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\frac{q1\cdot q2}{r^2} \\ F_{12}=(8.988\times10^9)\cdot(\frac{(2.6\times10^{-6})(3.75\times10^{-6})}{0.25^2}) \\ F_{12}=1.402N \end{gathered}$$\begin{gathered} F_{32}=K\cdot\frac{q3\cdot q2}{r^2} \\ F_{32}=(8.988\times10^9)\cdot(\frac{(2.6\times10^{-6})(3.75\times10^{-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