3 m apart in vacuum. The force between them is (k = 9 × 10⁹ N m²/C²)
A{'text': '0.4 mN', 'label': 'A'}
B{'text': '4 mN', 'label': 'B'}
C{'text': '4 N', 'label': 'C'}
D{'text': '40 N', 'label': 'D'}
Answer & Solution
Correct answer: B. {'text': '4 mN', 'label': 'B'}
1. F = k q₁ q₂ / r².
2. Substitute: F = 9 × 10⁹ × (2 × 10⁻⁶)² / 3².
3. = 9 × 10⁹ × 4 × 10⁻¹² / 9 = 4 × 10⁻³ N = 4 mN.
4. So the mutual force is 4 milli-newtons.
_Source: NCERT Class 12 Physics, Ch 1 "Electric Charges and Fields", §1.5_
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