Question

Two point dipoles $p\hat{k}and\frac{p}{2}\hat{k}$ are located at $(0,0,0)$ and $(1m,0,2m)$ respectively. The resultant electric field due to the dipoles at the point $(1m,0,0)$ is $\frac{x}{y}\times \frac{p}{4\pi {ϵ}_0}(-\hat{k}).$ The value of $xy$ is .

Answer 1

We know that,
The electric field due to an electric dipole at a distance $r$ from its center in axial position is $\overrightarrow{E_{axial}}=\frac{1}{4\pi {ϵ}_0}\times \frac{-2\overrightarrow{p}}{r^3}.$
The electric field due to an electric dipole at a distance $r$ from its center in equatorial position is $\overrightarrow{E_{eq}}=\frac{1}{4\pi {ϵ}_0}\times \frac{\overrightarrow{p}}{r^3}$


For dipole with moment $p\hat{k},\text{the point}(1,0,0)$ is an equatorial point at $r_1=1m$.
Electric field at point(1,0,0) because of dipole with moment $p\hat{k}$ is
$\overrightarrow{E_1}=\frac{1}{4\pi {ϵ}_0}\times \frac{p}{1^3}(-\hat{k})$

For dipole with moment $\frac{p}{2}\hat{k},\text{the point}(1,0,0)$ is an axial point at $r_2=2m$
Electric field at point (1,0,0) because of dipole with moment $\frac{p}{2}\hat{k}$ is

$\overrightarrow{E_2}=\frac{1}{4\pi {ϵ}_0}\times \frac{2\left(\frac{p}{2}\right)}{2^3}\hat{k}$

Now,
$\overrightarrow{E_{net}}=\overrightarrow{E_1}+\overrightarrow{E_2}$
$\overrightarrow{E_{net}}=\frac{1}{4\pi {ϵ}_0}\times p[-1+\frac{1}{8}]\hat{k}$
$\overrightarrow{E_{net}}=-\frac{7}{8}\times \frac{p}{4\pi {ϵ}_o}\hat{k}$
$\Rightarrow x=7,y=8andxy=56$