91Ó°ÊÓ

Columns law force is,

${\mathrm{F}}_{\mathrm{C}}={\mathrm{F}}_{1\text{on}2}={\mathrm{F}}_{2\text{on}1}$

$=\frac{\mathrm{K}\left|{\mathrm{q}}_1\right|\left|{\mathrm{q}}_2\right|}{{\mathrm{r}}^2}$

For charge ${\mathrm{q}}_1$and ${\mathrm{q}}_2$the force is,

${\stackrel{→}{\mathrm{F}}}_{12}=\frac{{\mathrm{Kq}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}\hat{\mathrm{j}}$

$=\frac{\left(9.0×{10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(-10×{10}^{-9}\mathrm{C}\right)\left(5×{10}^{-9}\mathrm{C}\right)}{(0.01\mathrm{m}{)}^2}\hat{\mathrm{j}}$

$=-4.5×{10}^{-3}\hat{\mathrm{j}}\mathrm{N}$

The distance of charges is,

${\mathrm{r}}_{13}=\sqrt{(1\mathrm{cm}{)}^2+(3\mathrm{cm}{)}^2}$

$=3.16\mathrm{cm}$

Force on charge ${\mathrm{q}}_1$and ${\mathrm{q}}_3$is,

${\mathrm{F}}_{13}=\frac{{\mathrm{Kq}}_1{\mathrm{q}}_3}{{\mathrm{r}}_{13}^2}$

$=\frac{\left(9.0×{10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(-10×{10}^{-9}\mathrm{C}\right)\left(-15×{10}^{-9}\mathrm{C}\right)}{(0.0316\mathrm{m}{)}^2}$

$=1.35×{10}^{-3}\mathrm{N}$

Angle for negative direction is,

$\mathrm{θ}={\tan }^{-1}\left(\frac{\text{opposite}}{\text{adjacent}}\right)$

$={\tan }^{-1}\left(\frac{1\mathrm{cm}}{3\mathrm{cm}}\right)$

$=18.4°$

Angle for negative direction is,

$\mathrm{θ}=180°-18.4°$

$=161.6°$

In vector form,

${\stackrel{→}{\mathrm{F}}}_{13}={\mathrm{F}}_{13}\cos 161.6°\hat{\mathrm{i}}+\mathrm{Esin}161.6^{\mathrm{o}}\hat{\mathrm{j}}$

$=\left(1.35×{10}^{-3}\mathrm{N}\right)\cos 161.6^{\mathrm{o}}\hat{\mathrm{i}}+\left(1.35×{10}^{-3}\mathrm{N}\right)\sin 161.6^{\mathrm{o}}\hat{\mathrm{j}}$

$=(-1.28\hat{\mathrm{i}}+0.43\hat{\mathrm{j}})×{10}^{-3}\mathrm{N}$

Net charge for all charges is,

${\mathrm{F}}_{\text{net}}=\sqrt{{\stackrel{→}{\mathrm{F}}}_{12}^2+{\stackrel{→}{\mathrm{F}}}_{13}^2}$

$=\sqrt{{\left(0\mathrm{N}+-1.28×{10}^{-3}\mathrm{N}\right)}^2\hat{\mathrm{i}}+{\left(-4.5×{10}^{-3}\mathrm{N}+0.43×{10}^{-3}\mathrm{N}\right)}^2\hat{\mathrm{j}}}$

$=4.3×{10}^{-3}\mathrm{N}$

Angle of direction is,

$\mathrm{Î\pm }={\tan }^{-1}\left(\frac{\left(-4.5×{10}^{-3}\mathrm{N}+0.43×{10}^{-3}\mathrm{N}\right)\hat{\mathrm{j}}}{\left(0\mathrm{N}+-1.28×{10}^{-3}\mathrm{N}\right)\hat{\mathrm{i}}}\right)$

$={73}^{°}$

Counter clockwise direction.