Question

A charge q2 is placed at x=.24 m from q1 which is at the origin (0). A third charge., q3= -1.0 nC is placed at x=.16 m. The charge q2 experiences no net electric force. What is q1 charge?

Please explain each step clearly. I don't understand how to find the answer. A diagram may help. Thanks so much in advance.

Answer 1

0.24 m q3 F2 F1 qi 0.16 m-

According to Coulomb's law electrostatic force between two charges can be given by

$F=\frac{1}{4\pi \epsilon _{0}}\frac{Q_{1}Q_{2}}{r^{2}}$

$\frac{1}{4\pi \epsilon _{0}}=9\times 10^{9}Nm^{2}/C^{2}$

According to given condition electric force on $q_{2}$ is zero

Hence

$\frac{1}{4\pi \epsilon _{0}}\frac{q_{1}q_{2}}{0.24^{2}}=\frac{1}{4\pi \epsilon _{0}}\frac{q_{3}q_{2}}{0.08^{2}}$

$q_{1}=\frac{0.24^{2}}{0.08^{2}}q_{3}$$(q_{3}=-1nC=-1\times 10^{-9}C)$

$q_{1}=(\frac{0.24^{2}}{0.08^{2}}\times 10^{-9})C=9\times 10^{-9}C=9nC$