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Electrostatics question

2023 · 8 Apr · Shift 1 · Q71
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Electrostatics question

2023 · 8 Apr · Shift 1 · Q71

JEE MainPhysicsElectrostaticsNumerical+4 / −1
An electric dipole of dipole moment is 6.0×10−6 Cm6.0 \times 10^{-6} ~\mathrm{C m}6.0×10−6 Cm placed in a uniform electric field of 1.5×103 NC−11.5 \times 10^{3} ~\mathrm{NC}^{-1}1.5×103 NC−1 in such a way that dipole moment is along electric field. The work done in rotating dipole by 180∘180^{\circ}180∘ in this field will be ‾\underline{\hspace{2cm}}​mJ\mathrm{m J}mJ.
Numerical answer
View written solutionFree

Correct answer: 18

  1. Potential energy of a dipole in a uniform electric field

The potential energy is

U=−pEcos⁡θU = -pE\cos\thetaU=−pEcosθ

where:

  • p=6.0×10−6 C mp = 6.0 \times 10^{-6}\,\text{C m}p=6.0×10−6C m
  • E=1.5×103 N C−1E = 1.5 \times 10^{3}\,\text{N C}^{-1}E=1.5×103N C−1
  1. Initial position

The dipole is along the field, so

θi=0∘\theta_i = 0^\circθi​=0∘

Hence,

Ui=−pEcos⁡0∘=−pEU_i = -pE\cos 0^\circ = -pEUi​=−pEcos0∘=−pE
  1. Final position after rotation by 180∘180^\circ180∘

Now,

θf=180∘\theta_f = 180^\circθf​=180∘

So,

Uf=−pEcos⁡180∘=+pEU_f = -pE\cos 180^\circ = +pEUf​=−pEcos180∘=+pE
  1. Work done in rotating the dipole

Work done by external agent in rotating it slowly equals increase in potential energy:

W=Uf−Ui=pE−(−pE)=2pEW = U_f - U_i = pE - (-pE) = 2pEW=Uf​−Ui​=pE−(−pE)=2pE
  1. Substitute values
W=2×(6.0×10−6)×(1.5×103)W = 2 \times (6.0 \times 10^{-6}) \times (1.5 \times 10^3)W=2×(6.0×10−6)×(1.5×103) W=2×9.0×10−3W = 2 \times 9.0 \times 10^{-3}W=2×9.0×10−3 W=18×10−3 JW = 18 \times 10^{-3}\,\text{J}W=18×10−3J W=18 mJW = 18\,\text{mJ}W=18mJ
  1. Final answer

The required integer is

18\boxed{18}18​
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