Chapter 14, Problem 14B.3AE

Interpretation Introduction

Interpretation:

The potential energy of the interaction between two collinear linear quadrupoles which are separated by a distance $r$ has to be calculated.

Concept introduction:

The dipole forces are attractive forces between two permanent dipoles.  The interaction is between the positive end of one polar molecule and the negative end of another polar molecule.  The induced dipole interaction occurs when an ion or a dipole induces a temporary dipole in an atom or a molecule with no dipole.  The induced dipole forces are the attraction forces between molecules having temporary dipoles.

Answer to Problem 14B.3AE

The potential energy of the interaction between two collinear linear quadrupoles which are separated by a distance $r$ is $\frac{6q^2{l}^4}{\text{π}{\epsilon }_0{r}^5}$.

Explanation of Solution

The two collinear linear quadrupoles which are separated by a distance $r$ are shown below.

Figure 1

The potential energy due to point dipole-point charge interaction is shown below.

    $V=-\frac{Q_1{Q}_{2}l}{4\text{π}{\epsilon }_0{r}^2}$

Where,

• $Q_1$ is the charge on each ion of dipole.
• $Q_2$ is the charge on point ion.
• $l$ is the length of the dipole.
• ${\epsilon }_0$ is the permittivity.
• $r$ is the distance between point dipole and point charge.

In case of two collinear linear quadrupoles which are separated by a distance $r$, there are total nine interactions are there.

The positive charge is $2q$.

The negative charge is $q$.

The potential energy of the interaction between two collinear linear quadrupoles which are separated by a distance $r$ calculated as shown below.

    $\begin{array}{c}V=\frac{1}{4\text{π}{\epsilon }_0}\left(\begin{array}{l}\frac{q^2}{r}-\frac{2q^2}{r+l}+\frac{q^2}{r+2l}-\frac{2q^2}{r-l}\\ +\frac{4q^2}{r}-\frac{2q^2}{r+l}+\frac{q^2}{r-2l}-\frac{2q^2}{r-l}+\frac{q^2}{r}\end{array}\right)\\ =\frac{q^2}{4\text{π}{\epsilon }_0}\left(\begin{array}{l}\frac{1}{r}-\frac{2}{r+l}+\frac{1}{r+2l}-\frac{2}{r-l}\\ +\frac{4}{r}-\frac{2}{r+l}+\frac{1}{r-2l}-\frac{2}{r-l}+\frac{1}{r}\end{array}\right)\end{array}$

The above equation can be written in terms of $\frac{l}{r}$.

    $\begin{array}{l}V=\frac{q^2}{4\text{π}{\epsilon }_0}\left(\begin{array}{l}\frac{1}{r}-\frac{2}{r\left(1+\frac{l}{r}\right)}+\frac{1}{r\left(1+\frac{2l}{r}\right)}-\frac{2}{r\left(1-\frac{l}{r}\right)}\\ +\frac{4}{r}-\frac{2}{r\left(1+\frac{l}{r}\right)}+\frac{1}{r\left(1-\frac{2l}{r}\right)}-\frac{2}{r\left(1-\frac{l}{r}\right)}+\frac{1}{r}\end{array}\right)\\ V=\frac{q^2}{4\text{π}{\epsilon }_{0}r}\left(\begin{array}{l}1-\frac{2}{\left(1+\frac{l}{r}\right)}+\frac{1}{\left(1+\frac{2l}{r}\right)}+\frac{2}{\left(1-\frac{l}{r}\right)}\\ +4-\frac{2}{\left(1+\frac{l}{r}\right)}+\frac{1}{\left(1-\frac{2l}{r}\right)}-\frac{2}{\left(1-\frac{l}{r}\right)}+1\end{array}\right)\end{array}$

The term $\frac{l}{r}$ can be supposed as $x$.

