One particle, of charge ${\mathit{q}}_{\mathbf{1}}$, is held at rest at the origin. Another particle, of charge ${\mathit{q}}_{\mathbf{2}}$, approaches along the x axis, in hyperbolic motion:

$\mathit{x}\left(t\right)\mathbf{=}\sqrt{{\mathbf{b}}^{\mathbf{2}}\mathbf{+}{\left(ct\right)}^{\mathbf{2}}}$

it reaches the closest point, b, at time $\mathit{t}\mathbf{=}\mathbf{0}$, and then returns out to infinity.

(a) What is the force ${\mathit{F}}_{\mathbf{2}}$on ${\mathit{q}}_{\mathbf{2}}$(due to ${\mathit{q}}_{\mathbf{1}}$ ) at time t?

(b) What total impulse $\left(I_2={\int }_{-\infty }^{\infty }{F}_{2}dt\right)$is delivered to ${\mathit{q}}_{\mathbf{2}}$by ${\mathit{q}}_{\mathbf{1}}$?

(c) What is the force ${\mathit{F}}_{\mathbf{1}}$on ${\mathit{q}}_{\mathbf{1}}$(due to ${\mathit{q}}_{\mathbf{2}}$ ) at time t?

(d) What total impulse $\left(I_1={\int }_{-\infty }^{\infty }{F}_{1}dt\right)$is delivered to ${\mathit{q}}_{\mathbf{1}}$by ${\mathit{q}}_{\mathbf{2}}$? [Hint: It might help to review Prob. 10.17 before doing this integral. Answer:${\mathit{I}}_{\mathbf{2}}\mathbf{=}\mathbf{-}{\mathit{I}}_{\mathbf{1}}\mathbf{=}\frac{{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}\mathbf{bc}}$ ]

Given data:

The distance along the x-axis in hyperbolic motion is$x\left(t\right)=\sqrt{b^2+{\left(ct\right)}^2}$ .

(a)

Write the expression for the force$F_2$ between two charged particles separated by a distance x.

${\mathit{F}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{{\mathbf{x}}^{\mathbf{2}}}$

Here, q is the charge,${\epsilon }_0$ is the permittivity of free space and x is the distance.

Substitute $x=\sqrt{b^2+{\left(ct\right)}^2}$in the above expression.

role="math" localid="1653892201740" $$\begin{gathered} F_2=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q_1{q}_2}{{\left(\sqrt{b^2+{\left(ct\right)}^2}\right)}^2} \\ {F}_2=\frac{q_1{q}_2}{4{\mathrm{\pi \epsilon }}_0}\frac{1}{\left(b^2+c^2{t}^2\right)} \end{gathered}$$

Therefore, the force$F_2$ on charge$q_2$ (due to$q_1$ ) at time t is$F_2=\frac{q_1{q}_2}{4{\mathrm{\pi \epsilon }}_0}\frac{}{\left(b^2+c^2{t}^2\right)}$ .

(b)

Write the expression for the total impulse$I_2$.

${\mathit{I}}_{\mathbf{2}}\mathbf{=}{\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}{\mathit{F}}_{\mathbf{2}}\mathit{d}\mathit{t}$

Substitute the known value of$F_2$in the above expression.

$$\begin{gathered} I_2={\int }_{-\infty }^{\infty }\left(\frac{q_1{q}_2}{4{\mathrm{\pi \epsilon }}_0}\frac{1}{\left(b^2+c^2{t}^2\right)}\right)dt \\ {I}_2=\frac{q_1{q}_2}{4{\mathrm{\pi \epsilon }}_0}{\int }_{-\infty }^{\infty }\left(\frac{1}{\left(b^2+c^2{t}^2\right)}\right)dt \\ {I}_2=\frac{q_1{q}_2}{4{\mathrm{\pi \epsilon }}_0}{\left(\frac{1}{bc}{\tan }^{-1}\left(\frac{ct}{b}\right)\right)}^{\infty } \\ {I}_2=\frac{q_1{q}_2}{4{\mathrm{\pi \epsilon }}_0}\frac{1}{bc}\left[{\tan }^{-1}\left(\infty \right)-{\tan }^{-1}\left(-\infty \right)\right] \end{gathered}$$

