An electric dipole with dipole moment $3 imes 10^{-6} \mathrm{Cm}$ is aligned in a stable equilibr

Question

An electric dipole with dipole moment $3 	imes 10^{-6} \mathrm{Cm}$ is aligned in a stable equilibrium position within a uniform electric field of magnitude $2 	imes 10^5 \mathrm{~N} / \mathrm{C}$. The work done in rotating the dipole to a position of unstable equilibrium is

Professor PC Thomas Classes · Answer

(a) : Stable equilibrium corresponds to an angle $	heta_1 = 0^{\circ}$, and unstable equilibrium corresponds to $	heta_2 = 180^{\circ}$. The work done to rotate the dipole is equal to the change in potential energy: $W = \Delta U = pE(\cos	heta_1 - \cos	heta_2)$. Substituting the angles: $W = pE(\cos 0^{\circ} - \cos 180^{\circ}) = pE(1 - (-1)) = 2pE$. Thus, $W = 2 	imes (3 	imes 10^{-6}) 	imes (2 	imes 10^5) = 1.2 \mathrm{~J}$.