Question

In the diagram, q₁ = +8.60 x 10-6 C,92 +5.10 x 10-6 C, and q3 = -3.30 x 10-6 C.=Find the magnitude of the net force on 92.|— 0.350 m →→→9192magnitude (N)↑0.155 m5+93(Make sure you know the direction of each force!Opposites attract, similar repel.)

Answer 1

Answer:

Fnet = 9.52 N

Step-by-step explanation:

Force exerted by q₁ on q₂

`\begin{gathered} F_(12)=(kq_1q_2)/(r_(12)^2) \\ \\ F_(12)=(9*10^9*8.6*10^(-6)*5.1*10^(-6))/(0.350^2) \\ \\ F_(12)=(0.39474)/(0.1225) \\ \\ F_(12)=3.22N \\ \end{gathered}`

Force exerted by q₃ on q₂

`\begin{gathered} F_(32)=(kq_2q_3)/(r_(32)^2) \\ \\ F_(32)=((9*10^9)(5.10*10^(-6))(-3.30*10^(-6)))/(0.155^2) \\ \\ F_(32)=(-0.15147)/(0.024025) \\ \\ F_(32)=-6.30N \end{gathered}`

The magnitude of the netforce on q₂

`\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}`