An electric dipole of dipole moment is $6.0 \times 10^{-6} ~\mathrm{C m}$ placed in a uniform electric field of $1.5 \times 10^{3} ~\mathrm{NC}^{-1}$ in such a way that dipole moment is along electric field. The work done in rotating dipole by $180^{\circ}$ in this field will be ___________ $\mathrm{m J}$.

<p>The work done $W$ in rotating an electric dipole in a uniform electric field is given by:</p> <p>$W = pE(1 - \cos\theta)$,</p> <p>where $p$ is the dipole moment, $E$ is the strength of the electric field, and $\theta$ is the angle the dipole is rotated through.</p> <p>In this case, the dipole moment $p$ is $6.0 \times 10^{-6} ~\mathrm{C m}$, the electric field $E$ is $1.5 \times 10^{3} ~\mathrm{NC}^{-1}$, and the angle $\theta$ is $180^{\circ}$.</p> <p>Substituting these values into the formula gives:</p> <p>$W = 6.0 \times 10^{-6} ~\mathrm{C m} \times 1.5 \times 10^{3} ~\mathrm{NC}^{-1} \times (1 - \cos180^{\circ})$.</p> <p>Since $\cos180^{\circ} = -1$, the equation becomes:</p> <p>$W = 6.0 \times 10^{-6} ~\mathrm{C m} \times 1.5 \times 10^{3} ~\mathrm{NC}^{-1} \times (1 - (-1))$,</p> <p>$W = 6.0 \times 10^{-6} ~\mathrm{C m} \times 1.5 \times 10^{3} ~\mathrm{NC}^{-1} \times 2$,</p> <p>$W = 18.0 \times 10^{-3} ~\mathrm{J} = 18.0 ~\mathrm{mJ}$.</p> <p>So the work done in rotating the dipole by $180^{\circ}$ in this field is 18.0 millijoules.</p>