91Ó°ÊÓ

The force attraction between charges is

Calculate the angle from triangle $âˆ$BAD by using trigonometric law as below:

$\begin{array}{l}\tan \mathrm{θ}=\frac{\text{opp}}{\text{adj}}=\frac{100\text{cm}}{141\text{cm}}\\ \mathrm{θ}={\tan }^{−1}\left(\frac{100\text{cm}}{141\text{cm}}\right)={35.3}^{°}\end{array}$

Force acting on the point charge B is given as,

${\stackrel{→}{F}}_B=−(3.0\text{N})\widehat{i}$

Force acting on the point charge C is given as,

${\stackrel{→}{F}}_C=−(6.0\text{N})\widehat{j}$

Force acting on the charge D is given as,

$\begin{array}{l}{\stackrel{→}{F}}_D=(2.0\text{N})\cos 35.3\widehat{i}−(2.0\text{N})\sin 35.3\widehat{j}\\ =(1.63\text{N})\widehat{i}−(1.15\text{N})\widehat{j}\end{array}$

Net force is calculated as,

${\stackrel{→}{F}}_{\text{net}}={\stackrel{→}{F}}_B+{\stackrel{→}{F}}_C+{\stackrel{→}{F}}_D$

Substitute the calculated forces at point charges $B, C$, and in above net force equation.

$\begin{array}{l}{\stackrel{→}{F}}_{\text{net}}=−(3.0\text{N})\widehat{i}+(−(6.0\text{N}\left)\widehat{j}\right)+(1.63\text{N})\widehat{i}−(1.15\text{N})\widehat{j}\\ =(−1.37\text{N})\widehat{i}−(7.2\text{N})\widehat{j}\end{array}$

Magnitude of the net force is calculated as,

$\left|{\stackrel{→}{F}}_{\text{net}}\right|=\sqrt{{(1.37\text{N})}^2+{(7.2\text{N})}^2}=7.3\text{N}$

Therefore, the magnitude of the force is .

$7.3\text{N}$

Direction of net force is calculated as,

$\begin{array}{l}\tan {\mathrm{θ}}_{\text{net}}=\frac{\left|{\left(F_{\text{net}}\right)}_y\right|}{\left|{\left(F_{\text{net}}\right)}_x\right|}\\ \tan {\mathrm{θ}}_{\text{net}}=\frac{7.2\text{N}}{1.37\text{N}}\\ {\mathrm{θ}}_{\text{net}}={\tan }^{−1}\left(\frac{7.2\text{N}}{1.37\text{N}}\right)\\ ={79}^{°}\end{array}$

Therefore, the net force$\stackrel{→}{F}(7.3\text{N})$ is oriented at an angle${79}^{°}$ of below the negative x-axis.