Question

91 +2.60 x 10-6 C, q2 = +3.75 x 10-6 C, and=93+1.30 x 10-6 C. Find the y-component ofthe net force on q₁. Include the correct+ or -sign to indicate direction.91=0.283 m0.200 m45.0°92930.200 m(Make sure you know the direction of each force!)y-component (N)

Answer 1

We can use Coulomb's law to solve the proble:

$\begin{gathered} F_{31}=K\cdot\frac{q1\cdot q3}{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}$$\begin{gathered} F_{21}=K\cdot\frac{q1\cdot q2}{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}$

Now, we can calculate the net force for the y-axis:

$\begin{gathered} F_y=0.759+1.094sin(45) \\ F_y=1.533 \end{gathered}$

Answer:

+ 1.533