Two equally charged tiny spheres of mass $1.0 \mathrm {~g}$ are placed $2.0 \mathrm {~cm}$ 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? $\left( k = 1 / 4 \pi \varepsilon _ { 0 } = 9.0 \times 10 ^ { 9 } \mathrm {~N} \cdot \mathrm { m } ^ { 2 } / \mathrm { C } ^ { 2 } \right)$

Question

Quizplus · Accepted Answer

The electric force $ F $ is given by $ F = ma $, where $ m = 1.0 \times 10^{-3} \mathrm{~kg} $ and $ a = 414 \mathrm{~m/s^2} $. Thus, $ F = 1.0 \times 10^{-3} \times 414 = 0.414 \mathrm{~N} $. The electric force is also given by $ F = \frac{kq^2}{r^2} $, where $ r = 0.02 \mathrm{~m} $. Solving for $ q $, we have $ 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} \mathrm{~C} = 140 \mathrm{~nC} $.

Two Equally Charged Tiny Spheres of Mass 1.0 g1.0 \mathrm {~g}1.0 g Are Placed

Two Equally Charged Tiny Spheres of Mass $1.0 \mathrm {~g}$ Are Placed