Chapter 8.8, Problem 29E

To determine

To Prove: The E is approximately proportional to $\frac{1}{D^3}$ for $P$ is far away from the dipole.

Explanation of Solution

Given:

The given equation for the electric field is,

  $E=\frac{q}{D^2}-\frac{q}{{\left(D+d\right)}^2}$

The given diagram is shown in Figure 1

  

Figure 1

Calculation:

Consider the given expression is,

  $E=\frac{q}{D^2}-\frac{q}{{\left(D+d\right)}^2}$

Then,

  $E=\frac{q}{D^2}\left[1-\left(1+{\left(\frac{d}{D}\right)}^{-2}\right)\right]$

Then, expand $\left(1+{\left(\frac{d}{D}\right)}^{-2}\right)$ as,

  $\left(1+{\left(\frac{d}{D}\right)}^{-2}\right)=\left[1-2\left(\frac{d}{D}\right)+\frac{2\cdot 3}{2!}{\left(\frac{d}{D}\right)}^2-\frac{2\cdot 3\cdot 4}{3!}{\left(\frac{d}{D}\right)}^3+\mathrm{...}\right]$

Then,

  $\begin{array}{c}E=\frac{q}{D^2}\left[1-\left[1-2\left(\frac{d}{D}\right)+\frac{2\cdot 3}{2!}{\left(\frac{d}{D}\right)}^2-\frac{2\cdot 3\cdot 4}{3!}{\left(\frac{d}{D}\right)}^3+\mathrm{...}\right]\right]\\ =\frac{q}{D^2}\left[2\left(\frac{d}{D}\right)-3{\left(\frac{d}{D}\right)}^2+4{\left(\frac{d}{D}\right)}^3-\mathrm{...}\right]\\ =2qd\frac{1}{D^3}\end{array}$

As, the value of $D$ is much more than $d$ , then ${\left(\frac{d}{D}\right)}^n\to 0$ as the value of $n$ increases so only the first term is concerned.

Thus, it is proved thatE is approximately proportional to $\frac{1}{D^3}$ for $P$ is far away from the dipole.