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

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Aicharya · Accepted Answer

Work done $\mathrm{W}=\mathrm{p} E\left[\cos 	heta_0-\cos 	hetaight]$ Given, $\mathrm{W}=\mathrm{pE}\left(1-\cos 180^{\circ}ight)$ $	herefore \quad \mathrm{W}=\mathrm{pE}\left(1-\cos 180^{\circ}ight)$ $\begin{aligned} & \mathrm{W}=\mathrm{pE}(1-(-1)) \ & \mathrm{W}=2 \mathrm{pE} \ & \mathrm{pE}=\frac{1}{2} \mathrm{~W}\end{aligned}$ $	herefore \quad$ When $	heta=60^{\circ}$ $\begin{aligned} & \mathrm{W}^{\prime}=\mathrm{pE}\left(1-\cos 60^{\circ}ight) \ & \mathrm{W}^{\prime}=\mathrm{pE}\left(1-\frac{1}{2}ight) \ & \mathrm{W}^{\prime}=\frac{1}{2} \mathrm{pE} \ & \mathrm{W}^{\prime}=\frac{1}{2} 	imes \frac{1}{2} \mathrm{~W} \ & \mathrm{~W}^{\prime}=\frac{1}{4} \mathrm{~W}\end{aligned}$ *

The work done in rotating a dipole placed parallel to the electric field through $180^{\circ}$ is 'w'. So…

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