Question

An electric dipole ( dipole moment p) is placed at a radial distance r>>a ( Where a is dipole length) from an infinite line of charge having linear charge density $+\lambda$. dipole moment vector is aligned along the radial vector $\overrightarrow{r}$ force experienced by the dipole is:-

• A. $\frac{\lambda p}{2\pi {\epsilon }_o{r}^2},\text{attractive}$
• B. $\frac{\lambda p}{2\pi {\epsilon }_o{r}^3},\text{attractive}$
• C. $\frac{\lambda p}{2\pi {\epsilon }_o{r}^2},\text{repulsive}$
• D. $\frac{\lambda p}{2\pi {\epsilon }_o{r}^3},\text{repulsive}$

Answer 1

The correct option is A $\frac{\lambda p}{2\pi {\epsilon }_o{r}^2},\text{attractive}$

We know,

$E=\frac{1}{4\pi {ϵ}_0}\frac{q}{{(r-a)}^2}+\frac{1}{4\pi {ϵ}_0}\frac{-q}{{(r+a)}^2}$

$=\frac{1}{4\pi {ϵ}_0}[\frac{q}{{(r-a)}^2}-\frac{q}{{(r+a)}^2}]$

$=\frac{1}{4\pi {ϵ}_0}4aq\frac{r}{{(r^2-a^2)}^2}$

The quality $2qa$ is defined as the dipole moment $\left(P\right)$ and so we can substitute $p=2qa$ with the substitution we now get

$E=\frac{1}{2\pi {ϵ}_0}\frac{Pr}{{(r^2-a^2)}^2}$

Consider the far field case (i.e the case where $r>>a$)

Then one can see that $r^2>>a^2$ and that $r^2-a^2$ approaches $r^2$, which allows us to simply simplify the equation in the far field.

$E=\frac{1}{2\pi {ϵ}_0}\frac{P}{r^3}$

$=\frac{1}{2\pi {ϵ}_0}\frac{\lambda P}{r^2}$