Question

An electric dipole consists of two electric charges of equal magnitude and opposite sign. If the charges are q and - q are located at a distance from each other, then the electric field E at the point Pin the figure is

By expanding this expression for E as a series in powers of d/D, show that is approximately proportional 1/D³ to when is far away from the dipole.

Short answer

HINT: Look for \[{\rm{E}}\]and utilise the power series for\[\frac{1}{{{{(1 - x)}^2}}}\].

Step-by-step solution

Concept Introduction

Taylor polynomials are function approximations that improve in quality as n rises. Taylor's theorem calculates the inaccuracy introduced by such approximations in terms of numbers. If a function's Taylor series is convergent, the sum of its Taylor polynomials is the limit of the infinite sequence of Taylor polynomials.

Consider the given values and simplify

Calculate the given statement,

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

The derivative of \[\frac{1}{{1 - x}}\] is \[\frac{1}{{{{(1 - x)}^2}}}\]. we can use the derivative of the power series for \[\frac{1}{{1 - x}}\]

\[\begin{array}{l}\frac{1}{{1 - x}} = \sum\limits_{n = 0}^\infty {{x^n}} \\\frac{1}{{{{(1 - x)}^2}}} = \sum\limits_{n = 0}^\infty n {x^{n - 1}}\end{array}\]

Using\[x = - \frac{d}{D}\]

\[\frac{1}{{{{\left( {1 + \frac{d}{D}} \right)}^2}}} = \frac{1}{{{{\left( {1 - \left( { - \frac{d}{D}} \right)} \right)}^2}}} = \sum\limits_{n = 0}^\infty n {\left( { - \frac{d}{D}} \right)^{n - 1}} = \sum\limits_{n = 1}^\infty {{{( - 1)}^{n - 1}}} n{\left( {\frac{d}{D}} \right)^{n - 1}}\]

Since the term for \[n = 0\] is 0, we may alter the series' start to \[n = 1\].

Calculation

Consider the following terms,

\[\sum\limits_{n = 1}^\infty {{{( - 1)}^{n - 1}}} n{\left( {\frac{d}{D}} \right)^{n - 1}} = 1 - 2\left( {\frac{d}{D}} \right) + 3{\left( {\frac{d}{D}} \right)^2} - 4{\left( {\frac{d}{D}} \right)^3} + \cdots \]

Substitute the following into the equation\[E\]

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

\[ \approx \frac{q}{{{D^2}}}\left( {1 - \left[ {1 - 2\left( {\frac{d}{D}} \right) + 3{{\left( {\frac{d}{D}} \right)}^2} - 4{{\left( {\frac{d}{D}} \right)}^3}} \right]} \right)\]

\[ \approx \frac{q}{{{D^2}}}\left( {2\left( {\frac{d}{D}} \right) - 3{{\left( {\frac{d}{D}} \right)}^2} + 4{{\left( {\frac{d}{D}} \right)}^3}} \right)\]

Factor out a \[\frac{d}{D}\]

\[E \approx \frac{{qd}}{{{D^3}}}\left( {2 - 3\left( {\frac{d}{D}} \right) + 4{{\left( {\frac{d}{D}} \right)}^2}} \right)\]

When P is far away from the dipole, D is very large compared to d and\[\frac{d}{D}\] becomes very small.

The equation can be simplified to

\[E \approx \frac{{2qd}}{{{D^3}}}\]

So, E is approximately proportional to 1/D³. When P is far away from the dipole.