Q76P - Fundamentals Of Physics

Question

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?

Step-by-Step Solution

Answer

The change in the dipole’s potential energy is.$-3.28x{10}^{-21}\text{J}$

Step 1: The given data

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,  $p=1.60x{10}^{-27}\text{C.m}$

Step 2: Understanding the concept of the electric dipole

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

Step 3: Calculation of the change in the potential energy

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.$-3.28x{10}^{-21}\text{J}$