The work done in rotating a dipole placed par | vedclass

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(D) The work done in rotating an electric dipole in an electric field is given by $W = pE(\cos 	heta_1 - \cos 	heta_2)$.Initially,the dipole is para

Vedclass · Accepted Answer

$\frac{w}{4}$

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(D) The work done in rotating an electric dipole in an electric field is given by $W = pE(\cos 	heta_1 - \cos 	heta_2)$.Initially,the dipole is parallel to the field,so $	heta_1 = 0^{\circ}$.For the first case,rotating through $180^{\circ}$ means $	heta_2 = 180^{\circ}$.$w = pE(\cos 0^{\circ} - \cos 180^{\circ}) = pE(1 - (-1)) = 2pE$.Therefore,$pE = \frac{w}{2}$.For the second case,rotating through $60^{\circ}$ means $	heta_2 = 60^{\circ}$.$W' = pE(\cos 0^{\circ} - \cos 60^{\circ}) = pE(1 - \frac{1}{2}) = pE(\frac{1}{2})$.Substituting $pE = \frac{w}{2}$ into the equation:$W' = (\frac{w}{2}) 	imes (\frac{1}{2}) = \frac{w}{4}$.

The work done in rotating a dipole placed parallel to the electric field through $180^{\circ}$ is $w$. What is the work done in rotating it through $60^{\circ}$? $(\cos 0^{\circ}=1, \cos 60^{\circ}=\frac{1}{2}, \cos 180^{\circ}=-1)$

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