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Question

A dipole with dipole moment 3.60 nCm is oriented at 60 to an electric field of magnitude 4×109 N/C, how much work is required to rotate the dipole until it is antiparallel to field.

A
21.6 J
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B
3.6 J
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C
14.4 J
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D
7.2 J
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Solution

The correct option is A 21.6 J
Potential energy of a dipole at an angle θ with electric field
U=p.E=pE cos θ
To rotate the dipole (slowly), we need to work on dipole so that the potential energy is increased. W=ΔU=UfUi
Ui=pE cos 60=pE2
Uf=pE cos 180=+pE
W=pE(pE2)=3pE2
=32×3.6×109×4×109
=3×3.6×2=7.2×3=21.6 J

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