Q57P - Fundamentals Of Physics

Question

An electric dipole consisting of charges of magnitude$\mathbf{1}\mathbf{.50}\mathbf{ }\mathbf{\text{nC}}$  separated by$\mathbf{6}\mathbf{.20}\mathbf{ }\mathbf{\text{mm}}$ .  is in an electric field of strength $\mathbf{1100}\mathbf{ }\mathbf{\text{N/C}}$.What are (a) the magnitude of the electric dipole moment and (b) the difference between the potential energies for dipole orientations parallel and anti-parallel to $\overrightarrow{\mathbf{E}}$? :

 

Step-by-Step Solution

Answer

a) The magnitude of the electric dipole moment is. $9.30\times {10}^{-15}\text{C.m}$

b) The difference between the potential energies for dipole orientations parallel and anti-parallel to $\overrightarrow{E}$is$2.05\times {10}^{-11}\text{J}$.

 

Step 1: The given data

a) Electric field,$E=1100\text{ N/C}$

b) Charges of magnitude, $q=1.50\text{ nC}$

c) Charges separated by,$r=6.20\text{ μm}$

Step 2: Understanding the concept of energy

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,$\overrightarrow{\mathbf{p}}\mathbf{=}\mathit{q}\overrightarrow{\mathbf{d}}$ where $\mathit{q}$,is the magnitude of the charge, and  is the separation between the two charges. When placed in an electric field, the potential energy of the dipole is given by

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

 

Formulae:

Therefore, if the initial angle between p and E is ${\theta }_o$and 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)

 

Step 3: a) Calculation of magnitude of electric dipole moment

Substituting the given data in equation (ii), we find the magnitude of the dipole moment to be:

$\begin{array}{c}p=(1.50\times {10}^{-6}\text{C})(6.20\times {10}^{-6}\text{m})\\ = 9.30\times {10}^{-15}\text{C.m}\end{array}$

Hence, the value of the dipole moment is.$9.30\times {10}^{-15}\text{C.m}$

 

Step 4: b) Calculation of the difference between potential energies for parallel and antiparallel fields

The initial and the final angles are ${\theta }_o=0^o$(parallel) and, ${\theta }_o=\text{ 18}{0}^o$so the difference between the potential energies is given using equation (i) as follows:

$\begin{array}{c}\Delta U= U\left({180}^o\right)- U_o\left(0\right)\\ =2pE\\ =2(9.30\times {10}^{-15}\text{C.m})\left(\frac{1100\text{N}}{\text{C}}\right)\\ =2.05\times {10}^{-11}\text{J}\end{array}$

The potential energy is a maximum ($U_{max}=+pE$) when the dipole is oriented antiparallel to E, and is a minimum ($U_{\min }=-pE$) when it is parallel to E.

Hence, the value of the difference is.$2.05\times {10}^{-11}\text{J}$