Question

An electric dipole has a fixed dipole moment $\overrightarrow{p}$, which makes angle $\theta$ with respect to x-axis. When subjected to an electric field${\overrightarrow{E}}_1=E\hat{i}$, it experiences a torque $\overrightarrow{T_1}=\tau \hat{k}$. When subjected to another electric field $\overrightarrow{E_2}=\sqrt{3}{E}_1\hat{j}$, it experiences torque $\overrightarrow{T_2}=-\overrightarrow{T_1}$. The angle $\theta$ is:
(a) 90${}^{\circ }$
(b) 30${}^{\circ }$
(c) 45${}^{\circ }$
(d) 60${}^{\circ }$

Answer 1

Hint: Use the given Electric field vectors and resultant Torque to find the components of dipole moment. This can then be used to find the angle it makes with x-axis.

Formula used:
Torque:
$\overrightarrow{T}=\overrightarrow{p}\times \overrightarrow{E}$ …… (1)
where,
$\overrightarrow{p}$ is the dipole moment.
$\overrightarrow{E}$ is the Electric field.
Angle made by vector with x-axis:
$\theta ={\tan }^{-1}{\displaystyle \frac{p_y}{p_x}}$ …… (2)
where,
$p_y$ is the y component of the vector $\overrightarrow{p}$
$p_x$ is the x component of the vector $\overrightarrow{p}$

Step-by-step answer:
Given:
1. Electric field (1) ${\overrightarrow{E}}_1=E\hat{i}$
2. Torque (1) $\overrightarrow{T_1}=\tau \hat{k}$
3. Electric field (2) $\overrightarrow{E_2}=\sqrt{3}{E}_1\hat{j}$
4. Torque (2) $\overrightarrow{T_2}=-\overrightarrow{T_1}$

To find: The angle $\overrightarrow{p}$makes with x-axis.

Step 1 of 5:
Let $\overrightarrow{p}$ be the following:
$$\overrightarrow{p}=p_x\hat{i}+p_y\hat{j}$$

Step 2 of 5:
Use eq (1) to find Torque (1):
$\tau \hat{k}=(p_x\hat{i}+p_y\hat{j})\times (E\hat{i})$
$$\begin{gathered} \tau \hat{k}=(p_x\hat{i})\times (E\hat{i})+(p_y\hat{j})\times (E\hat{i}) \\ \tau \hat{k}=p_{y}E(\hat{j}\times \hat{i}) \\ \tau \hat{k}=-p_{y}E\hat{k} \end{gathered}$$
Compare the magnitudes of unit vectors on LHS and RHS:
$\tau =-p_{y}E$
Rearrange to find $p_y$:
$p_y=-{\displaystyle \frac{\tau }{E}}$ ……(3)

Step 3 of 5:
Find Electric field (2):
$\overrightarrow{E_2}=\sqrt{3}{E}_1\hat{j}$
$\overrightarrow{E_2}=\sqrt{3}E\hat{j}$
Find Torque (2):
$$\begin{gathered} \overrightarrow{T_2}=-\overrightarrow{T_1} \\ \overrightarrow{T_2}=-\tau \hat{k} \end{gathered}$$

Step 4 of 5:
Use eq (1) to find Torque (2):
$-\tau \hat{k}=(p_x\hat{i}+p_y\hat{j})\times (\sqrt{3}E\hat{j})$
$$\begin{gathered} -\tau \hat{k}=(p_x\hat{i})\times (\sqrt{3}E\hat{j})+(p_y\hat{j})\times (\sqrt{3}E\hat{j}) \\ -\tau \hat{k}=\sqrt{3}{p}_{x}E(\hat{i}\times \hat{j}) \\ \tau \hat{k}=-\sqrt{3}{p}_{x}E\hat{k} \end{gathered}$$
Compare the magnitudes of unit vectors on LHS and RHS:
$\tau =-\sqrt{3}{p}_{x}E$
Rearrange to find $p_x$:
$p_x=-{\displaystyle \frac{\tau }{\sqrt{3}E}}$ …… (4)

Step 5 of 5:
Use eq (2) to find the angle $\theta$ made by $\overrightarrow{p}$ with the x-axis:
$\theta ={\tan }^{-1}{\displaystyle \frac{p_y}{p_x}}$
$p_y$ and $p_x$are given in eq (4) and (3) respectively:
$$\begin{gathered} \theta ={\tan }^{-1}{\displaystyle \frac{({\displaystyle \frac{-\tau }{E}})}{({\displaystyle \frac{-\tau }{\sqrt{3}E}})}} \\ \theta ={\tan }^{-1}\sqrt{3} \\ \theta ={60}^{\circ } \end{gathered}$$

Correct Answer:
The angle $\theta$ is: (d) 60${}^{\circ }$

Additional Information: In dipole moment we approximate charge to be separated by very small and finite distance which lead us to calculate torque and force acting on the dipole altogether. Otherwise, we would have to use coulomb's law for each individual charge of dipole and superposition of fields produced by them.

Note: In questions like these, Assume a general expression for $\overrightarrow{p}$ (dipole moment). Obtain the expressions for $\overrightarrow{T}$. Compare the magnitudes of unit vectors to find the x and y components of $\overrightarrow{p}$. This can be used to find the angle $\theta$.