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Electrostatics question

2025 · 23 Jan · Shift 2 · Q60
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Electrostatics question

2025 · 23 Jan · Shift 2 · Q60

JEE MainPhysicsElectrostaticsMCQ+4 / −1
Two point charges −4μc-4 \mu \mathrm{c}−4μc and 4μc4 \mu \mathrm{c}4μc, constituting an electric dipole, are placed at (−9,0,0)cm(-9,0,0) \mathrm{cm}(−9,0,0)cm and (9,0,0)cm(9,0,0) \mathrm{cm}(9,0,0)cm in a uniform electric field of strength 104NC−110^4 \mathrm{NC}^{-1}104NC−1. The work done on the dipole in rotating it from the equilibrium through 180∘180^{\circ}180∘ is :
  1. A
    18.4 mJ
  2. B
    12.4 mJ
  3. C
    16.4 mJ
  4. D
    14.4 mJ
View written solutionFree

Correct answer: D

  1. Given data
  • Charges: −4 μC-4\,\mu C−4μC and +4 μC+4\,\mu C+4μC
  • Positions: (−9,0,0) cm(-9,0,0)\,cm(−9,0,0)cm and (9,0,0) cm(9,0,0)\,cm(9,0,0)cm
  • Uniform electric field: E=104 N/CE=10^4\,N/CE=104N/C

So the separation between the charges is 2a=18 cm=0.18 m2a = 18\,cm = 0.18\,m2a=18cm=0.18m

  1. Dipole moment

Magnitude of dipole moment is p=q×(separation)p = q \times (\text{separation})p=q×(separation) p=4×10−6×0.18p = 4\times 10^{-6} \times 0.18p=4×10−6×0.18 p=7.2×10−7 C mp = 7.2\times 10^{-7}\,C\,mp=7.2×10−7Cm

  1. Potential energy of a dipole in uniform electric field

The potential energy is U=−pEcos⁡θU = -pE\cos\thetaU=−pEcosθ

  • In stable equilibrium, dipole is aligned with the field, so θ=0\theta=0θ=0. Ui=−pEU_i = -pEUi​=−pE

  • After rotation through 180∘180^\circ180∘, θ=180∘\theta=180^\circθ=180∘. Uf=−pEcos⁡180∘=+pEU_f = -pE\cos 180^\circ = +pEUf​=−pEcos180∘=+pE

  1. Work done in rotating from equilibrium through 180∘180^\circ180∘

The work done by an external agent is the increase in potential energy: W=Uf−UiW = U_f - U_iW=Uf​−Ui​ W=pE−(−pE)=2pEW = pE - (-pE) = 2pEW=pE−(−pE)=2pE

Now substitute: W=2×7.2×10−7×104W = 2 \times 7.2\times 10^{-7} \times 10^4W=2×7.2×10−7×104 W=14.4×10−3 JW = 14.4\times 10^{-3}\,JW=14.4×10−3J W=14.4 mJW = 14.4\,mJW=14.4mJ

  1. Option check

Thus the correct option is: D  :  14.4 mJ\boxed{D\; :\; 14.4\,mJ}D:14.4mJ​

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