An electric dipole consists of two point charges(±q), each of massm, fixed to the ends of a (massless) rod of lengthd. (Donotassumedis small.)

(a) Find the net self-force on the dipole when it undergoes hyperbolic motion (Eq. 12.61) along a line perpendicular to its axis. [Hint:Start by appropriately modifying Eq. 11.90.]

$\begin{array}{l}x\left(t\right)=\frac{F}{m}\frac{t\text{'}}{\sqrt{1+{\left(\frac{Ft\text{'}}{mc}\right)}^2}}dt\text{'}=\frac{m{c}^2}{F}{\sqrt{1+{\left(\frac{Ft\text{'}}{mc}\right)}^2}|}_0^t=\frac{m{c}^2}{F}\sqrt{1+{\left(\frac{Ft}{mc}\right)}^2}-1...(12.61)\\ \end{array}$

$F_{self}=\frac{q}{2}(E_1+E_2)=\frac{q^2}{8\pi {\epsilon }_0{c}^2}\frac{(l{c}^2-a{d}^2)}{{(l^2+d^2)}^{3/2}}\widehat{x}...(11.90)$

(b) Notice that this self-force is constant (t drops out), and points in the direction of motion—just right to produce hyperbolic motion. Thus it is possible for the dipole to undergo self-sustaining accelerated motion with no external force at all !! [Where do you suppose the energy comes from?] Determine the self-sustaining force, F, in terms of m, q, and d.

When a small distance separates two charges of opposite polarity, this is called a dipole. For example, consider a positively charged particle and a negatively charged particle separated by a certain distance.

We know that

$x\left(t\right)=\frac{c}{\alpha }[\sqrt{1+{\left(\alpha t\right)}^2}-1]$

Here,

$\alpha =\frac{F}{mc}$

Here, F is the force that has been applied by +q over –q is the same force that will be applied by –q over +q, so the net force will only be available in the x direction, so we have to find the x component of E.

Therefore, from the equation (10.72), we can write the electric field at +q due to –q is:

$E(r,t)=\frac{q}{4\pi {\epsilon }_0}\frac{r}{{(r-u)}^3}\left[(c^2-v^2)u+r\times (u\times a)\right]$

Modifying equation 10.72, we can write:

$E(r,t)=-\frac{q}{4\pi {\epsilon }_0}\frac{r}{{(r-u)}^3}\left[(c^2-v^2)u+u(r.a)-a(r.u)\right]\mathrm{...}\left(i\right)$

As we know:

$u=cr-v$

Therefore,

$\begin{array}{c}{u}_x=c\frac{l}{r}-v\\ =\frac{1}{r}(cl-vr)\end{array}$

We also know that,

$u=cr-lv$

And,

$r.a=la$

Therefore, putting the values in equation (i), we get:

$\begin{array}{c}{E}_x=-\frac{q}{4\pi {\epsilon }_0}\frac{r}{{(cr-vl)}^3}\left[(cl-vr)(c^2-v^2)\frac{1}{r}+u(r.a)-a(r.u)\right]\\ =-\frac{q}{4\pi {\epsilon }_0}\frac{r}{{(cr-vl)}^3}\left(\frac{1}{r}(cl-vr)la-a(cr-lv)\right)\\ =-\frac{q}{4\pi {\epsilon }_0}\frac{r}{{(cr-vl)}^3}\left[(cl-vr)(c^2-v^2)-ca{d}^2\right]\end{array}$

Therefore, force on +q is $q{E}_x$, as mentioned earlier, there will be the same force on –q.

Therefore, the total magnitude of the force can be calculated as:

$F=-\frac{2q^2}{4\pi {\epsilon }_0}\frac{r}{{(cr-vl)}^3}\left[(cl-vr)(c^2-v^2)-ca{d}^2\right]\widehat{x}$

We know that,

$\begin{array}{c}v\left(t\right)=\frac{dx}{dt}\\ =\frac{c}{\alpha }\frac{1}{2}\frac{1}{\sqrt{1+{\left(\alpha t\right)}^2}}2{\alpha }^{2}t\\ =\frac{c\alpha t}{\sqrt{1+{\left(\alpha t\right)}^2}}\end{array}$

Therefore,

$v\left(t_r\right)=\frac{c\alpha t_r}{T}$

Here,

$T=\sqrt{1+{\left(\alpha t_r\right)}^2}$

As we know, acceleration is :

