Question

Charges $-q$and $+2q$in $FIGUREP23.39$are located at $x=\pm a$. Determine the electric field at points $1-4$. Write each field in component form.

Short answer

The Electric field component is ,$\stackrel{\rightharpoonup }{E}=\frac{1}{4\pi {\epsilon }_0}\left(\frac{q}{5\sqrt{5}{a}^2}\right)(3\widehat{i}-2\widehat{j}).$

Step-by-step solution

Step: 1 Electric field:

The Electric field charge is

$E=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}$

The following diagram depicts the electromagnetic lines at node one due to these two charges, which can be used to compute the electric field.

Step: 2 Finding distance:

Charge from node one as

$\begin{array}{l}{r}_1^2=a^2+(2a{)}^2\\ r_1^2=a^2+4a^2\\ r_1^2=5a^2\end{array}$

The distance from charge $-q$is

$\begin{array}{l}{r}_2^2=a^2+(2a{)}^2\\ r_2^2=a^2+4a^2\\ r_2^2=5a^2\end{array}$

Step: 3 Finding angle:

The value of sine angle is

$\begin{array}{l}\sin \theta =\frac{a}{r_2}\\ \sin \theta =\frac{a}{a\sqrt{5}}\\ \sin \theta =\frac{1}{\sqrt{5}}\end{array}$

The value of cosine angle is

$\begin{array}{l}\cos \theta =\frac{2a}{r_2}\\ \cos \theta =\frac{2a}{a\sqrt{5}}\\ \cos \theta =\frac{2}{\sqrt{5}}\end{array}$

Step; 4 Magnitude at node one:

The magnitude at electric field is at node one is

$E_1=\frac{1}{4\pi {\epsilon }_0}\frac{2q}{r_1^2}$

$E_1=\frac{1}{4\pi {\epsilon }_0}\frac{2q}{5a^2}$

The components as

$E_{1x}=E_1\sin \theta$

Substituting$\sin \theta$

$\begin{array}{l}{E}_{1x}=\frac{1}{4\pi {\epsilon }_0}\left(\frac{2q}{5a^2}\right)\left(\frac{1}{\sqrt{5}}\right)\\ E_{1x}=\frac{1}{4\pi {\epsilon }_0}\frac{2q}{5\sqrt{5}{a}^2}\end{array}$

Step: 5 Equating:

The component $y$as

$\begin{array}{l}{E}_{1y}=\left(\frac{1}{4\pi {\epsilon }_0}\frac{2q}{5a^2}\right)\left(\frac{2}{\sqrt{5}}\right)\\ E_{1y}=\frac{1}{4\pi {\epsilon }_0}\frac{4q}{5\sqrt{5}{a}^2}\end{array}$

$E_{1y}=E_1\cos \theta$

Substituting$\cos \theta$

$\begin{array}{l}{E}_{1y}=\left(\frac{1}{4\pi {\epsilon }_0}\frac{2q}{5a^2}\right)\left(\frac{2}{\sqrt{5}}\right)\\ E_{1y}=\frac{1}{4\pi {\epsilon }_0}\frac{4q}{5\sqrt{5}{a}^2}\end{array}$

Step: 6 Finding component:

The $x$component at field as

$\begin{array}{l}{E}_{2x}=E_2\sin \theta \\ E_{2x}=\frac{1}{4\pi {\epsilon }_0}\frac{q}{5a^2}\left(\frac{1}{\sqrt{5}}\right)\\ E_{2x}=\frac{1}{4\pi {\epsilon }_0}\frac{q}{5\sqrt{5}{a}^2}\end{array}$

The $y$component at field as

$\begin{array}{l}{E}_{2y}=-E_2\cos \theta \\ E_{2y}=-\frac{1}{4\pi {\epsilon }_0}\frac{q}{5a^2}\left(\frac{2}{\sqrt{5}}\right)\\ E_{2y}=-\frac{1}{4\pi {\epsilon }_0}\frac{2q}{5\sqrt{5}{a}^2}\end{array}$

The negative sign denotes field in downward side.

Step: 7 Finding nodes:

The field at node two $+2q$is

$E_{x,2q}=-\frac{1}{4\pi {\epsilon }_0}\left(\frac{2q}{a^2}\right)$

The field at node two $-q$is

$\begin{array}{l}{E}_{x,-q}=\frac{1}{4\pi {\epsilon }_0}\left(\frac{q}{(3a{)}^2}\right)\\ E_{x,-q}=\frac{1}{4\pi {\epsilon }_0}\left(\frac{q}{9a^2}\right)\end{array}$

Because these two variables are solely on the $x$axis, their $y$components are zero.

