A dipole is located at the origin and is composed of charged particles with charge$\mathbf{+}\mathit{e}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{-}\mathit{e}$, separated by a distance $\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{}\mathbf{\text{m}}$along the x axis. Calculate the magnitude of electric field due to this dipole at a location$\left(0,3\times {10}^{-8},0\right)\mathbf{}\mathit{m}\mathbf{.}$

The given data is listed below,

• Distance between the charged particles is, $p=2\times {10}^{-10}\text{m}$
• Location of charged particles is, $r=\left(0,3\times {10}^{-8},0\right)\text{m}$

The electric field is a region around a charged particle where its electric effect exists. When other charged particles are placed in that region then, there is a generation of electric force.

The expression for the magnitude of the net electric field is as follows,

$$\begin{gathered} E_{net}=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{qp}{r^3} \\ =k\frac{qp}{r^3} \end{gathered}$$

Here, $k=\frac{1}{4\pi {\epsilon }_o}$is Coulomb’s constant and with value $9x{10}^{9}N.m^2/C^2,p$is the distance through which charged particles are separated, $q$is the charge on electron and with value is $1.6\times {10}^{-19}C,r$is the location of charge from centre point.

For $k=9\times {10}^9\text{N}·{\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$, $p=2\times {10}^{-10}\text{m}$, $q=1.6\times {10}^{-19}\text{C}$, and $r=3\times {10}^{-8}\text{m}$.

$$\begin{gathered} E_{net}=\left(9x{10}^{9}N.m^2/C^2\right)\frac{\left(1.6\times {10}^{-19}C\right)\times \left(2\times {10}^{-10}m\right)}{{\left(3x{10}^{-8}m\right)}^3} \\ =1.067\times {10}^{4}N/C \end{gathered}$$

Thus, the net electric field due to dipole is $1.067\times {10}^4\text{N}/\text{C}$.