91Ó°ÊÓ

Two forces are exerted on the charge $q_1\left(F_{2}and{F}_Q\right)$

So, $F_2$is the force of $q_2$and $q_1$and $F_Q$is the force of Q on $q_1$

.

Therefore, force $F_2$is given by

$F_2=K\frac{\left|q_2{q}_1\right|}{r^2}$

Here,

localid="1668414015667" $$\begin{gathered} K=9.0×{10}^9{\mathrm{Nm}}^2/{\mathrm{C}}^2 \\ \mathrm{r}=0.60\mathrm{m},{\mathrm{q}}_2={\mathrm{q}}_1=2.0\mathrm{μ°ä} \\ {\mathrm{F}}_2={\mathrm{F}}_{\mathrm{y}}=\left(9×{10}^9\right)\frac{{\left(2.0×{10}^{−6}\right)}^2}{(0.60{)}^2} \\ =0.10\mathrm{N} \end{gathered}$$

Now for $F_Q$+as charge Q exert force on $q_1$into direction $\left(-xand-y\right)$, here distance $r=0.5m$in the x-direction with angle $\left(\mathrm{Î\pm }\right)$, and distance y=0.5m, $Q=-4.0\mathrm{μ}C$.

Therefore,

localid="1668156853761" $$\begin{gathered} F_Q=−F_x\cos \mathrm{Î\pm }+F_y\sin \mathrm{Î\pm } \\ =K\left|Q{q}_1\right|\left(\frac{−\cos \mathrm{Î\pm }}{(0.5{)}^2}+\frac{\sin \mathrm{Î\pm }}{(0.5{)}^2}\right) \\ =\left(9×{10}^9\right)\left|\left(4×{10}^{−6}\right)\left(2×{10}^{−6}\right)\right|\left[\frac{−\cos \mathrm{Î\pm }}{(0.5{)}^2}+\frac{\sin \mathrm{Î\pm }}{(0.5{)}^2}\right] \\ =0.57N \end{gathered}$$

Now here,

$$\begin{gathered} f_x=-0.23,f_y=0.17 \\ \cos \mathrm{Î\pm }=\frac{0.4}{0.5},\sin \mathrm{Î\pm }=\frac{0.3}{0.5} \end{gathered}$$

Therefore, the magnitude of the net force $F_1$exerted on $q_1$is

$$\begin{gathered} \left|F_1\right|=\sqrt{{\left({\mathrm{F}}_{\mathrm{x}}\right)}^2+{\left({\mathrm{F}}_{\mathrm{y}\mathrm{of}2}+{\mathrm{F}}_{\mathrm{y}\mathrm{of}\mathrm{Q}}\right)}^2} \\ =\sqrt{(−0.23{)}^2+(0.10+0.17{)}^2} \\ =0.35\mathrm{N} \end{gathered}$$

$$\begin{gathered} \tan \mathrm{θ}=\frac{F_y}{F_X} \\ \mathrm{θ}={\tan }^{−1}\frac{F_y}{F_x} \\ \mathrm{θ}=\left(\frac{0.10+0.17}{0.23}\right) \\ \mathrm{θ}={40.0}^{∘} \end{gathered}$$

The net force on charge $q_1$is 0.35 N at the direction $40.0°$from y-axis.