In Fig. 22-67, an electric dipole swings from an initial orientation i(${\mathit{\theta }}_{\mathbf{i}}\mathbf{}\mathbf{=}\mathbf{20}\mathbf{.0}\mathbf{°}$) to a final orientation f($\mathbf{⟨}\mathbf{f}\mathbf{}\mathbf{=}\mathbf{}\mathbf{20}\mathbf{.0}\mathbf{°}$) in a uniform external electric field $\overrightarrow{\mathbf{E}}$. The electric dipole moment is$\mathbf{1}\mathbf{.60}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{27}}\mathbf{\text{C.m}}$; the field magnitude is$\mathbf{3}\mathbf{.00}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{6}}\mathbf{\text{N/C}}$. What is the change in the dipole’s potential energy?

a) Initial orientation of the electric dipole i,${\theta }_i={20.0}^0$

b) Final orientation of the electric dipole f, ${\theta }_f={20.0}^o$

c) Strength of the uniform external electric field $\overrightarrow{E}=3.00x{10}^6\text{N/C}.$

d) The electric dipole moment, $p=1.60x{10}^{-27}\text{C.m}$

Using the concept of an electric dipole, we can get the potential energy of the dipole. This determines the change in the potential energy of the system.

Formula:

Potential Energy of an Electric Dipole is given by:$U=-\overrightarrow{p}.\overrightarrow{E}=-pEcos\theta$ (i)

Where,

$\overrightarrow{p}$and $\overrightarrow{E}$are the dipole moment and Electric field

From the given data and the figure, we know that the initial and final angles of orientation ${\theta }_i={70.0}^0$are${\theta }_f={110.0}^o$, we can get the change of the potential energy using equation (i) as follows:

$\begin{array}{c}\Delta U=-pE(cos70.0°-cos110°)\\ =-\left(1.60x{10}^{-27}\text{C.m}\right)\left(3.00x{10}^6\text{N/C}\right)\left(0.684\right)\\ =-3.28x{10}^{-21}\text{J}\end{array}$

Hence, the value of the potential energy is.$-3.28x{10}^{-21}\text{J}$