Answer:
Fnet = 9.52 N
Explanation:
Force exerted by q₁ on q₂
[tex]\begin{gathered} F_{12}=\frac{kq_1q_2}{r_{12}^2} \\ \\ F_{12}=\frac{9\times10^9\times8.6\times10^{-6}\times5.1\times10^{-6}}{0.350^2} \\ \\ F_{12}=\frac{0.39474}{0.1225} \\ \\ F_{12}=3.22N \\ \end{gathered}[/tex]
Force exerted by q₃ on q₂
[tex]\begin{gathered} F_{32}=\frac{kq_2q_3}{r_{32}^2} \\ \\ F_{32}=\frac{(9\times10^9)(5.10\times10^{-6})(-3.30\times10^{-6)}}{0.155^2} \\ \\ F_{32}=\frac{-0.15147}{0.024025} \\ \\ F_{32}=-6.30N \end{gathered}[/tex]
The magnitude of the netforce on q₂
[tex]\begin{gathered} F_{net}=|F_{12}|+|F_{32}| \\ \\ F_{net}=|3.22|+|-6.30| \\ \\ F_{net}=3.22+6.30 \\ \\ F_{net}=9.52N \end{gathered}[/tex]
