91Ó°ÊÓ

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

$x\left(t\right)=\frac{c}{\mathrm{Î\pm }}[\sqrt{1+{\left(\mathrm{Î\pm }t\right)}^2}−1]$

Here,

$\mathrm{Î\pm }=\frac{F}{mc}$

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:

$E(r,t)=\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{r}{{(r−u)}^3}\left[(c^2−v^2)u+r×(u×a)\right]$

Modifying equation 10.72, we can write:

$E(r,t)=−\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{r}{{(r−u)}^3}\left[(c^2−v^2)u+u(r.a)−a(r.u)\right]\mathrm{...}\left(i\right)$

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:

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

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:

$F=−\frac{2q^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{r}{{(cr−vl)}^3}\left[(cl−vr)(c^2−v^2)−ca{d}^2\right]\widehat{x}$

We know that,

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

Therefore,

$v\left(t_r\right)=\frac{c\mathrm{Î\pm }{t}_r}{T}$

Here,

$T=\sqrt{1+{\left(\mathrm{Î\pm }{t}_r\right)}^2}$

As we know, acceleration is :

$\begin{array}{c}a=\frac{dv}{dt}\\ =\frac{c\mathrm{Î\pm }}{T}+c\mathrm{Î\pm }{t}_r(−\frac{1}{2})\frac{2{\mathrm{Î\pm }}^2{t}_r}{T^3}\\ =\frac{c\mathrm{Î\pm }}{T^3}(1+{\left(\mathrm{Î\pm }{t}_r\right)}^2−{\left(\mathrm{Î\pm }{t}_r\right)}^2)\\ =\frac{c\mathrm{Î\pm }}{T^3}\end{array}$

Now, we have to calculate $t_r$

We know that,

$c^2{(t−t_r\text{​})}^2=r^2=l^2+d^2$

And,

$l=x\left(t\right)−x\left(t_r\right)=\frac{c}{\mathrm{Î\pm }}\sqrt{1+{\left(\mathrm{Î\pm }{t}_r\right)}^2}−\sqrt{1+{\left(\mathrm{Î\pm }{t}_r\right)}^2}$

Hence,

$t^2−2t{t}_r+t_r^2=\frac{1}{{\mathrm{Î\pm }}^2}[1+{\left(\text{​}\mathrm{Î\pm }t\right)}^2+1+{\left(\text{​}\mathrm{Î\pm }{t}_r\right)}^2−2\sqrt{1+{\left(\text{​}\mathrm{Î\pm }t\right)}^2}\sqrt{1+{\left(\text{​}\mathrm{Î\pm }t\right)}^2}]+{\left(\frac{d}{c}\right)}^2$

As we know,

Squaring both the sides, we get:

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

Therefore,

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

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

We know that,

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

Where,

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

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

$F=\frac{q^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{\mathrm{Î\pm }}{cd{[1+{\left(\frac{ad}{2c}\right)}^2]}^{3/2}\widehat{x}}$

(b)

As we know from step1 that$\mathrm{Î\pm }=\frac{F}{mc}$

Hence,

$\begin{array}{c}F=mc\mathrm{Î\pm }\\ =\frac{q^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{\mathrm{Î\pm }}{cd{[1+{\left(\frac{ad}{2c}\right)}^2]}^{3/2}\widehat{x}}\\ ={[1+{\left(\frac{ad}{2c}\right)}^2]}^{3/2}\\ =\frac{q^2}{8\mathrm{Ï€}{\mathrm{Î\mu }}_0\text{​}m{c}^{2}d\text{​}}\\ =\frac{{\mathrm{μ}}_0{q}^2}{8\mathrm{Ï€}\text{​}md\text{​}}\end{array}$

Hence, this is the force on one end,

Therefore,

$\mathrm{Î\pm }=\frac{2c}{d}\sqrt{{\left(\frac{{\mathrm{μ}}_0{q}^2}{8\mathrm{Ï€}\text{​}md\text{​}}\right)}^{2/3}−1}$

So, the self-sustained force is:

$F=\frac{2m{c}^2}{d}\sqrt{{\left(\frac{{\mathrm{μ}}_0{q}^2}{8\mathrm{π}md}\right)}^{2/3}−1}$

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