Question

Starting from Eq. 24-30, derive an expression for the electric field due to a dipole at a point on the dipole axis.

Short answer

$\mathrm{E}=\frac{1}{2{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\cdot \frac{\mathrm{p}}{{\mathrm{r}}^3}$

Step-by-step solution

Understanding the concept

After reading the question, consider an electric dipole of charge $\pm \mathrm{q}$ to be separated by ‘d’.

The electric potential at a distance ‘r’ with angle $\mathrm{\theta }$from the middle of the electric dipole is given as:

$\mathrm{V}=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\cdot \frac{\mathrm{p}\text{\hspace{0.17em}cos\hspace{0.17em}}\mathrm{\theta }}{{\mathrm{r}}^2},\text{where}\mathrm{p}=\mathrm{qd}$

$\text{If\hspace{0.17em}}\mathrm{\theta }=0,\text{cos\hspace{0.17em}}0=1$

$\mathrm{V}=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\cdot \frac{\mathrm{p}}{{\mathrm{r}}^2}$

Step 2: Derive an expression for the electric field due to a dipole at a point on the dipole axis

According to the definition of a potential gradient,

$\begin{array}{c}\mathrm{E}=\frac{-\partial \mathrm{V}}{\partial \mathrm{r}}\\ \mathrm{E}=-\frac{\mathrm{p}}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{\mathrm{d}\left({\mathrm{r}}^{-2}\right)}{\mathrm{dr}}\\ \mathrm{E}=+2\times \frac{\mathrm{p}}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}{\mathrm{r}}^{-3}\\ \mathrm{E}=\frac{1}{2{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\cdot \frac{\mathrm{p}}{{\mathrm{r}}^3}\end{array}$

Hence, the expression is$\mathrm{E}=\frac{1}{2{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\cdot \frac{\mathrm{p}}{{\mathrm{r}}^3}$