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An electric dipole with dipole moment \(p\) is placed with its axis at \(30^{\circ}\) to a uniform electric field. The work done in rotating the dipole to a position where its axis is perpendicular to the field is
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Verified Answer
The correct answer is:
\(\frac{\sqrt{3} p E}{2}\)
Given, for an electric dipole,
\(\begin{aligned}
& \theta_1=30^{\circ} \\
& \theta_2=90^{\circ}
\end{aligned}\)
\(\therefore\) Work done in rotating the dipole from \(\theta_1=30^{\circ}\) to a position, where its axis is perpendicular \(\left(\theta_2=90^{\circ}\right)\) to electric field is given as
\(\begin{aligned}
W & =p E\left(\cos \theta_1-\cos \theta_2\right) \\
& =p E\left(\cos 30^{\circ}-\cos 90^{\circ}\right) \\
& =p E\left(\frac{\sqrt{3}}{2}-0\right)=\frac{\sqrt{3}}{2} p E
\end{aligned}\)
\(\begin{aligned}
& \theta_1=30^{\circ} \\
& \theta_2=90^{\circ}
\end{aligned}\)
\(\therefore\) Work done in rotating the dipole from \(\theta_1=30^{\circ}\) to a position, where its axis is perpendicular \(\left(\theta_2=90^{\circ}\right)\) to electric field is given as
\(\begin{aligned}
W & =p E\left(\cos \theta_1-\cos \theta_2\right) \\
& =p E\left(\cos 30^{\circ}-\cos 90^{\circ}\right) \\
& =p E\left(\frac{\sqrt{3}}{2}-0\right)=\frac{\sqrt{3}}{2} p E
\end{aligned}\)
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