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 = 1/4πε₀ = 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 given by $ F = ma $, where $ m = 1.0 \, \text{g} = 0.001 \, \text{kg} $ and $ a = 414 \, \text{m/s}^2 $. So, $ F = 0.001 \times 414 = 0.414 \, \text{N} $. The electric force is also given by $ F = \frac{kq^2}{r^2} $, where $ r = 0.02 \, \text{m} $. Solving for $ q $, we have $ q^2 = \frac{Fr^2}{k} = \frac{0.414 \times (0.02)^2}{9 \times 10^9} $. Solving this gives $ q \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 $414 \mathrm {~m} / \mathrm { s } ^ { 2 }$