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Two point particles of masses 3 mg and 1.8 mg are separated by a distance r. An equal number of electrons are removed from both particles until the electrostatic repulsion exactly balances the gravitational attraction. How many electrons n were removed from each particle? (G = 20/3 * 10⁻¹¹ N m² kg⁻²)
- 1250
- 1250
- 1250
- 1250
Correct answer: 1250
Solution
Gravitational attraction: F_g = G*m1*m2/r². Electrostatic repulsion: Fₑ = k*(ne)²/r². Setting F_g = Fₑ (r cancels): G*m1*m2 = k*(ne)². Plugging in: (20/3)*10⁻¹¹ * 3*10⁻⁶ * 1.8*10⁻⁶ = 9*10⁹ * n² * (1.6*10⁻¹⁹)². LHS = (20/3)*10⁻¹¹ * 5.4*10⁻¹² = 36*10⁻²³ = 3.6*10⁻²². RHS = 9*10⁹ * 2.56*10⁻³⁸ * n² = 2.304*10⁻²⁸ * n². n² = 3.6*10⁻²² / 2.304*10⁻²⁸ = 1.5625*10⁶. n = 1250.
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