Question

Two identical permanent dipoles, each consisting of charges +qand -qseparated by a distance s, are aligned along the xaxis, a distance rfrom each other, where$\mathbf{r}\mathbf{\gg }\mathbf{s}$(Figure 13.75). Show all of the steps in your work, and briefly explain each step. (a) Draw a diagram showing all individual forces acting on each particle, and draw heavier vectors showing the net force on each dipole. (b) Show that the magnitude of the net force exerted on one dipole by the other dipole is this:

$\mathbf{F}\mathbf{\approx }\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{6}{\mathbf{q}}^{\mathbf{2}}{\mathbf{s}}^{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{4}}}$

Short answer

(a)

(b) The magnitude of the net force exerted on one dipole by the other is proved

$\mathrm{F}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{6{\mathrm{q}}^2{\mathrm{s}}^2}{{\mathrm{r}}^4}\right)$.

Step-by-step solution

Identification of the given data

The given data can be listed below as:

• The charges of the permanent dipoles are,$+qand-q$
• The distance between the charges of the dipoles is, s
• The distance between the dipoles is, r

Significance of the magnitude of the net force

The magnitude of the net force is directly proportional to the product of the charges and inversely proportional to the square of the distances.

(a) Determination of the diagram of the forces and the vectors

The diagram showing all the forces and also the heavier vectors has been drawn below-

(b) Prove of the magnitude of the net force exerted on one dipole by the other dipole

The equation of the force exerted due to the negative second charge on the first negative charge is expressed as:

$$\begin{gathered} F_1=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{{\left(r+{\displaystyle \frac{s}{2}}-{\displaystyle \frac{s}{2}}\right)}^2} \\ =\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r^2} \end{gathered}$$

Here, localid="1656922836036" $\frac{1}{4\pi {\epsilon }_0}$is the electric field constant, r is the distance between the dipoles and s is the distance between the charges of the dipoles.

The equation of the force exerted due to the positive second charge on the first negative charge is expressed as:

${\mathrm{F}}_2=\frac{1}{4{\mathrm{\pi \epsilon }}_0{\displaystyle \frac{{\mathrm{q}}^2}{{\left(\mathrm{r}+\mathrm{s}\right)}^2}}}$

Here, $\frac{1}{4\pi {\epsilon }_0}$is the electric field constant, r is the distance between the dipoles and s is the distance between the charges of the dipoles.

The equation of the net force on the first negative charge is expressed as:

$F_3=F_1-F_2$

Here, $F_3$is the equation of the net force on the first negative charge, F is the force exerted due to the negative second charge on the first negative charge and $F_2$is the force exerted due to the positive second charge on the first negative charge.

Substitute the values in the above equation.

$$\begin{gathered} F_3=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{{\left(r\right)}^2}-\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{{\left(r+s\right)}^2} \\ =\frac{q^2}{4\pi {\epsilon }_0}\left(\frac{1}{{\left(r\right)}^2}-\frac{1}{{\left(r+s\right)}^2}\right) \\ =\frac{1}{4\pi {\epsilon }_0}\frac{q^{2}s}{{\left(r\right)}^2}\left(\frac{2r+s}{{\left(r+s\right)}^2}\right) \end{gathered}$$

The equation of the net force on the first positive charge is expressed as:

$F=F_4-F_5$ …(i)

Here, F is the net force on the first positive charge, $F_4$is the force exerted on the first positive charge due to the second positive charge and $F_5$is the force exerted on the second negative charge due to the first positive charge.

The equation of the force exerted on the first positive charge due to the second positive charge is expressed as:

$$\begin{gathered} F_4=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{{\left(r-{\displaystyle \frac{s}{2}}-{\displaystyle \frac{s}{2}}\right)}^2} \\ =\frac{q^2}{4\pi {\epsilon }_0}\frac{1}{{\left(r-s\right)}^2} \end{gathered}$$

Substitute the values in the equation (i).

$$\begin{gathered} F=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{{\left(r-{\displaystyle \frac{s}{2}}-{\displaystyle \frac{s}{2}}\right)}^2}-\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{{\left(r\right)}^2} \\ =\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{{\left(r-s\right)}^2}-\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{{\left(r\right)}^2} \\ =\frac{q^2}{4\pi {\epsilon }_0}\left(\frac{1}{{\left(r-s\right)}^2}-\frac{1}{r^2}\right) \\ =\frac{q^{2}s}{4\pi {\epsilon }_0}\left(\frac{2r-s}{{\left(r-s\right)}^2}\right)-\frac{q^{2}s}{4\pi {\epsilon }_0{r}^2}\left(\frac{s+2r}{{\left(r+s\right)}^2}\right) \end{gathered}$$

Hence, further as it is solved as,

$$\begin{gathered} \mathrm{F}=\frac{{\mathrm{q}}^2\mathrm{s}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}\left(\frac{{\left(\mathrm{r}+\mathrm{s}\right)}^2\left(2\mathrm{r}-\mathrm{s}\right)-{\left(\mathrm{r}-\mathrm{s}\right)}^2\left(\mathrm{s}+2\mathrm{r}\right)}{{\left(\mathrm{r}-\mathrm{s}\right)}^2{\left(\mathrm{r}+\mathrm{s}\right)}^2}\right) \\ =\frac{{\mathrm{q}}^2\mathrm{s}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}\left(\frac{8{\mathrm{r}}^8-2{\mathrm{sr}}^2-2{\mathrm{s}}^3}{\left({\mathrm{r}}^2-{\mathrm{s}}^2\right)}\right) \\ =\frac{2{\mathrm{q}}^2{\mathrm{s}}^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}\left(\frac{3{\mathrm{r}}^2-{\mathrm{s}}^2}{{\left({\mathrm{r}}^2-{\mathrm{s}}^2\right)}^2}\right) \end{gathered}$$

As the distance between the dipoles is way more than the charge on the dipoles, then$r\gg s$.

$$\begin{gathered} F=\frac{2q^2{s}^2}{4\pi {\epsilon }_0{r}^2}\left(\frac{3r^2}{{\left(r^2\right)}^2}\right) \\ =\frac{1}{4\pi {\epsilon }_0}\left(\frac{6q^2{s}^2}{r^4}\right) \end{gathered}$$

Thus, the magnitude of the net force exerted on one dipole by the other is proved .

$F=\frac{1}{4\pi {\epsilon }_0}{\left(\frac{6q^2{s}^2}{r^4}\right)}^2$.