JEE MAIN - Physics (2017 (Offline) - No. 10)
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 $$\mathop {{E_1}}\limits^ \to = E\widehat i$$ , it experiences a torque $$\overrightarrow {{T_1}} = \tau \widehat k$$ . When subjected to another electric
field $$\mathop {{E_2}}\limits^ \to = \sqrt 3 {E_1}\widehat j$$ it experiences a torque $$\mathop {{T_2}}\limits^ \to = \mathop { - {T_1}}\limits^ \to $$ . The angle $$\theta$$ is:
90o
45o
30o
60o
Explanation
Torque experienced by the dipole in an electric field,
$$T $$ = pE sin$$\theta $$
$$\overrightarrow T = \overrightarrow p \times \overrightarrow E $$
$$\overrightarrow p = p\cos \theta \widehat i + p\sin \theta \widehat j$$
$$\mathop {{E_1}}\limits^ \to = E\widehat i$$
$$\overrightarrow {{T _1}} = \overrightarrow P \times {\overrightarrow E _1}$$
= ($$p\cos \theta \widehat i + p\sin \theta \widehat j$$) $$ \times $$ $$E\left( {\widehat i} \right)$$
= pE sin$$\theta $$$$\left( { - \widehat k} \right)$$
$$\mathop {{E_2}}\limits^ \to = \sqrt 3 {E_1}\widehat j$$
$$\overrightarrow {{T _2}} = $$($$p\cos \theta \widehat i + p\sin \theta \widehat j$$) $$ \times $$ $$\sqrt 3 {E_1}\widehat j$$
= $$\sqrt 3 pE\cos \theta \left( {\widehat k} \right)$$
Now given, $$\overrightarrow {{T _2}}$$ = $$-\overrightarrow {{T _1}}$$
$$ \Rightarrow $$ $$\sqrt 3 pE\cos \theta \left( {\widehat k} \right)$$ = -pE sin$$\theta $$$$\left( { - \widehat k} \right)$$
$$ \Rightarrow $$ $$\tan \theta = \sqrt 3 $$
$$ \Rightarrow $$ $$\theta $$ = 60o
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