Two tiny particles carrying like charges of the same magnitude are $1.0 \mathrm {~mm}$ apart. If the electric force on one of them is $5.0 \mathrm {~N} \text {, }$ what is the magnitude of the charge on each of these particles?(k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²) A) 2.4 × 10^-2 C B) 2.4 × 10^-8 C C) 4.5 × 10⁴ C D) 2.4 × 10^-5 C

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Quizplus · Accepted Answer

The electric force $ F $ between two charges is given by Coulomb's law: $ F = \frac{k \cdot q^2}{r^2} $. Rearranging for $ q $, we get $ q = \sqrt{\frac{F \cdot r^2}{k}} $. Substituting $ F = 5.0 \, \text{N} $, $ r = 1.0 \times 10^{-3} \, \text{m} $, and $ k = 9.0 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 $, we find $ q = \sqrt{\frac{5.0 \times (1.0 \times 10^{-3})^2}{9.0 \times 10^9}} = 2.4 \times 10^{-8} \, \text{C} $.

Two Tiny Particles Carrying Like Charges of the Same Magnitude 1.0 mm1.0 \mathrm {~mm}1.0 mm

Two Tiny Particles Carrying Like Charges of the Same Magnitude $1.0 \mathrm {~mm}$