Find the amount of work done in rotating an electric dipole of dipole moment 3 x 10^–3 cm from its position of stable equilibrium to the position unstable equlibrium in a uniform electric field of intensities 10⁴ NC^–1.

Given, dipole moment of an electric dipole, p= 3 x 10^–3
Electric field intensity of dipole, E= 10⁴ NC^–1

Initially, dipole is in a stable equilibrium.
This implies, ${\mathbf{\theta }}_{\mathbf{1}}\mathbf{=}{\mathbf{0}}^{\mathbf{\circ }}$

and, ${\mathbf{\theta }}_2\mathbf{=}{\mathbf{180}}^{\mathbf{\circ }}\mathbf{ }\mathbf{[}\mathbf{ }\mathbf{dipole}\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{in}\mathbf{ }\mathbf{unstable}\mathbf{ }\mathbf{equilibrium}\mathbf{]}$

Therefore,
Work done in rotating the dipole from an angle of θ₁ to θ_2, W = pE (cos θ₁ – cos θ₂)
             W = 3 x 10^–3 x 10⁴ [1 – (– 1)]
                 = 60 J.