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JEE Main 202523 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Evening Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2025 · 23 January · Evening Shift · Physics · Electrostatics

Two point charges -4 μ c and 4 μ c , constituting an electric...

Two point charges −4μc-4 \mu c and 4μc4 \mu c, constituting an electric dipole, are placed at (−9,0,0)cm(-9,0,0) \mathrm{cm} and (9,0,0)cm(9,0,0) \mathrm{cm} in a uniform electric field of strength 104NC−110^4 \mathrm{NC}^{-1}. The work done on the dipole in rotating it from the equilibrium through 180∘180^{\circ} is :

Options

  1. A18.4 mJ
  2. B14.4 mJ
  3. C12.4 mJ
  4. D16.4 mJ

Correct answer

B. 14.4 mJ

Step-by-step solution

U=−PEcos⁡θwext=ΔU=Uf−Ui=−PEcos⁡180∘+PEcos⁡0∘wext=2PE=2×(4×10−6)(18)×104=144×10−2=14.4 mJ\begin{aligned} & \mathrm{U}=-\mathrm{PE} \cos \theta \\ & \mathrm{w}_{\mathrm{ext}}=\Delta \mathrm{U}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}=-\mathrm{PE} \cos 180^{\circ}+\mathrm{PE} \cos 0^{\circ} \\ & \mathrm{w}_{\mathrm{ext}}=2 \mathrm{PE} \\ & =2 \times\left(4 \times 10^{-6}\right)(18) \times 10^4 \\ & =144 \times 10^{-2} \\ & =14.4 \mathrm{~mJ}\end{aligned}
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