Question

Calculate electric field due to dipole of dipole momentum ‘$P$’ at distance ‘$r$’ and angle $\theta$.

Answer 1

$E_1=\frac{1}{4\pi {\epsilon }_0}\times \frac{2p\cos \theta }{r^3}$ (along OC) .... (i)

Simlialry, $P$ lies on the equatorial line of dipole with moment ${\overrightarrow{p}}_2$. Hence, the magnitude of electric field intensity $]vec{E}_2$ at $P$ due to ${\overrightarrow{p}}_2$ is

$E_2=\frac{1}{4\pi {\epsilon }_0}-\frac{p\sin \theta }{r^3}$ (opposite to $p_2$) ...(ii)

Hence, the resultant intensity at $P$ is $\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2$.

The magnitude of $\overrightarrow{E}$ is $E=\sqrt{(E_1^2+E_2^2)}$ (as ${\overrightarrow{E}}_1$ and ${\overrightarrow{E}}_2$ are mutally perpendicular).

or $E=\sqrt{{\left(\frac{2p\cos \theta }{4\pi {\epsilon }_0{r}^3}\right)}^2+{\left(\frac{p\sin \theta }{4\pi {\epsilon }_0{r}^3}\right)}^2}$

$=\frac{p}{4\pi {\epsilon }_0{r}^3}\sqrt{4{\cos }^2\theta +{\sin }^2\theta }=\frac{p}{4\pi {\epsilon }_0{r}^3}\sqrt{1+3{\cos }^2\theta }$