An electric dipole of dipole moment is $6.0 \times 10^{-6}\,Cm$ placed in a uniform electric field of $1.5 \times 10^3\,NC ^{-1}$ in such a way that dipole moment is along electric field. The work done in rotating dipole by $180^{\circ}$ in this field will be $.........\,mJ$
JEE MAIN 2023, Medium
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The work done in rotating the electric dipole $=\Delta U$
$= U _{ f }- U _{ i }$
$=\left(- pE \cos \left(180^{\circ}\right)\right)-\left(- pE \cos \left(0^{\circ}\right)\right)$
$= pE + pE =2 pE$
$=2 \times 6 \times 10^{-6} \times 1.5 \times 10^3=18\,mJ$
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