On further solving, the above equation becomes,

$$\begin{gathered} I_2=\frac{q_1{q}_2}{4{\mathrm{\pi \epsilon }}_0}\frac{1}{bc}\left[\frac{\mathrm{\pi }}{2}-\left(-\frac{\mathrm{\pi }}{2}\right)\right] \\ {I}_2=\frac{q_1{q}_2}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{\pi }}{bc} \\ {I}_2=\frac{q_1{q}_2}{4{\epsilon }_{0}bc} \end{gathered}$$

Therefore, the total impulse $I_2$delivered to $q_2$by $q_1$isrole="math" localid="1653892889493" $I_2=\frac{q_1{q}_2}{4{\epsilon }_{0}bc}$ .

(c)

$$\begin{gathered} v\left(t\right)=\frac{1}{2}\frac{2c^{2}t}{\sqrt{b^2+c^2{t}^2}} \\ v\left(t\right)=\frac{c^{2}t}{x} \end{gathered}$$

Write the equation to calculate the force$F_1$.

$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}\mathbf{=}\mathit{q}{\overrightarrow{\mathbf{E}}}_{\mathbf{1}}$ …… (1)

Here, $E_1$is the electric field due to charge$q_2$.

Write the expression for an electric field due to charge$q_2$.

$\overrightarrow{E}=-\frac{q_2}{4{\mathrm{\pi \epsilon }}_0}\frac{1}{{\left(x\left(t_r\right)\right)}^2}\left(\frac{c-v\left(t_r\right)}{c+v\left(t_r\right)}\right)\hat{x}$ …… (2)

Here, $v\left(t_r\right)$is the velocity, $x\left(t_r\right)$is the distance which is given as:

$x\left(t_r\right)=c\left(t-t_r\right)$ .....(3)

Substitute $x\left(t\right)=\sqrt{b^2+{\left(ct\right)}^2}$in the above expression.

$$\begin{gathered} \sqrt{b^2+{\left(c{t}_r\right)}^2}=c\left(t-t_r\right) \\ {\left(\sqrt{b^2+{\left(c{t}_r\right)}^2}\right)}^2={\left(c\left(t-t_r\right)\right)}^2 \\ {b}^2+c^2{t}_r^2=c^2{t}^2+c^2{t}_r^2-2c^{2}t{t}_r \\ {t}_r=\frac{c^2{t}^2-b^2}{2c^{2}t} \end{gathered}$$

Substitute$t_r=\frac{c^2{t}^2-b^2}{2c^{2}t}$in equation (3).

$$\begin{gathered} x\left(t_r\right)=c\left(t-\frac{c^2{t}^2-b^2}{2c^{2}t}\right) \\ x\left(t_r\right)=\frac{2c^2{t}^2-c^2{t}^2+b^2}{2ct} \\ x\left(t_r\right)=\frac{c^2{t}^2+b^2}{2ct} \end{gathered}$$

For $t>0$the value of $v\left(t\right)$will be,

$$\begin{gathered} v\left(t\right)=\frac{1}{2}\frac{2c^{2}t}{\sqrt{b^2+c^2{t}^2}} \\ v\left(t\right)=\frac{c^{2}t}{x} \end{gathered}$$

Calculate the value of$v\left(t_r\right)$.

$$\begin{gathered} v\left(t_r\right)=\frac{c^2{t}_r}{x} \\ v\left(t_r\right)=\frac{{\displaystyle \frac{c^2\left(c^2{t}^2-b^2\right)}{2c^{2}t}}}{x} \\ v\left(t_r\right)=c\left(\frac{c^2{t}^2-b^2}{c^2{t}^2+b^2}\right) \end{gathered}$$

Substitute all the known values in equation (2).

