Q60P - Fundamentals Of Physics

Question

A certain electric dipole is placed in a uniform electric field of magnitude $\mathbf{40}\mathbf{ }\mathbf{\text{N/C}}$

. Figure 22-63 gives the magnitude t of the torque on the dipole versus the angle  between field and the dipole moment .The vertical axis scale is set by${\mathit{t}}_{\mathbf{s}}\mathbf{=}\mathbf{ }\mathbf{1}\mathbf{.00}\mathbf{ }\mathbf{\times }\mathbf{ }{\mathbf{10}}^{\mathbf{-}\mathbf{28}}\mathbf{\text{N.m}}$. What is the magnitude of  $\overrightarrow{\mathbf{p}}$?

 

Step-by-Step Solution

Answer

The magnitude of the dipole moment $\overrightarrow{p}$ is.$2.5\times {10}^{-30}\text{C.m}$

Step 1: The given data

 

1. Electric field of magnitude, $E=40\text{N/C}$
2. Vertical axis scale, ${\tau }_s=1.00\times {10}^{-28}\text{Nm}$

Step 2: Understanding the concept of electric dipole

Using the equation of the torque in the electric dipole, we can get the required magnitude of the torque with the given data.

 

Formula:

The torque acting on a dipole tends to rotate the dipole: (hence the dipole) into the direction of field,E  is given by:   $\tau =-pEsin \theta$                                                (i)

where,$\theta$ is the angle between p and E.

Step 3: Calculation of the dipole moment

From equation (i), considering $\theta$ as a variable, we note that it reaches its maximum value when, :

$\theta =-{90}^o$

${\tau }_{max}=pE$

Thus, with $E=40\text{N/C}$and (${\tau }_{\max }=1.00\times {10}^{-28}\text{Nm}$determined from the graph), the dipole moment is given using above equation as:

$\begin{array}{c}1.00\times {10}^{-28}\text{Nm}=p\left(40\text{N/C}\right)\\ p=2.5\times {10}^{-30}\text{C.m}\end{array}$

Hence, the value of the dipole moment is.$2.5\times {10}^{-30}\text{C.m}$