An electric dipole of length 4 cm, when placed with its axis making an angle of 60° with a uniform electric field, experiences a torque of $4\sqrt{3}\mathrm{Nm}$. Calculate the potential energy of the dipole, if it has charge ± 8 nC.

$$\begin{gathered} \mathrm{Given}, 2\mathrm{a}=4\mathrm{cm}=4\times {10}^{-2}\mathrm{m}, \mathrm{\theta }=60° \\ \mathrm{\tau }=4\times \sqrt{3}\mathrm{Nm}, \mathrm{q}=\pm 8 \mathrm{nC}=\pm 8\times {10}^{-9}\mathrm{C} \\ \therefore \mathrm{\tau }=\mathrm{pE} \sin \mathrm{\theta } \\ \therefore \mathrm{E}=\frac{\mathrm{\tau }}{\mathrm{p} \sin \mathrm{\theta }} \mathrm{p}=\mathrm{q}\times 2\mathrm{a} \\ \therefore \mathrm{E}=\frac{4\times \sqrt{3}}{8\times {10}^{-9}\times 4\times {10}^{-2}\times {\displaystyle \frac{\sqrt{3}}{2}}}=\frac{1}{4}\times {10}^{11} \\ \mathrm{Now} \mathrm{P}.\mathrm{E}.=\left|\mathrm{p}\right| \left|\mathrm{E}\right| \cos \mathrm{\theta } \\ =(\mathrm{q}\times 2\mathrm{a}) \left(\mathrm{E}\right) \cos \mathrm{\theta } \\ =(8\times {10}^{-9}\times 4\times {10}^{-2})\times \left(\frac{1}{4}\times {10}^{11}\right)\times \frac{1}{2} \\ =-4 \mathrm{Joules} [\therefore \cos 60°=\frac{1}{2} \end{gathered}$$