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

### Related Formula The potential energy of an electric dipole aligned at an \angle theta$\theta$ inside a uniform electric field is given by: U = -p E cos theta$$U = -p E \cos \theta$$ The work done by an external agent to rotate the dipole equals its change in potential energy: W = Delta U = -pE (cos theta_f - cos theta_i)$$W = \Delta U = -pE (\cos \theta_f - \cos \theta_i)$$ ### Core Logic Given parameters [cite: 805, 806]: * Dipole moment, p = 6 times 10^-6 text Cm$p = 6 \times 10^{-6} \text{ Cm}$ * Electric field, E = 10^6 text V/m$E = 10^6 \text{ V/m}$ * Initial \angle (\parallel state), theta_i = 0^circ$\theta_i = 0^{\circ}$ * Final \angle (opposite state), theta_f = 180^circ$\theta_f = 180^{\circ}$ Substitute the values into the work equation [cite: 807, 808]: W = -pE (cos 180^circ - cos 0^circ)$$W = -pE (\cos 180^{\circ} - \cos 0^{\circ})$$ W = -pE (-1 - 1) = 2 p E quad text$$W = -pE (-1 - 1) = 2 p E \quad \text{}$$ W = 2 times (6 times 10^-6) times 10^6 = 12 text J quad text$$W = 2 \times (6 \times 10^{-6}) \times 10^6 = 12 \text{ J} \quad \text{}$$ ### Pattern Recognition Rotating a dipole from its most stable configuration (\parallel, theta=0^circ$\theta=0^{\circ}$) to its most unstable position (anti-\parallel, theta=180^circ$\theta=180^{\circ}$) always requires a maximum work value of exactly 2pE$2pE$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics