NEET 2017 · PhysicsCoulomb forcePrevious Year Question
Suppose the charges of a proton and an electron differ slightly. One of them is −e, the other is (e + Δe). If the net of the electrostatic force and the gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then Δe is of the order of [Given: mass of hydrogen mₕ = 1.67 × 10⁻²⁷ kg]
Answer: (C) 10⁻³⁷ C. Answer: (c) Solution: Each H atom carries net charge Δe (= (e+Δe) − e).
- A.10⁻²⁰ C
- B.10⁻²³ C
- C.10⁻³⁷ C✓
- D.10⁻⁴⁷ C
Correct Answer
(C) 10⁻³⁷ C
Solution & Explanation
Answer: (c) Solution: Each H atom carries net charge Δe (= (e+Δe) − e). The two atoms repel electrostatically and attract gravitationally; setting the magnitudes equal: (1/4πε₀)·(Δe)²/d² = G·mₕ²/d² The d² cancels: (Δe)² = 4πε₀·G·mₕ² (Δe)² = G·mₕ² / k, where k = 1/4πε₀ = 8.99 × 10⁹ G·mₕ² = 6.67 × 10⁻¹¹ × (1.67 × 10⁻²⁷)² = 6.67 × 10⁻¹¹ × 2.79 × 10⁻⁵⁴ = 1.86 × 10⁻⁶⁴ (Δe)² = 1.86 × 10⁻⁶⁴ / 8.99 × 10⁹ = 2.07 × 10⁻⁷⁴ Δe ≈ 1.44 × 10⁻³⁷ C Order of magnitude: 10⁻³⁷ C.
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