Question

Question: In Fig. 24-41a, a particle of elementary charge +eis initially at coordinate z = 20 nmon the dipole axis (here a zaxis) through an electric dipole, on the positive side of the dipole. (The origin of zis at the center of the dipole.) The particle is then moved along a circular path around the dipole center until it is at coordinate z = -20 nm, on the negative side of the dipole axis. Figure 24-41bgives the work done by the force moving the particle versus the angle u that locates the particle relative to the positive direction of the z-axis. The scale of the vertical axis is set by${\mathit{W}}_{\mathbf{a}\mathbf{s}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{30}}\mathbf{\text{J}}$.What is the magnitude of the dipole moment?

Short answer

Answer:

The magnitude of the dipole moment is $5.56\times {10}^{-37}\text{C.m}$.

Step-by-step solution

The given data

1. A particle of charge +e is initially at z = 20 nm on the dipole axis.
2. The particle is moved along a circular path around the dipole center until it is at z = -20nm on the negative side of the dipole axis.
3. The scale of the vertical axis,$W_{as}=4\times {10}^{-30}\text{J}$
4. The origin of z is at the center of the dipole.

Understanding the concept of the electric field

Using the concept of electric potential energy, we can get the value of the energy. Using this value of energy, we can get the value of work done by the system. This determines the dipole moment of the system by using the given values.

Formulae:

The potential energy of the particle due to dipole, $U=\frac{k{q}_1{q}_2}{d}$ (i)

The magnitude of the dipole moment, p = q d (ii)

The work done by a body to change in energy, $W=U\text{'}-U$ (iii)

Calculation of the magnitude of the dipole moment

When, $\theta =0^o$. The initial energy of the system is given using equation (i) as follows:

$$\begin{gathered} U=\left[\frac{kqe}{\left(z-\frac{d}{2}\right)}\right]+\left[\frac{k\left(-q\right)e}{\left(z+\frac{d}{2}\right)}\right] \\ =kqe\left[\frac{1}{\left(z-\frac{d}{2}\right)}-\frac{1}{\left(z+\frac{d}{2}\right)}\right]...................\left(a\right) \end{gathered}$$

At, $\theta ={180}^o$ the final potential energy of the system is given using equation (i) as follows:

$$\begin{gathered} U\text{'}=\left[\frac{kqe}{\left(z+\frac{d}{2}\right)}\right]+\left[\frac{k\left(-q\right)e}{\left(z-\frac{d}{2}\right)}\right] \\ =-kqe\left[\frac{1}{\left(z-\frac{d}{2}\right)}-\frac{1}{\left(z+\frac{d}{2}\right)}\right]........................\left(b\right) \end{gathered}$$

Thus, using the equations (a) and (b) in equation (iii), we can get the value of the work as follows:

$$\begin{gathered} W=-2kqe\frac{\left[\frac{1}{\left(z-\frac{d}{2}\right)}-1\right]}{\left(z+\frac{d}{2}\right)} \\ =\frac{-2kqed}{\left(z^2-\frac{d^2}{4}\right)} \\ (\because d\ll z) \\ =-\frac{2kep}{z^2}\text{(from equation (ii))} \end{gathered}$$

Rearranging the above equation for dipole moment, we can get the value of the dipole moment by substituting the given values in the equation as follows:

$$\begin{gathered} p=\frac{-W{z}^2}{\left(2ke\right)} \\ =\frac{-\left(-4.0\times {10}^{-30}\text{J}\right){(20\times {10}^{-9}\text{m})}^2}{2\left(8.99\times {10}^9{\text{Nm}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(1.6\times {10}^{-19}\text{C}\right)} \\ =5.56\times {10}^{-37}\text{C.m} \end{gathered}$$

Therefore, the magnitude of dipole momentum is$5.56\times {10}^{-37}\text{C.m}$.