An electric dipole of dipole moment $$6 \times 10^{-6}Cm$$ is placed in uniform electric field of magnitude $$10^{6}V/m$$. 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.

We need to find the work done to rotate an electric dipole from parallel to antiparallel to the electric field.

The potential energy of a dipole with moment $$\vec{p}$$ in a uniform electric field $$\vec{E}$$ is $$U = -\,pE\cos\theta$$, where $$\theta$$ is the angle between $$\vec{p}$$ and $$\vec{E}$$.

The work done by an external agent to rotate the dipole from angle $$\theta_i$$ to $$\theta_f$$ is $$W = U_f - U_i = \bigl(-pE\cos\theta_f\bigr) - \bigl(-pE\cos\theta_i\bigr) = pE\bigl(\cos\theta_i - \cos\theta_f\bigr)\,.$$

For the given situation, initially parallel: $$\theta_i = 0^\circ$$ so $$\cos 0^\circ = 1$$. Finally antiparallel: $$\theta_f = 180^\circ$$ so $$\cos 180^\circ = -1$$. Therefore $$W = pE\bigl(\cos 0^\circ - \cos 180^\circ\bigr) = pE\bigl(1 - (-1)\bigr) = 2pE\,. $$

Substituting $$p = 6\times 10^{-6}\text{ C·m}$$ and $$E = 10^6\text{ V/m}$$ gives $$W = 2 \times 6\times 10^{-6} \times 10^6 = 12\text{ J}\,. $$

The answer is 12 J.