What are the strength and direction of the electric field at the position indicated by the dot in $FIGUREP23.36$? Give your answer $\left(a\right)$in component form and $\left(b\right)$as a magnitude and angle measured $cw$or $ccw$(specify which) from the positive $x$-axis.

Find the net electrostatic force by dividing the components of the electrostatic potential due to all positive ions at the target point.

In the diagram below, the placement of the electrons and the vector of the related fields are depicted

The electric filed intensity as

$E=k\frac{q}{d^2}$

$E_1=k\frac{q_1}{d_1^2}$

Solving as,

The vector form as

${\overrightarrow{E}}_1=\left(5.625\times {10}^4\mathrm{N}/\mathrm{C}\right)(-\widehat{i}).$

The Field intensity as

$E_2=k\frac{q_2}{r^2}$

Calculate radius as

$\begin{array}{l}r=\sqrt{(4.0\mathrm{cm}{)}^2+(2.0\mathrm{cm}{)}^2}\\ r=4.472\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)\\ r=0.04472\mathrm{m}.\end{array}$

The angle as

$\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{4.0\mathrm{cm}}{2.0\mathrm{cm}}\right)\\ \theta ={63.435}^{\circ }.\end{array}$

From the above equation,

Solving as

$\begin{array}{l}{E}_2=\left(9\times {10}^9\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{C}}^2\right)\frac{10\mathrm{nC}\left(\frac{{10}^{-9}\mathrm{C}}{1\mathrm{nC}}\right)}{(0.04472\mathrm{m}{)}^2}\\ E_2=4.5\times {10}^4\mathrm{N}/\mathrm{C}.\end{array}$

The component in field as

$E_{2x}=E_2\sin \theta$

Solving as

$\begin{array}{l}{E}_{2x}=\left(4.5\times {10}^4\mathrm{N}/\mathrm{C}\right)\sin {63.435}^{\circ }\\ E_{2x}=4.0249\times {10}^4\mathrm{N}/\mathrm{C}.\end{array}$

The intensity field as

$E_3=k\frac{q_3}{d_3^2}$

$\begin{array}{l}{E}_3=\left(9\times {10}^9\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{C}}^2\right)\frac{-5.0\mathrm{nC}\left(\frac{{10}^{-9}\mathrm{C}}{1\mathrm{nC}}\right)}{{\left(2.0\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)\right)}^2}\\ E_3=1.125\times {10}^5\mathrm{N}/\mathrm{C}\end{array}$

The net electric field is

role="math" localid="1651417476230" $\begin{array}{l}\overrightarrow{E}=\left(5.625\times {10}^4\mathrm{N}/\mathrm{C}\right)(-\widehat{i})+\left(4.0249\times {10}^4\mathrm{N}/\mathrm{C}\right)(-\widehat{i})+\left(2.0125\times {10}^4\mathrm{N}/\mathrm{C}\right)(-\widehat{j})\\ +\left(1.125\times {10}^5\mathrm{N}/\mathrm{C}\right)\left(\widehat{j}\right)\end{array}$

$\stackrel{\rightharpoonup }{E}=\left(9.6499\times {10}^4\mathrm{N}/\mathrm{C}\right)(-\widehat{i})+\left(9.2375\times {10}^4\mathrm{N}/\mathrm{C}\right)\left(\widehat{j}\right).$

The net field component as

$\stackrel{\rightharpoonup }{E=}\left(-9.7\times {10}^4\mathrm{N}/\mathrm{C}\right)\left(\widehat{i}\right)+\left(9.2\times {10}^4\mathrm{N}/\mathrm{C}\right)\left(\widehat{j}\right).$

The magnitude in field as

$E=\sqrt{E_x^2+E_y^2}$

Solving as

$\begin{array}{l}E=\sqrt{{\left(-9.7\times {10}^4\mathrm{N}/\mathrm{C}\right)}^2+{\left(9.2\times {10}^4\mathrm{N}/\mathrm{C}\right)}^2}\\ E=1.335\times {10}^5\mathrm{N}/\mathrm{C}.\end{array}$

$E=1.34\times {10}^5\mathrm{N}/\mathrm{C}.$

The angle as

$\theta ={\tan }^{-1}\left(\frac{E_y}{E_x}\right)$

Solving as

$\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{9.2\times {10}^4\mathrm{N}/\mathrm{C}}{-9.7\times {10}^4\mathrm{N}/\mathrm{C}}\right)\\ \theta =-{43.5}^{\circ }.\end{array}$

As a result, the net field is $43.5^{\circ }$degrees around from the negative $x$-axis, or${136.5}^{\circ }\left(180-{43.5}^{\circ }\right)$ counterclockwise from the positive $x$-axis.

As a result, the net field's relative positive $x$-axis rotates almost ${136}^{\circ }$ counterclockwise.