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Electrostatics question

2017 · Shift 0 · Q53
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Electrostatics question

2017 · Shift 0 · Q53

JEE MainPhysicsElectrostaticsMCQ+4 / −1
An electric dipole has a fixed dipole moment p→\overrightarrow pp​, which makes angle θ\thetaθ with respect to x-axis. When subjected to an electric field E1→=Ei^\mathop {{E_1}}\limits^ \to = E\widehat iE1​→​=Ei, it experiences a torque T1→=τk^\overrightarrow {{T_1}} = \tau \widehat kT1​​=τk. When subjected to another electric field E2→=3E1j^\mathop {{E_2}}\limits^ \to = \sqrt 3 {E_1}\widehat jE2​→​=3​E1​j​ it experiences a torque T2→=−T1→\mathop {{T_2}}\limits^ \to = \mathop { - {T_1}}\limits^ \toT2​→​=−T1​→​. The angle θ\thetaθ is:
  1. A
    90o
  2. B
    45o
  3. C
    30o
  4. D
    60o
View written solutionFree

Correct answer: D

  1. Write the dipole moment in component form

Since the dipole moment p⃗\vec pp​ makes an angle θ\thetaθ with the x-axis,

p⃗=pcos⁡θ i^+psin⁡θ j^\vec p = p\cos\theta\,\hat i + p\sin\theta\,\hat jp​=pcosθi^+psinθj^​
  1. Use torque formula

Torque on an electric dipole in an electric field is

τ⃗=p⃗×E⃗\vec \tau = \vec p \times \vec Eτ=p​×E
  1. First electric field: E⃗1=Ei^\vec E_1 = E\hat iE1​=Ei^

Then

T⃗1=p⃗×E⃗1\vec T_1 = \vec p \times \vec E_1T1​=p​×E1​

Substitute:

T⃗1=(pcos⁡θ i^+psin⁡θ j^)×(Ei^)\vec T_1 = (p\cos\theta\,\hat i + p\sin\theta\,\hat j) \times (E\hat i)T1​=(pcosθi^+psinθj^​)×(Ei^)

Using cross products:

  • i^×i^=0\hat i \times \hat i = 0i^×i^=0
  • j^×i^=−k^\hat j \times \hat i = -\hat kj^​×i^=−k^

So,

T⃗1=pcos⁡θE(0)+psin⁡θE(−k^)\vec T_1 = p\cos\theta E(0) + p\sin\theta E(-\hat k)T1​=pcosθE(0)+psinθE(−k^) T⃗1=−pEsin⁡θ k^\vec T_1 = -pE\sin\theta\,\hat kT1​=−pEsinθk^

Given that T⃗1=τk^\vec T_1 = \tau \hat kT1​=τk^, its magnitude is

τ=pEsin⁡θ\tau = pE\sin\thetaτ=pEsinθ

(We will use the vector relation with the second case.)

  1. Second electric field: E⃗2=3Ej^\vec E_2 = \sqrt{3}E\hat jE2​=3​Ej^​

Then

T⃗2=p⃗×E⃗2\vec T_2 = \vec p \times \vec E_2T2​=p​×E2​

Substitute:

T⃗2=(pcos⁡θ i^+psin⁡θ j^)×(3Ej^)\vec T_2 = (p\cos\theta\,\hat i + p\sin\theta\,\hat j) \times (\sqrt{3}E\hat j)T2​=(pcosθi^+psinθj^​)×(3​Ej^​)

Using cross products:

  • i^×j^=k^\hat i \times \hat j = \hat ki^×j^​=k^
  • j^×j^=0\hat j \times \hat j = 0j^​×j^​=0

So,

T⃗2=pcos⁡θ3E k^\vec T_2 = p\cos\theta\sqrt{3}E\,\hat kT2​=pcosθ3​Ek^

Thus,

T⃗2=3pEcos⁡θ k^\vec T_2 = \sqrt{3}pE\cos\theta\,\hat kT2​=3​pEcosθk^
  1. Use the given condition T⃗2=−T⃗1\vec T_2 = -\vec T_1T2​=−T1​

We have

3pEcos⁡θ k^=−(−pEsin⁡θ k^)\sqrt{3}pE\cos\theta\,\hat k = -(-pE\sin\theta\,\hat k)3​pEcosθk^=−(−pEsinθk^) 3pEcos⁡θ=pEsin⁡θ\sqrt{3}pE\cos\theta = pE\sin\theta3​pEcosθ=pEsinθ

Cancel pEpEpE:

3cos⁡θ=sin⁡θ\sqrt{3}\cos\theta = \sin\theta3​cosθ=sinθ tan⁡θ=3\tan\theta = \sqrt{3}tanθ=3​

Therefore,

θ=60∘\theta = 60^\circθ=60∘
  1. Check options
  • A: 90∘90^\circ90∘ ❌
  • B: 45∘45^\circ45∘ ❌
  • C: 30∘30^\circ30∘ ❌
  • D: 60∘60^\circ60∘ ✅

So the correct answer is D.

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