  $\begin{array}{c}V=\frac{q^2}{4\text{π}{\epsilon }_{0}r}\left(\begin{array}{l}1-\frac{2}{\left(1+x\right)}+\frac{1}{\left(1+2x\right)}-\frac{2}{\left(1-x\right)}\\ +4-\frac{2}{\left(1+x\right)}+\frac{1}{\left(1-2x\right)}-\frac{2}{\left(1-x\right)}+1\end{array}\right)\\ =\frac{q^2}{4\text{π}{\epsilon }_{0}r}\left(6-\frac{4}{\left(1+x\right)}+\frac{1}{\left(1+2x\right)}-\frac{4}{\left(1-x\right)}+\frac{1}{\left(1-2x\right)}\right)\end{array}$

The above equation can be further simplified as shown below.

  $V=\frac{q^2}{4\text{π}{\epsilon }_{0}r}\left(\frac{\begin{array}{l}\left(6\right)\left(1+x\right)\left(1+2x\right)\left(1-x\right)\left(1-2x\right)-4\left(1+2x\right)\left(1-x\right)\left(1-2x\right)\\ -4\left(1+x\right)\left(1+2x\right)\left(1-2x\right)+\left(1+x\right)\left(1-x\right)\left(1+2x\right)\\ +\left(1+x\right)\left(1-x\right)\left(1-2x\right)\end{array}}{\left(1+x\right)\left(1-x\right)\left(1+2x\right)\left(1-2x\right)}\right)$

The value of $r$ is very much greater than $l$.  Therefore, $x\ll 1$ and the denominator can be assumed as $1$.

    $\begin{array}{c}V=\frac{q^2}{4\text{π}{\epsilon }_{0}r}\left(\frac{\begin{array}{l}\left(6\right)\left(1+x\right)\left(1+2x\right)\left(1-x\right)\left(1-2x\right)-4\left(1+2x\right)\left(1-x\right)\left(1-2x\right)\\ -4\left(1+x\right)\left(1+2x\right)\left(1-2x\right)+\left(1+x\right)\left(1-x\right)\left(1+2x\right)\\ +\left(1+x\right)\left(1-x\right)\left(1-2x\right)\end{array}}{1}\right)\\ =\frac{q^2}{4\text{π}{\epsilon }_{0}r}\left(\frac{\begin{array}{l}\left(6\right)\left(1-x^2\right)\left(1-4x^2\right)-4\left(1-4x^2\right)\left(1-x\right)\\ -4\left(1+x\right)\left(1-4x^2\right)+\left(1-x^2\right)\left(1+2x\right)+\left(1-x^2\right)\left(1-2x\right)\end{array}}{1}\right)\end{array}$

The above equation after expanding the terms is shown below.

  $V=\frac{q^2}{4\text{π}{\epsilon }_{0}r}\left(\frac{\begin{array}{l}\left(6\right)\left(1\left(1-4x^2\right)-x^2\left(1-4x^2\right)\right)-4\left(1\left(1-x\right)-4x^2\left(1-x\right)\right)\\ -4\left(1\left(1-4x^2\right)+x\left(1-4x^2\right)\right)\\ +\left(1\left(1+2x\right)-x^2\left(1+2x\right)\right)+\left(1\left(1-2x\right)-x^2\left(1-2x\right)\right)\end{array}}{1}\right)$

The above equation after expanding the terms is shown below.

    $\begin{array}{c}V=\frac{q^2}{4\text{π}{\epsilon }_{0}r}\left(\frac{\begin{array}{l}6-24x^2-6x^2+24x^4-4+4x+16x^2-16x^3\\ -4+16x^2-4x+16x^3+1+2x-x^2+2x^3+\\ 1-2x-x^2-2x^3\end{array}}{1}\right)\\ =\frac{q^2}{4\text{π}{\epsilon }_{0}r}\left(\frac{24x^4}{1}\right)\\ =\frac{6q^2{x}^4}{\text{π}{\epsilon }_{0}r}\end{array}$

Substitute the value of $x=\frac{l}{r}$ in the above expression.

    $\begin{array}{c}V=\frac{6q^2{\left(\frac{l}{r}\right)}^4}{\text{π}{\epsilon }_{0}r}\\ =\frac{6q^2{l}^4}{\text{π}{\epsilon }_0{r}^5}\end{array}$

Therefore, the potential energy of the interaction between two collinear linear quadrupoles which are separated by a distance $r$ is $\frac{6q^2{l}^4}{\text{π}{\epsilon }_0{r}^5}$.