$$\begin{gathered} \overrightarrow{E}=-\frac{q_2}{4{\mathrm{\pi \epsilon }}_0}\frac{1}{{\left({\displaystyle \frac{c^2{t}^2+b^2}{2ct}}\right)}^2}\left(\frac{c-c\left({\displaystyle \frac{c^2{t}^2-b^2}{c^2{t}^2+b^2}}\right)}{c+c\left({\displaystyle \frac{c^2{t}^2-b^2}{c^2{t}^2+b^2}}\right)}\right) \\ \stackrel{\rightharpoonup }{E}=-\frac{q_2}{4{\mathrm{\pi \epsilon }}_0}\frac{1}{{\left({\displaystyle \frac{c^2{t}^2+b^2}{2ct}}\right)}^2}\left(\frac{b^2}{c^2{t}^2}\right)\hat{x} \\ \stackrel{\rightharpoonup }{E}=-\frac{q_2}{4{\mathrm{\pi \epsilon }}_0}\frac{4c^2{t}^2}{{\left(c^2{t}^2+b^2\right)}^2}\left(\frac{b^2}{c^2{t}^2}\right)\hat{x} \\ \stackrel{\rightharpoonup }{E}=-\frac{q_2}{4{\mathrm{\pi \epsilon }}_0}\frac{4b^2}{{\left(c^2{t}^2+b^2\right)}^2}\hat{x} \end{gathered}$$

Substitute$\stackrel{\rightharpoonup }{E}=-\frac{q_2}{4{\mathrm{\pi \epsilon }}_0}\frac{4b^2}{{\left(c^2{t}^2+b^2\right)}^2}\hat{x}$in equation (1).

$$\begin{gathered} {\stackrel{\rightharpoonup }{F}}_1=q_1\stackrel{\rightharpoonup }{E} \\ {\overrightarrow{F}}_1=q_1\left(-\frac{q_2}{4{\mathrm{\pi \epsilon }}_0}\frac{4b^2}{{\left(c^2{t}^2+b^2\right)}^2}\hat{x}\right) \\ {\stackrel{\rightharpoonup }{F}}_1=\frac{-q_1{q}_2}{4{\mathrm{\pi \epsilon }}_0}\frac{4b^2}{{\left(b^2+c^2{t}^2\right)}^2}\hat{x} \end{gathered}$$

Therefore, the force$F_1$on$q_1$is${\stackrel{\rightharpoonup }{F}}_1=\frac{-q_1{q}_2}{4{\mathrm{\pi \epsilon }}_0}\frac{4b^2}{{\left(b^2+c^2{t}^2\right)}^2}\hat{x}$.

(d)

Write the expression for the total impulse$I_1$ .

${\mathit{I}}_{\mathbf{1}}\mathbf{=}{\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}{\mathit{F}}_{\mathbf{1}}\mathit{d}\mathit{t}$

Substitute the known value of$F_1$ in the above expression.

$$\begin{gathered} I_1={\int }_{-\infty }^{\infty }\left(\frac{-q_1{q}_2}{4{\mathrm{\pi \epsilon }}_0}\frac{4b^2}{{\left(b^2+c^2{t}^2\right)}^2}\right)dt \\ {I}_1=-\frac{q_1{q}_2}{4{\mathrm{\pi \epsilon }}_0}{\int }_{-\infty }^{\infty }\left(\frac{4b^2}{b^2+c^2{t}^2}\right)dt \\ {I}_1=-\frac{q_1{q}_2}{4{\mathrm{\pi \epsilon }}_0}{\left(\frac{1}{bc}{\tan }^{-1}\left(\frac{ct}{b}\right)\right)}_{-\infty }^{\infty } \\ {I}_1=-\frac{q_1{q}_2}{4{\mathrm{\pi \epsilon }}_0}\frac{1}{bc}\left[{\tan }^{-1}\left(\infty \right)-{\tan }^{-1}\left(-\infty \right)\right] \end{gathered}$$

On further solving, the above equation becomes,

$$\begin{gathered} I_1=-\frac{q_1{q}_2}{4{\mathrm{\pi \epsilon }}_0}\frac{1}{bc}\left[\frac{\mathrm{\pi }}{2}-\left(-\frac{\mathrm{\pi }}{2}\right)\right] \\ {I}_1=-\frac{q_1{q}_2}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{\pi }}{bc} \\ {I}_1=-\frac{q_1{q}_2}{4{\epsilon }_{0}bc} \end{gathered}$$

Therefore, the total impulse$I_1$ delivered to$q_1$ by$q_2$ isrole="math" localid="1653895416777" $I_1=-\frac{q_1{q}_2}{4{\epsilon }_{0}bc}$ .