Two equally charged tiny spheres of mass 1.0 g are placed 2.0 cm apart. When released, they begin to accelerate away from each other at $414 \mathrm {~m} / \mathrm { s } ^ { 2 }$ What is the magnitude of the charge on each sphere, assuming only that the electric force is present? (k = 1/4??₀ = 9.0 × 10⁹ N ? m²/C²) A)140 nC B)120 nC C)95 nC D)75 nC

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Quizplus · Accepted Answer

The electric force $ F $ is equal to the product of mass $ m $ and acceleration $ a $, $ F = ma $. Given $ m = 1.0 \, \text{g} = 0.001 \, \text{kg} $ and $ a = 414 \, \text{m/s}^2 $, we find $ F = 0.001 \times 414 = 0.414 \, \text{N} $. The electric force is also given by Coulomb's law: $ F = \frac{kq^2}{r^2} $, where $ k = 9.0 \times 10^9 \, \text{N m}^2/\text{C}^2 $, $ r = 0.02 \, \text{m} $, and $ q $ is the charge. Solving for $ q $, we get $ q = \sqrt{\frac{Fr^2}{k}} = \sqrt{\frac{0.414 \times (0.02)^2}{9.0 \times 10^9}} \approx 1.4 \times 10^{-7} \, \text{C} = 140 \, \text{nC} $.

Two Equally Charged Tiny Spheres of Mass 1 414 m/s2414 \mathrm {~m} / \mathrm { s } ^ { 2 }414 m/s2

Two Equally Charged Tiny Spheres of Mass 1 $414 \mathrm {~m} / \mathrm { s } ^ { 2 }$