Question

An equilateral triangle of side0.250 m has charges91 +2.60 x 10-6 C,9192= -3.75 x 10-6 C, and93 +2.60 x 10-6 C.92Find the x-component of the net force on 92.Include the correct + or - sign to indicate direction.(Make sure you know the direction of each force!Remember, equilateral triangles have 60 degree interiorangles.)x-component (N)Enter93

Answer 1

Using Coulomb's law:

`\begin{gathered} F_(12)=K\cdot(q1\cdot q2)/(r^2) \\ F_(12)=(8.988*10^9)\cdot(((2.6*10^(-6))(3.75*10^(-6)))/(0.25^2)) \\ F_(12)=1.402N \end{gathered}`
`\begin{gathered} F_(32)=K\cdot(q3\cdot q2)/(r^2) \\ F_(32)=(8.988*10^9)\cdot(((2.6*10^(-6))(3.75*10^(-6)))/(0.25^2)) \\ F_(32)=1.402N \end{gathered}`

Now, we can calculate the net force:

`\begin{gathered} F_x=F_(12)cos(60)+F_(32) \\ F_x=2.103N \end{gathered}`

Answer:

+ 2.103 N