A $–12nC$charge is located at $1x,y2=11.0cm,0cm2.$What are the electric fields at the position $1x,y2=15.0cm,0cm2,1-5.0cm,0cm2,$and$10cm,5.0cm2?$ Write each electric field vector in component form.

The electrostatic force is the amount of electrostatic forces exerted on most any charge $q$, with the magnitude of the charge determining the force. That observational point seems to be where we measure the electric field, while the charge that causes the electric field is known as the source charge at the source location. Equation $(22.8)$in the form gives the electrostatic force at the observing location due to the point charge.

Equation $1$

$\overrightarrow{E}=\frac{1}{4\pi {\epsilon }_o}\frac{q}{r^2}\widehat{r}$

The electric current is $E$, the distance here between charges and the position is $r$, and the charge is $q$. The expression $\left(\frac{1}{4\pi {\epsilon }_o}\right)$is equal $9.0\times {10}^9\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{C}}^2$. $(x,y)=(1\mathrm{cm},0\mathrm{cm})$is where the charge is positioned. At $(x,y)=(5\mathrm{cm},0\mathrm{cm})$

The electric field is a long way away.

$r=\sqrt{(5\mathrm{cm}-1\mathrm{cm}{)}^2+(0\mathrm{cm}-0\mathrm{cm}{)}^2}=4\mathrm{cm}$

As a result, the intensity of the electric field is solely in the $x$direction and can be determined using

${\overrightarrow{E}}_{5,0}=\frac{1}{4\pi {\epsilon }_o}\frac{q}{r^2}(-\widehat{i})$

$=\left(9.0\times {10}^9\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(\frac{-12\times {10}^{-9}\mathrm{C}}{{\left(4\times {10}^{-2}\mathrm{m}\right)}^2}\right)$

$=-6.8\times {10}^4\widehat{i}\mathrm{N}/\mathrm{C}$

At position $(x,y)=(-5cm,0cm)$

The distance of electric field is

$r=\sqrt{(-5\mathrm{cm}-1\mathrm{cm}{)}^2+(0\mathrm{cm}-0\mathrm{cm}{)}^2}=6\mathrm{cm}$

As a result, the intensity of the electric field can only be determined in the $x$direction.

${\overrightarrow{E}}_{-5,0}=\frac{1}{4\pi {\epsilon }_o}\frac{q}{r^2}(-\widehat{i})$

$=\left(9.0\times {10}^9\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(\frac{-12\times {10}^{-9}\mathrm{C}}{{\left(6\times {10}^{-2}\mathrm{m}\right)}^2}\right)$

$=3\times {10}^4\widehat{i}\mathrm{N}/\mathrm{C}$

At position $(x,y)=(0cm,5cm)$

The distance of electric field is

$r=\sqrt{(0\mathrm{cm}-1\mathrm{cm}{)}^2+(5\mathrm{cm}-0\mathrm{cm}{)}^2}=5.1\mathrm{cm}$

The electric field magnitude is

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

$=\left(9.0\times {10}^9\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(\frac{12\times {10}^{-9}\mathrm{C}}{{\left(5.1\times {10}^{-2}\mathrm{m}\right)}^2}\right)$

$=-4.15\times {10}^4\mathrm{N}/\mathrm{C}$

The angle formed by the electric field and the $x$axis is determined as follows:

$\theta ={\tan }^{-1}\left(\frac{\text{opposite}}{\text{adjacent}}\right)={\tan }^{-1}\frac{5\mathrm{cm}}{1\mathrm{cm}}={78.7}^{\circ }$

As a result, we can write the electric field as a vector.

${\overrightarrow{E}}_{0,5}=-E\cos {78.7}^o\widehat{i}+E\sin {78.7}^o\widehat{j}$

$=-\left(-4.5\times {10}^4\mathrm{N}/\mathrm{C}\right)\cos {78.7}^o\widehat{i}+\left(-4.5\times {10}^4\mathrm{N}/\mathrm{C}\right)\sin {78.7}^{\circ }\widehat{j}$

$=(8.81\widehat{i}-4.41\widehat{j})\times {10}^3\mathrm{N}/\mathrm{C}$