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JEE Main 202528 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
28 January
Shift
Evening Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2025 · 28 January · Evening Shift · Physics · Electrostatics

An electric dipole of dipole moment 6 × 10^ -6 Cm is placed in...

An electric dipole of dipole moment 6×10−6Cm6 \times 10^{-6} \mathrm{Cm} is placed in uniform electric field of magnitude 106 V/m10^6 \mathrm{~V} / \mathrm{m}. Initially, the dipole moment is parallel to electric field. The work that needs to be done on the dipole to make its dipole moment opposite to the field, will be ____ J.

Correct answer

0

Step-by-step solution

Work done in rotating a dipole =ΔU or  =\Delta U \text{\text{ or } }W=(−PEcos⁡θf)−(−PEcos⁡θi)=2PE(∵θf=180∘ and θi=0∘)=(2×6×10−6×106)J=12J\begin{aligned} W & =\left(-P E \cos \theta_f\right)-\left(-P E \cos \theta_i\right) \\ & =2 P E \quad\left(\because \theta_f=180^{\circ} \text{ and } \theta_i=0^{\circ}\right) \\ & =\left(2 \times 6 \times 10^{-6} \times 10^6\right) \mathrm{J}=12 \mathrm{ J} \end{aligned}$$
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