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 = 1/4πε₀ = 9.0 10⁹ N ∙ m²/C²)

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Quizplus · Accepted Answer

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

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