91Ó°ÊÓ

The given data can be listed below,

• Location of the electric field is,$⟨0,\text{ }0,\text{ }0âŸ\copyright$
• The value of electric field is,$\mathrm{E}=⟨0,\text{ }4104,\text{ }0âŸ\copyright \text{N/C}$ .

An electric field can be defined as the accumulation of electric influence from an infinite encirclement of charge-hosting bodies, even at great distances.

The component of electric field is only in the y-direction so from this it can be concluded that the source and observation point have same value on x and z-axis. The value of electric field on x and z direction is zero so the source and observation point must lie on y-axis.

The distance vector for the electric filed is given by,

$\begin{array}{c}\stackrel{→}{\mathrm{r}}=⟨0,\text{ }{\mathrm{y}}_{\mathrm{f}},0âŸ\copyright −⟨0,\text{ }{\mathrm{y}}_{\mathrm{i}},\text{ }0âŸ\copyright \\ =⟨0,\text{ }\mathrm{Δ²â},\text{ }0âŸ\copyright \end{array}$

Here, $\mathrm{Δ²â}$ is the displacement of y components.

The magnitude of the distance vector is given by,

$\begin{array}{c}\left|\stackrel{→}{\mathrm{r}}\right|=\sqrt{{\left(0\right)}^2+{\left(\mathrm{Δ²â}\right)}^2+{\left(0\right)}^2}\\ =\mathrm{Δ²â}\text{ }\end{array}$

The unit vector is given by,

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

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

Substitute values in the above,

$\begin{array}{c}\widehat{\mathrm{r}}=\frac{⟨0,\mathrm{Δ²â},\text{ }0âŸ\copyright }{\mathrm{Δ²â}}\\ =⟨0,\text{ }1,\text{ }0âŸ\copyright \end{array}$

The position of proton is determined by electric field expression that is given by,

$\mathrm{E}=\frac{\mathrm{Kq}}{{\mathrm{r}}^2}\widehat{\mathrm{r}}$

Here,q is the charge on the proton,K is the coulomb constant whose value is$9.1×{10}^9\text{ N}â‹\dots {\text{m}}^2{\text{/C}}^2$,$\widehat{\mathrm{r}}$is the unit vector.

Substitute all the values in the above equation.

$\begin{array}{c}⟨0,\text{ }4104,\text{ }0âŸ\copyright \text{N/C}=\frac{(9.1×{10}^9\text{ N}â‹\dots {\text{m}}^2{\text{/C}}^2)(1.6×{10}^{−19}\text{ C})}{{\left(\mathrm{Δ²â}\right)}^2}⟨0,\text{ }1,\text{ }0âŸ\copyright \\ \mathrm{Δ²â}=\sqrt{\frac{(9.1×{10}^9\text{ N}â‹\dots {\text{m}}^2{\text{/C}}^2)(1.6×{10}^{−19}\text{ C})}{⟨0,\text{ }4104,\text{ }0âŸ\copyright \text{N/C}}⟨0,\text{ }1,\text{ }0âŸ\copyright }\\ =5.92×{10}^{−7}\text{″¾}\end{array}$

Thus, the proton is placed below the observation point origin at a displacement of $5.92×{10}^{−7}\text{″¾}$on the -y-axis.

The electric filed is going inward due to the negative charge on the electron so the electron must be placed on y-direction above the observation point i.e., origin.

The distance vector is given by,

$\begin{array}{c}\stackrel{→}{\mathrm{r}}=⟨0,\text{ }{\mathrm{y}}_{\mathrm{f}},0âŸ\copyright −⟨0,\text{ }{\mathrm{y}}_{\mathrm{i}},\text{ }0âŸ\copyright \\ =⟨0,\text{ }\mathrm{Δ²â},\text{ }0âŸ\copyright \end{array}$

Here, the initial distance is greater than final distance so the displacement in y direction will be negative.

The unit vector is given y,

$\begin{array}{c}\widehat{\mathrm{r}}=\frac{⟨0,\mathrm{Δ²â},\text{ }0âŸ\copyright }{\mathrm{Δ²â}}\\ =⟨0,\text{ }−1,\text{ }0âŸ\copyright \end{array}$

The distance at which the electron is placed is given by,

$\mathrm{E}=\frac{\mathrm{Kq}}{{\mathrm{r}}^2}\widehat{\mathrm{r}}$

Here,q is the charge on the electron,K is the coulomb constant whose value is$9.1×{10}^9\text{ N}â‹\dots {\text{m}}^2{\text{/C}}^2$,$\widehat{\mathrm{r}}$is the unit vector.

Substitute all the values in the above,

$\begin{array}{c}⟨0,\text{ }4104,\text{ }0âŸ\copyright \text{N/C}=\frac{(9.1×{10}^9\text{ N}â‹\dots {\text{m}}^2{\text{/C}}^2)(−1.6×{10}^{−19}\text{ C})}{{\left(\mathrm{Δ²â}\right)}^2}⟨0,\text{ }−1,\text{ }0âŸ\copyright \\ \mathrm{Δ²â}=\sqrt{\frac{(9.1×{10}^9\text{ N}â‹\dots {\text{m}}^2{\text{/C}}^2)(−1.6×{10}^{−19}\text{ C})}{⟨0,\text{ }4104,\text{ }0âŸ\copyright \text{N/C}}⟨0,\text{ }−1,\text{ }0âŸ\copyright }\\ =5.92×{10}^{−7}\text{″¾}\end{array}$

Thus, the electron is placed above the origin on the y-direction with a displacement of $5.92×{10}^{−7}\text{″¾}$.