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(A) The torque $	au$ on an electric dipole in a uniform electric field is given by $	au = pE \sin 	heta$.Given $	au = 6 \ N-m$ and $	heta = 60^{\

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$-2 \sqrt{3} \ J$

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(A) The torque $	au$ on an electric dipole in a uniform electric field is given by $	au = pE \sin 	heta$.Given $	au = 6 \ N-m$ and $	heta = 60^{\circ}$,we have $6 = pE \sin 60^{\circ}$.Since $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,we get $6 = pE \left( \frac{\sqrt{3}}{2} ight)$,which implies $pE = \frac{12}{\sqrt{3}}$.The potential energy $U$ of an electric dipole is given by $U = -pE \cos 	heta$.Substituting the values,$U = -\left( \frac{12}{\sqrt{3}} ight) \cos 60^{\circ}$.Since $\cos 60^{\circ} = \frac{1}{2}$,we have $U = -\left( \frac{12}{\sqrt{3}} ight) \left( \frac{1}{2} ight) = -\frac{6}{\sqrt{3}} = -2\sqrt{3} \ J$.

An electric dipole of length $20 \ cm$ having $\pm 3 \times 10^{-3} \ C$ charge is placed at $60^{\circ}$ with respect to a uniform electric field and experiences a torque of magnitude $6 \ N-m$. The potential energy of the dipole is:

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