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# 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${\overrightarrow E _1} = E\widehat i$, it experiences a torque $\overrightarrow {{T_1}} = \tau \widehat k$. When subjected to another electric field $\overrightarrow {{E_2}} = \sqrt 3 {E_1}\widehat j$, it experiences torque $\overrightarrow {{T_2}} = - \overrightarrow {{T_1}}$. The angle $\theta$ is:(a) 90$^\circ$(b) 30$^\circ$(c) 45$^\circ$(d) 60$^\circ$

Last updated date: 18th Jul 2024
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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:
$\theta = {\tan ^{ - 1}}\dfrac{{{p_y}}}{{{p_x}}}$ …… (2)
where,
${p_y}$ is the y component of the vector $\overrightarrow p$
${p_x}$ is the x component of the vector $\overrightarrow p$

Given:
1. Electric field (1) ${\overrightarrow E _1} = E\widehat i$
2. Torque (1) $\overrightarrow {{T_1}} = \tau \widehat k$
3. Electric field (2) $\overrightarrow {{E_2}} = \sqrt 3 {E_1}\widehat j$
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:
$\overrightarrow p = {p_x}\widehat i + {p_y}\widehat j$

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

Step 3 of 5:
Find Electric field (2):
$\overrightarrow {{E_2}} = \sqrt 3 {E_1}\widehat j$
$\overrightarrow {{E_2}} = \sqrt 3 E\widehat j$
Find Torque (2):
$\overrightarrow {{T_2}} = - \overrightarrow {{T_1}} \\ \overrightarrow {{T_2}} = - \tau \widehat k \\$

Step 4 of 5:
Use eq (1) to find Torque (2):
$- \tau \widehat k = ({p_x}\widehat i + {p_y}\widehat j) \times (\sqrt 3 E\widehat j)$
$- \tau \widehat k = ({p_x}\widehat i) \times (\sqrt 3 E\widehat j) + ({p_y}\widehat j) \times (\sqrt 3 E\widehat j) \\ - \tau \widehat k = \sqrt 3 {p_x}E(\widehat i \times \widehat j) \\ \tau \widehat k = - \sqrt 3 {p_x}E\widehat k \\$
Compare the magnitudes of unit vectors on LHS and RHS:
$\tau = - \sqrt 3 {p_x}E$
Rearrange to find ${p_x}$:
${p_x} = - \dfrac{\tau }{{\sqrt 3 E}}$ …… (4)

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

The angle $\theta$ is: (d) 60$^\circ$
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 $\theta$.