An electric dipole of dipole moment $6 \times 10^{-6} \mathrm{Cm}$ is placed in uniform electric field of magnitude $10^{6} \mathrm{V} / \mathrm{m}$. Initially, the dipole moment is parallel to electric field. The work that needs to be done on the dipole to make its dipole moment opposite to the field, will be _______ J.

<p><b>Given values:</b></p> <p>The dipole moment is $ p = 6 \times 10^{-6} \ \mathrm{Cm} $.</p> <p>The electric field is $ E = 10^6 \ \mathrm{V/m} $.</p> <p><b>Formula for work done:</b></p> <p>Work required to rotate the dipole is $ W = \Delta U = -pE(\cos \theta_f - \cos \theta_i) $ where $ \theta_i $ is the initial angle between the dipole and the electric field, and $ \theta_f $ is the final angle.</p> <p><b>Angles:</b></p> <p>At the start, the dipole is parallel to the field ($ \theta_i = 0^\circ $), so $\cos \theta_i = 1$.</p> <p>At the end, the dipole is opposite the field ($ \theta_f = 180^\circ $), so $\cos \theta_f = -1$.</p> <p><b>Calculation:</b></p> <p> $ W = -pE (\cos 180^\circ - \cos 0^\circ) = -pE(-1 - 1) = 2pE $ </p> <p><b>Plug in the values:</b></p> <p> $ W = 2 \times (6 \times 10^{-6}) \times (10^6) = 2 \times 6 = 12\ \mathrm{J} $ </p>