MHT CET · Physics · Electrostatics
An electric dipole of moment \(\vec{P}\) is lying along a uniform electric field \(\overrightarrow{\mathrm{E}}\). The work done in rotating the dipole through \(\frac{\pi^c}{3}\) is \([\sin 30^{\circ}=\cos 60^{\circ}=0 \cdot 5, \cos 30^{\circ}=\) \(\sin 60^{\circ}=\sqrt{3} / 2]\)
- A 3 pE
- B \(\sqrt{2} \mathrm{pE}\)
- C pE
- D \(\frac{\mathrm{pE}}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{pE}}{2}\)
Step-by-step Solution
Detailed explanation
The P.E. of a dipole in an electric field is
\(\mathrm{W}=-\mathrm{pE} \cos \theta\)
When \(\theta=0^{\circ}\),
\(\mathrm{W}_1=-\mathrm{pE} \cos (0)=-\mathrm{pE}\)
When \(\theta=\frac{\pi}{3}=60^{\circ}\),
\(\mathrm{W}_2=-\mathrm{pE} \cos \left(60^{\circ}\right)=\frac{-\mathrm{pE}}{2}\)
\(\begin{aligned}
\therefore \quad \text { Work done }=\mathrm{W}_2-\mathrm{W}_1 & =-\frac{\mathrm{pE}}{2}-(-\mathrm{pE}) \\
& =\frac{\mathrm{pE}}{2}
\end{aligned}\)
\(\mathrm{W}=-\mathrm{pE} \cos \theta\)
When \(\theta=0^{\circ}\),
\(\mathrm{W}_1=-\mathrm{pE} \cos (0)=-\mathrm{pE}\)
When \(\theta=\frac{\pi}{3}=60^{\circ}\),
\(\mathrm{W}_2=-\mathrm{pE} \cos \left(60^{\circ}\right)=\frac{-\mathrm{pE}}{2}\)
\(\begin{aligned}
\therefore \quad \text { Work done }=\mathrm{W}_2-\mathrm{W}_1 & =-\frac{\mathrm{pE}}{2}-(-\mathrm{pE}) \\
& =\frac{\mathrm{pE}}{2}
\end{aligned}\)
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