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QJEEM17Ph6e8a: JEE Main 2017 Physics - Class 12 JEEM17Ph6e8a Solution: An electric dipole has a fixed dipole moment , which makes…

An electric dipole has a fixed dipole moment p→\overrightarrow p , which makes angle θ\theta with respect to x-axis. When subjected to an electric field E1→=Ei^\mathop {{E_1}}\limits^ \to = E\widehat i , it experiences a torque T1→=τk^\overrightarrow {{T_1}} = \tau \widehat k . When subjected to another electric field E2→=3E1j^\mathop {{E_2}}\limits^ \to = \sqrt 3 {E_1}\widehat j it experiences a torque T2→=−T1→\mathop {{T_2}}\limits^ \to = \mathop { - {T_1}}\limits^ \to . The angle θ\theta is:

Multiple Choice Options

  • A 90o
  • B 45o
  • C 30o
  • D 60o

Step-by-Step Verified Solution



Torque experienced by the dipole in an electric field,

TT = pE sin θ\theta

T→=p→×E→\overrightarrow T = \overrightarrow p \times \overrightarrow E

p→=pcos⁡θi^+psin⁡θj^\overrightarrow p = p\cos \theta \widehat i + p\sin \theta \widehat j

E1→=Ei^\mathop {{E_1}}\limits^ \to = E\widehat i

T1→=P→×E→1\overrightarrow {{T _1}} = \overrightarrow P \times {\overrightarrow E _1}

= ( pcos⁡θi^+psin⁡θj^p\cos \theta \widehat i + p\sin \theta \widehat j ) ×\times E(i^)E\left( {\widehat i} \right)

= pE sin θ\theta (−k^)\left( { - \widehat k} \right)

E2→=3E1j^\mathop {{E_2}}\limits^ \to = \sqrt 3 {E_1}\widehat j

T2→=\overrightarrow {{T _2}} = ( pcos⁡θi^+psin⁡θj^p\cos \theta \widehat i + p\sin \theta \widehat j ) ×\times 3E1j^\sqrt 3 {E_1}\widehat j

= 3pEcos⁡θ(k^)\sqrt 3 pE\cos \theta \left( {\widehat k} \right)

Now given, T2→\overrightarrow {{T _2}} = −T1→-\overrightarrow {{T _1}}

⇒\Rightarrow 3pEcos⁡θ(k^)\sqrt 3 pE\cos \theta \left( {\widehat k} \right) = -pE sin θ\theta (−k^)\left( { - \widehat k} \right)

⇒\Rightarrow tan⁡θ=3\tan \theta = \sqrt 3

⇒\Rightarrow θ\theta = 60o

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