Question

91+2.60 x 10-6 C, q2 = +3.75 x 10-6 C, and93 +1.30 x 10-6 C. Find the x-component ofthe net force on q₁. Include the correct + or -sign to indicate direction.91=0.283 m0.200 m9245.0°0.200 m93(Make sure you know the direction of each force!)

Answer 1

Using coulomb's law:

$\begin{gathered} F_{21}=K\cdot\frac{q_1\cdot q_2}{r^2} \\ F_{21}=(8.988\times10^9)\cdot\frac{(2.6\times10^{-6})(3.75\times10^{-6})}{0.283^2} \\ F_{21}=1.094N \end{gathered}$$\begin{gathered} F_{31}=K\cdot\frac{q_1\cdot q_3}{r^2} \\ F_{31}=(8.988\times10^9)\cdot\frac{(2.6\times10^{-6})(1.3\times10^{-6})}{0.2^2} \\ F_{31}=0.759N \end{gathered}$

Now, let's calculate the net force on each component:

$\begin{gathered} F_x=F_{21}cos(45)=0.774N \\ F_y=F_{31}+F_{21}sin(45)=1.533N \end{gathered}$

Now, we can calculate the net force and the direction:

$\begin{gathered} F_{net}=\sqrt{F_x^2+F_y^2} \\ F_{net}=\sqrt{(0.774^2)+(1.533^2)} \\ F_{net}=+1.717N \\ \end{gathered}$

Answer:

+ 1.717 N