Step: 8 Resultant field at node two:

The resultant field at node two is

$\overrightarrow{E}=\left(E_{x,2q}+E_{x,-q}\right)\widehat{i}$

Substituting

$\begin{array}{l}\overrightarrow{E}=\left(-\frac{1}{4\pi {\epsilon }_0}\left(\frac{2q}{a^2}\right)+\frac{1}{4\pi {\epsilon }_0}\left(\frac{q}{9a^2}\right)\right)\widehat{i}\\ \overrightarrow{E}=-\frac{1}{4\pi {\epsilon }_0}\left(\frac{q}{a^2}\right)\left(\frac{17}{9}\right)\widehat{i}\\ \overrightarrow{E}=\left[-\frac{17}{36\pi {\epsilon }_0}\left(\frac{q}{a^2}\right)\right]\widehat{i}\end{array}$

Step: 9 Resultant field at node three:

The resultant charges is

$\overrightarrow{E}=\left(E_{x,2q}+E_{x,-q}\right)\widehat{i}$

Substituting

$\begin{array}{l}\overrightarrow{E}=\left(\frac{1}{4\pi {\epsilon }_0}\left(\frac{2q}{9a^2}\right)-\frac{1}{4\pi {\epsilon }_0}\left(\frac{q}{a^2}\right)\right)\widehat{i}\\ E=\left(\frac{1}{4\pi {\epsilon }_0}\left(\frac{q}{a^2}\right)\left[\frac{2}{9}-1\right]\right)\widehat{i}\\ E=\left[-\frac{7}{36\pi {\epsilon }_0}\left(\frac{q}{a^2}\right)\right]\widehat{i}\end{array}$

The Diagram as

Step: 10 Finding charge:

The distance from charge $2q$is

$\begin{array}{l}{r}_1^2=a^2+(2a{)}^2\\ r_1^2=a^2+4a^2\\ r_1^2=5a^2\end{array}$

The distance from charge $-q$is

$\begin{array}{l}{r}_2^2=a^2+(2a{)}^2\\ r_2^2=a^2+4a^2\\ r_2^2=5a^2\end{array}$

Step: 11 Finding angle:

The sine angle as

$\begin{array}{l}\sin \theta =\frac{a}{r_2}\\ \sin \theta =\frac{a}{a\sqrt{5}}\\ \sin \theta =\frac{1}{\sqrt{5}}\end{array}$

The cosine angle as

$\begin{array}{l}\cos \theta =\frac{2a}{r_2}\\ \cos \theta =\frac{2a}{a\sqrt{5}}\\ \cos \theta =\frac{2}{\sqrt{5}}\end{array}$

Step: 12 Finding component:

Te field at component is

$E_x=E_{1x}+E_{2x}$

Substituting

$\begin{array}{l}{E}_x=\frac{1}{4\pi {\epsilon }_0}\frac{2q}{5\sqrt{5}{a}^2}+\frac{1}{4\pi {\epsilon }_0}\frac{q}{5\sqrt{5}{a}^2}\\ E_x=\frac{1}{4\pi {\epsilon }_0}\left(\frac{3q}{5\sqrt{5}{a}^2}\right)\end{array}$

Similarlly,

$\begin{array}{l}{E}_y=-\frac{1}{4\pi {\epsilon }_0}\frac{4q}{5\sqrt{5}{a}^2}+\frac{1}{4\pi {\epsilon }_0}\frac{2q}{5\sqrt{5}{a}^2}\\ E_y=-\frac{1}{4\pi {\epsilon }_0}\left(\frac{2q}{5\sqrt{5}{a}^2}\right)\end{array}$

Step: 13 Finding electric field: 

The net field in component as

$\overrightarrow{E}=E_x\widehat{i}+E_y\widehat{j}$

Substituting the values

$\begin{array}{l}\overrightarrow{E}=\left[\frac{1}{4\pi {\epsilon }_0}\left(\frac{3q}{5\sqrt{5}{a}^2}\right)\right]\widehat{i}+\left[-\frac{1}{4\pi {\epsilon }_0}\left(\frac{2q}{5\sqrt{5}{a}^2}\right)\right]\widehat{j}\\ \overrightarrow{E}=\frac{1}{4\pi {\epsilon }_0}\left(\frac{q}{5\sqrt{5}{a}^2}\right)(3\widehat{i}-2\widehat{j}).\end{array}$