How much work is required to turn an electric dipole$\mathbf{180}\mathbf{°}$in a uniform electric field of magnitude$\mathit{E}\mathbf{}\mathbf{=}\mathbf{}\mathbf{46}\mathbf{.0}\mathbf{}\mathbf{\text{N/C}}\mathbf{}$if the dipole moment has a magnitude of$\mathit{p}\mathbf{}\mathbf{=}\mathbf{}\mathbf{3}\mathbf{.02}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{25}}\mathbf{\text{C.m}}$and the initial angle is$\mathbf{64}\mathbf{°}$?

1. Dipole moment,$p=3.02x{10}^{-25}\text{C.m}$
2. Electric field of magnitude,$E=46.0\text{N/C}$
3. Initial angle,${\theta }_i={64}^o$
4. Final angle of the dipole,${\theta }_f={180}^o$

The potential energy of the electric dipole placed in an electric field depends on its orientation relative to the electric field. The magnitude of the electric dipole moment is$\overrightarrow{p}=q\overrightarrow{d}$, where,qis the magnitude of the charge, and dis the separation between the two charges. When placed in an electric field, the potential energy of the dipole is given by

$\begin{array}{c}\mathbf{U}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathbf{=}\mathbf{-}\overrightarrow{\mathbf{p}}\mathbf{.}\overrightarrow{\mathbf{E}}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{-}\mathbf{p}\mathbf{E}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{\theta }\end{array}$

Formulae:

Therefore, if the initial angle between p and E isand the final angle is$\theta$, then the change in potential energy would be

$\begin{array}{c}\Delta U=U\left(\theta \right)-U_o\left(\theta \right)\\ =-pE(cos\theta -cos{\theta }_o)\end{array}$ (i)

The electric dipole moment of a system,$p=qd$ (ii)

“Torque and energy of an electric dipole in an electric field,” thus we can find the amount of work can be given as the net potential difference and using equation (i) it is given as follows

$\begin{array}{c}W=U({\theta }_o+\pi )-U_o\left({\theta }_o\right)\\ =-pE\left(cos({\theta }_o+\pi )-cos\left({\theta }_o\right)\right)\\ =2pEcos{\theta }_o\\ =2(3.02\times {10}^{-25}\text{C.m})\left(\frac{\text{46.0N}}{\text{C}}\right)cos{64.0}^o\\ =1.22\times {10}^{-23}\text{J}\end{array}$

Hence, the value of the amount of work done for the required change in angle is.$1.22\times {10}^{-23}\text{J}$