Question

Derive Equation $23.11$for the field ${\stackrel{\rightharpoonup }{E}}_{dipole}$in the plane that bisects an electric dipole.

Short answer

The Equitorial time at field point is $\frac{P}{4\pi {\epsilon }_0{r}^3}$.

Step-by-step solution

Step: 1 Electric field:

The electric field radial component by

$\begin{array}{l}{E}_r=\frac{-\mathrm{\partial }}{\mathrm{\partial }r}\\ E_r=\frac{-\mathrm{\partial }}{\mathrm{\partial }r}\left(\frac{P\cos \theta }{4\pi {\epsilon }_0{r}^2}\right)\\ E_r=\frac{-P\cos \theta }{4\pi {\epsilon }_0}\left(\frac{-L}{r^3}\right)\\ E_r=\frac{2P\cos \theta }{4\pi {\epsilon }_0{r}^3}\end{array}$

Step; 2 Transverse electric field:

The component transverse of electric field by

$\begin{array}{l}{E}_{\theta }=\frac{-1}{r}\left(\frac{\mathrm{\partial }E}{\mathrm{\partial }\theta }\right)\\ E_{\theta }=\frac{-1}{r}\frac{\mathrm{\partial }}{\mathrm{\partial }\theta }\left(\frac{P\cos \theta }{4\pi {\epsilon }_0{r}^2}\right)\\ E_{\theta }=\frac{P\sin \theta }{4\pi {\epsilon }_0{r}^3}\end{array}$

Step: 3 Resultant field:

The resultant field by,

$\begin{array}{l}\left.E=\sqrt{{\left(\frac{2P\cos \theta }{4\pi {\epsilon }_0{r}^3}\right)}^2+{\left(\frac{P\sin \theta }{4\pi {\epsilon }_0{r}^3}\right)}^2}\right]\\ E=\frac{P}{4\pi {\epsilon }_0{r}^3}\sqrt{4{\cos }^2\theta +{\sin }^2\theta }\\ E=\frac{P}{4\pi {\epsilon }_0{r}^3}\sqrt{1+3{\cos }^2\theta }\dots \dots \dots \dots \dots \dots \end{array}$

Step; 4 Field angle:

The angle at field,

$\begin{array}{l}\tan \varphi =\frac{E_{\theta }}{E_r}\\ \tan \varphi =\left[\frac{P\sin \theta }{4\pi {\epsilon }_0{r}^3}\right]\\ \tan \varphi =\left[\frac{2P\cos \theta }{4\pi {\epsilon }_0{r}^3}\right]\\ \tan \varphi =\frac{1}{2}\frac{\sin \theta }{\cos \theta }\dots \dots \dots \dots \dots \dots \dots .\end{array}$

Because of magnitude and direction in dipole,the angle on equitorial time at $\theta ={90}^{\circ }$

Therefore,the equitorial time on field point is$\frac{P}{4\pi {\epsilon }_0{r}^3}.$