$\begin{array}{c}a=\frac{dv}{dt}\\ =\frac{c\alpha }{T}+c\alpha t_r(-\frac{1}{2})\frac{2{\alpha }^2{t}_r}{T^3}\\ =\frac{c\alpha }{T^3}(1+{\left(\alpha t_r\right)}^2-{\left(\alpha t_r\right)}^2)\\ =\frac{c\alpha }{T^3}\end{array}$

Now, we have to calculate $t_r$

We know that,

$c^2{(t-t_r\text{})}^2=r^2=l^2+d^2$

And,

$l=x\left(t\right)-x\left(t_r\right)=\frac{c}{\alpha }\sqrt{1+{\left(\alpha t_r\right)}^2}-\sqrt{1+{\left(\alpha t_r\right)}^2}$

Hence,

$t^2-2t{t}_r+t_r^2=\frac{1}{{\alpha }^2}[1+{\left(\text{}\alpha t\right)}^2+1+{\left(\text{}\alpha t_r\right)}^2-2\sqrt{1+{\left(\text{}\alpha t\right)}^2}\sqrt{1+{\left(\text{}\alpha t\right)}^2}]+{\left(\frac{d}{c}\right)}^2$

As we know,

Squaring both the sides, we get:

$$\begin{gathered} =\frac{c\alpha }{T^3}(1+{\left(\alpha t_r\right)}^2-{\left(\alpha t_r\right)}^2) \\ \begin{array}{c}1+{\left(\alpha t\right)}^2+{\left(\alpha t_r\right)}^2+{\alpha }^4{t}^2{t}_r^2=1+{\alpha }^4{t}^2{t}_r^2+\frac{1}{4}{\left(\frac{ad}{c}\right)}^4+2{\alpha }^{2}t{t}_r\text{}+{\left(\frac{ad}{c}\right)}^2+{\alpha }^{2}t{t}_r{\left(\frac{ad}{c}\right)}^2\\ t^2+t_r^2-2t{t}_r-t{t}_r{\left(\frac{ad}{c}\right)}^2-{\left(\frac{d}{c}\right)}^2-\frac{{\alpha }^2}{4}({\frac{d}{c})}^4=0\end{array} \end{gathered}$$

Therefore,

$t=t_r[1+\frac{1}{2}{\left(\frac{ad}{c}\right)}^2]\pm \sqrt{[1+{\left(\text{}\alpha t\right)}^2]{\left(\frac{d}{c}\right)}^2[1+\text{}\left(\frac{\alpha d}{2c}\right)]}$

As we know, the value is small; hence$t\approx t_r+d/c$we need the + sign only.

We know that,

$l=\alpha (d\frac{d}{2c}T+t_{r}D)$

Where,

$D=\sqrt{1+{\left(\frac{\alpha d}{2c}\right)}^2}$

Hence, by putting the respective values in the force (F) equation, we get:

$F=\frac{q^2}{4\pi {\epsilon }_0}\frac{\alpha }{cd{[1+{\left(\frac{ad}{2c}\right)}^2]}^{3/2}\widehat{x}}$

(b)

As we know from step1 that$\alpha =\frac{F}{mc}$

Hence,

$\begin{array}{c}F=mc\alpha \\ =\frac{q^2}{4\pi {\epsilon }_0}\frac{\alpha }{cd{[1+{\left(\frac{ad}{2c}\right)}^2]}^{3/2}\widehat{x}}\\ ={[1+{\left(\frac{ad}{2c}\right)}^2]}^{3/2}\\ =\frac{q^2}{8\pi {\epsilon }_0\text{}m{c}^{2}d\text{}}\\ =\frac{{\mu }_0{q}^2}{8\pi \text{}md\text{}}\end{array}$

Hence, this is the force on one end,

Therefore,

$\alpha =\frac{2c}{d}\sqrt{{\left(\frac{{\mu }_0{q}^2}{8\pi \text{}md\text{}}\right)}^{2/3}-1}$

So, the self-sustained force is:

$F=\frac{2m{c}^2}{d}\sqrt{{\left(\frac{{\mu }_0{q}^2}{8\pi md}\right)}^{2/3}-1}$

Hence, the self-sustained force has been determined in terms of m, q, and d.