When $1.0 - \mu \mathrm { C }$ point charge is $15 \mathrm {~m}$ from a second point charge, the force each one experiences a force of $1.0 \mu \mathrm { N }$. What is the magnitude of the second charge? $\left( k = 1 / 4 \pi \varepsilon _ { 0 } = 9.0 \times 10 ^ { 9 } \mathrm {~N} \cdot \mathrm { m } ^ { 2 } / \mathrm { C } ^ { 2 } \right)$

Question

Quizplus · Accepted Answer

Using Coulomb's law, $ F = k \frac{|q_1 q_2|}{r^2} $, where $ F = 1.0 \mu \mathrm{N} = 1.0 \times 10^{-6} \mathrm{N} $, $ q_1 = 1.0 \mu \mathrm{C} = 1.0 \times 10^{-6} \mathrm{C} $, $ r = 15 \mathrm{m} $, and $ k = 9.0 \times 10^9 \mathrm{N \cdot m^2/C^2} $. Solving for $ q_2 $, we get $ q_2 = \frac{F \cdot r^2}{k \cdot q_1} = \frac{1.0 \times 10^{-6} \cdot 15^2}{9.0 \times 10^9 \cdot 1.0 \times 10^{-6}} = 25 \times 10^{-9} \mathrm{C} = 25 \mathrm{nC} $.

When 1.0−μC1.0 - \mu \mathrm { C }1.0−μC Point Charge Is 15 m15 \mathrm {~m}15 m

When $1.0 - \mu \mathrm { C }$ Point Charge Is $15 \mathrm {~m}$