A dipole is located at the origin and is composed of charged particle with charge $\mathbf{+}\mathbf{e}$and $\mathbf{-}\mathbf{e}$, separated by a distance $\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{\text{\hspace{0.17em}m}}$along the $\mathbf{x}$axis. The charge $\mathbf{+}\mathbf{e}$is on the $\mathbf{+}\mathbf{x}$axis. Calculate the electric field due to this dipole at a location $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{-}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{\text{\hspace{0.17em}m}}$ .

The given data can be listed below as,

• The distance of separation between charged particles is, $\mathrm{p}=6\times {10}^{-10\text{}}\text{\hspace{0.17em}m}$.
• The dipole location of charged particles is, $\mathrm{r}=-5\times {10}^{-8}\text{\hspace{0.17em}m}$.

An electric field is a way to show the characteristic of the electrical environment having charges in the particular system.The electric field in space at any point shows the force acted on the unit's positive charge if it is situated at that specific point.

The expression for the electric field is given as,

${\mathrm{E}}_{}=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{\mathrm{qp}}{{\mathrm{r}}^3}$

Here,$\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}$ which is Coulomb’s constant and its value $9\times {10}^9\text{\hspace{0.17em}N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$, $p$is the distance through which charged particles are separated and its value is $6\times {10}^{-10}\text{\hspace{0.17em}m}$, $q$is the charge on electron and its value is $1.6\times {10}^{-19}\text{\hspace{0.17em}C}$, $r$is the location of charge from centre point and its value is $-5\times {10}^{-8}\text{\hspace{0.17em}m}$.

Substitute $9\times {10}^9\text{\hspace{0.17em}N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$for $\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}$, $6\times {10}^{-10}\text{\hspace{0.17em}m}$for $p$, and $-5\times {10}^{-8}\text{\hspace{0.17em}m}$for $r$,$1.6\times {10}^{-19}\text{\hspace{0.17em}C}$for $q$.

$\begin{array}{c}\mathrm{E}=(9\times {10}^9\text{\hspace{0.17em}N}\cdot {\text{m}}^2{\text{/C}}^2)\frac{1.6\times {10}^{-19}\text{\hspace{0.17em}C}\times 6\times {\text{10}}^{-10}\text{\hspace{0.17em}m}}{{(-5\times {10}^{-8}\text{\hspace{0.17em}m)}}^3}\\ \mathrm{E}=-6912\times {10}^{-3}\text{\hspace{0.17em}N/C}\\ =-6.912\text{\hspace{0.17em}N/C}\end{array}$

Thus, the electric field due to this dipole is $-6.912\text{\hspace{0.17em}N/C}$.