Question

The potential energy U of an electric dipole of moment $\overrightarrow{p}$ in an electric field $\overrightarrow{E}$ is given by $U=-\overrightarrow{p}.\overrightarrow{E}$ and electric field due to $\overrightarrow{p}$ at a point of position vector $\overrightarrow{r}$ is given by $\overrightarrow{E}=\frac{1}{4\pi {ϵ}_o{r}^3}\left[3\right(\overrightarrow{p_1}.\hat{r})\hat{r}-\overrightarrow{p_1}],$ where $\overrightarrow{r}$ = $r\hat{r}$ Now, consider two dipoles $\overrightarrow{p_1}$ and $\overrightarrow{p_2}$ ; the dipole $\overrightarrow{p_1}$ is aligned along x - axis at the origin O (0, 0) and the dipole $\overrightarrow{p_2}$ is kept at A (a,a). The minimum energy in terms of $U_o=\frac{p_1{p}_2}{8\pi {\epsilon }_o{r}^3}$, of the dipole system is:

• A. $\sqrt{5}{U}_o$
• B. $U_o$
• C. $\sqrt{2}{U}_o$
• D. $\sqrt{10}{U}_o$

Answer 1

The correct option is D $\sqrt{10}{U}_o$

Given that the dipole $\overrightarrow{p_1}$ is placed at $(0,0)$ and $\overrightarrow{p_2}$ is placed at $(a,a)$.

As $\overrightarrow{p_1}$ is aligned along x-axis, we have $\overrightarrow{p_1}=p_1\hat{i}$

Position vector from O to A is $\overrightarrow{r}=a(\hat{i}+\hat{j})$

Thus unit vector from O to A $\hat{r}=\frac{\overrightarrow{r}}{|\overrightarrow{r}|}=\frac{a(\hat{i}+\hat{j})}{a\sqrt{2}}=\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}$

The electric field at $(a,a)$ due to the dipole at origin is

$\overrightarrow{E}=\frac{1}{4\pi {ϵ}_0{r}^3}[3(p_1\hat{i}\cdot (\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}))(\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j})-p_1\hat{i}]=\frac{1}{4\pi {ϵ}_0{r}^3}[\frac{3}{2}{p}_1\hat{i}+\frac{3}{2}{p}_1\hat{j}-p_i\hat{i}]$

$\Rightarrow \overrightarrow{E}=\frac{p_1}{8\pi {ϵ}_0{r}^3}(\hat{i}+3\hat{j})$

Thus, potential energy of the system of two dipoles is $U=-\overrightarrow{p_2}\cdot \overrightarrow{E}$

But, minimum of dot product of two vectors is the product of their magnitudes.

Thus, $U_{min}=|\overrightarrow{p_2}||\overrightarrow{E}|=p_2\frac{p_1}{8\pi {ϵ}_0{a}^3}|\hat{i}+3\hat{j}|=\frac{p_1{p}_2}{8\pi {ϵ}_0{r}^3}\sqrt{10}=\sqrt{10}{U}_0$