You want to create an electric field $\mathbf{⟨}\mathbf{0}\mathbf{,}\mathbf{4104}\mathbf{,}\mathbf{0}\mathbf{⟩}\mathbf{\text{\hspace{0.17em}N/C}}$at location $\mathbf{⟨}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{⟩}$.

(a) Where would you place a proton to produce this field at the origin?

(b) Instead of a proton, where would you place an electron to produce this field at the origin?

The given data can be listed below,

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

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,

$\begin{array}{c}\overrightarrow{\mathrm{r}}=⟨0,{\mathrm{y}}_{\mathrm{f}},0⟩-⟨0,{\mathrm{y}}_{\mathrm{i}},0⟩\\ =⟨0,\mathrm{\Delta y},0⟩\end{array}$

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

The magnitude of the distance vector is given by,

$\begin{array}{c}\left|\overrightarrow{\mathrm{r}}\right|=\sqrt{{\left(0\right)}^2+{\left(\mathrm{\Delta y}\right)}^2+{\left(0\right)}^2}\\ =\mathrm{\Delta y}\end{array}$

The unit vector is given by,

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

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

Substitute values in the above,

$\begin{array}{c}\widehat{\mathrm{r}}=\frac{⟨0,\mathrm{\Delta y},0⟩}{\mathrm{\Delta y}}\\ =⟨0,1,0⟩\end{array}$

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$9.1\times {10}^9\text{\hspace{0.17em}N}\cdot {\text{m}}^2{\text{/C}}^2$,$\widehat{\mathrm{r}}$is the unit vector.

Substitute all the values in the above equation.

$\begin{array}{c}⟨0,4104,0⟩\text{N/C}=\frac{(9.1\times {10}^9\text{\hspace{0.17em}N}\cdot {\text{m}}^2{\text{/C}}^2)(1.6\times {10}^{-19}\text{\hspace{0.17em}C})}{{\left(\mathrm{\Delta y}\right)}^2}⟨0,1,0⟩\\ \mathrm{\Delta y}=\sqrt{\frac{(9.1\times {10}^9\text{\hspace{0.17em}N}\cdot {\text{m}}^2{\text{/C}}^2)(1.6\times {10}^{-19}\text{\hspace{0.17em}C})}{⟨0,4104,0⟩\text{N/C}}⟨0,1,0⟩}\\ =5.92\times {10}^{-7}\text{\hspace{0.17em}m}\end{array}$

Thus, the proton is placed below the observation point origin at a displacement of $5.92\times {10}^{-7}\text{\hspace{0.17em}m}$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,

$\begin{array}{c}\overrightarrow{\mathrm{r}}=⟨0,{\mathrm{y}}_{\mathrm{f}},0⟩-⟨0,{\mathrm{y}}_{\mathrm{i}},0⟩\\ =⟨0,\mathrm{\Delta y},0⟩\end{array}$

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

The unit vector is given y,

$\begin{array}{c}\widehat{\mathrm{r}}=\frac{⟨0,\mathrm{\Delta y},0⟩}{\mathrm{\Delta y}}\\ =⟨0,-1,0⟩\end{array}$

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$9.1\times {10}^9\text{\hspace{0.17em}N}\cdot {\text{m}}^2{\text{/C}}^2$,$\widehat{\mathrm{r}}$is the unit vector.

Substitute all the values in the above,

$\begin{array}{c}⟨0,4104,0⟩\text{N/C}=\frac{(9.1\times {10}^9\text{\hspace{0.17em}N}\cdot {\text{m}}^2{\text{/C}}^2)(-1.6\times {10}^{-19}\text{\hspace{0.17em}C})}{{\left(\mathrm{\Delta y}\right)}^2}⟨0,-1,0⟩\\ \mathrm{\Delta y}=\sqrt{\frac{(9.1\times {10}^9\text{\hspace{0.17em}N}\cdot {\text{m}}^2{\text{/C}}^2)(-1.6\times {10}^{-19}\text{\hspace{0.17em}C})}{⟨0,4104,0⟩\text{N/C}}⟨0,-1,0⟩}\\ =5.92\times {10}^{-7}\text{\hspace{0.17em}m}\end{array}$

Thus, the electron is placed above the origin on the y-direction with a displacement of $5.92\times {10}^{-7}\text{\hspace{0.17em}m}$.