91Ó°ÊÓ

The given data can be listed below,

• The location of electron is,
• The observation location of electric field is,

The charge causes the formation of an electric field. It's basically a depiction of the electric force that a unit test charge feels when it's close to the source charge.

The source location is the location of the electron i.e.,

The observation location is the location where the electric field due to electron is to be found i.e.,

The distance vector is given by,

$\stackrel{→}{r}=\text{final}\text{location}-\text{initial}\text{location}$

Substitute values in the above,

The magnitude of distance vector is given by,

$\left|\stackrel{→}{r}\right|=\sqrt{{\left(x\right)}^2+{\left(y\right)}^2+{\left(z\right)}^2}$

Here, $x$ is the x component of the distance vector,$y$ is the y component of the distance vector, and$z$ is the z components of the distance vector

Substitute all the values in the above,

$$\begin{gathered} \left|\stackrel{→}{r}\right|=\sqrt{{\left(-0.3\right)}^2+{\left(0.3\right)}^2+{\left(0.3\right)}^2} \\ =\frac{3\sqrt{3}}{10}\text{m} \end{gathered}$$

Thus, the value of $\left|\stackrel{→}{r}\right|$is $\frac{3\sqrt{3}}{10}\text{m}$.

The unit vector in direction of distance vector is given by,

$\hat{r}=\frac{\stackrel{→}{r}}{\left|\stackrel{→}{r}\right|}$

Here, $\stackrel{→}{r}$is the distance vector and $\left|\stackrel{→}{r}\right|$is the magnitude of distance vector.

Substitute all the values in the above,

Thus, the value of$\hat{r}$ is .

The value of electric field is given by,

$E=\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^2}$

Here,q is the charge on the electron whose value is $1.6×{10}^{-19}\text{C}$, $\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}$is the coulomb constant whose value is$9×{10}^9\text{N}·{\text{m}}^2{\text{/C}}^2$ and r is the distance between source and observation location.

Substitute all the values in the above equation.

$$\begin{gathered} E=\frac{\left(9×{10}^9\text{N}·{\text{m}}^2{\text{/C}}^2\right)\left(1.6×{10}^{-19}\text{C}\right)}{{\left(\frac{3\sqrt{3}}{10}\text{m}\right)}^2} \\ =5.33×{10}^{-9}\text{N/C} \end{gathered}$$

Thus, the value of $\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^2}$is $5.33×{10}^{-9}\text{N/C}$.

The electric field in vector form is given by,

$\stackrel{→}{E}=\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{\left|\stackrel{→}{r}\right|}^2}\hat{r}$

Here,q is the charge on the electron whose value is$1.6×{10}^{-19}\text{C}$, $\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}$is the coulomb constant whose value is$9×{10}^9\text{N}·{\text{m}}^2{\text{/C}}^2$, $\left|\stackrel{→}{r}\right|$is the magnitude of distance between source and observation location and $\hat{r}$is the unit vector.

Substitute all the values in the above equation.

Thus, the value of the electric field in vector form is $3.07×{10}^{-9}\text{N/C}$.