Two electrons are $20.0 \mathrm {~mm}$ apart at closest approach. What is the magnitude of the maximum electric force that they exert on each other? (e = 1.60 × 10^-19 C, k = 1/4??₀ = 9.0 10⁹ N ? m²/C²) A)5.8 × 10^-25 N B)2.3 × 10¹⁰ N C)2.3 N D)5.8 × 10^-27 N

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Quizplus · Accepted Answer

The electric force between two charges is given by Coulomb's law: $ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} $. Substituting the given values: $ k = 9.0 \times 10^9 \, \text{N m}^2/\text{C}^2 $, $ q_1 = q_2 = 1.60 \times 10^{-19} \, \text{C} $, and $ r = 20.0 \, \text{mm} = 0.020 \, \text{m} $, we get $ F = \frac{9.0 \times 10^9 \times (1.60 \times 10^{-19})^2}{(0.020)^2} = 5.8 \times 10^{-25} \, \text{N} $.

Two Electrons Are 20.0 mm20.0 \mathrm {~mm}20.0 mm Apart at Closest Approach

Two Electrons Are $20.0 \mathrm {~mm}$ Apart at Closest Approach