91影视

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.

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

Here, q is the charge,${蔚}_0$ 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" $$\begin{gathered} F_2=\frac{1}{4{\mathrm{蟺蔚}}_0}\frac{q_1{q}_2}{{\left(\sqrt{b^2+{\left(ct\right)}^2}\right)}^2} \\ {F}_2=\frac{q_1{q}_2}{4{\mathrm{蟺蔚}}_0}\frac{1}{\left(b^2+c^2{t}^2\right)} \end{gathered}$$

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

(b)

Write the expression for the total impulse$I_2$.

${\mathit{I}}_{\mathbf{2}}\mathbf{=}{\mathbf{鈭�}}_{\mathbf{-}\mathbf{鈭�}}^{\mathbf{鈭�}}{\mathit{F}}_{\mathbf{2}}\mathit{d}\mathit{t}$

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

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

On further solving, the above equation becomes,

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

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

(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$F_1$.

$\stackrel{\mathbf{鈬赌}}{\mathbf{F}}\mathbf{=}\mathit{q}{\stackrel{\mathbf{鈫�}}{\mathbf{E}}}_{\mathbf{1}}$ 鈥︹�� (1)

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

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

$\stackrel{鈫�}{E}=-\frac{q_2}{4{\mathrm{蟺蔚}}_0}\frac{1}{{\left(x\left(t_r\right)\right)}^2}\left(\frac{c-v\left(t_r\right)}{c+v\left(t_r\right)}\right)\hat{x}$ 鈥︹�� (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).

$$\begin{gathered} \stackrel{鈫�}{E}=-\frac{q_2}{4{\mathrm{蟺蔚}}_0}\frac{1}{{\left({\displaystyle \frac{c^2{t}^2+b^2}{2ct}}\right)}^2}\left(\frac{c-c\left({\displaystyle \frac{c^2{t}^2-b^2}{c^2{t}^2+b^2}}\right)}{c+c\left({\displaystyle \frac{c^2{t}^2-b^2}{c^2{t}^2+b^2}}\right)}\right) \\ \stackrel{鈬赌}{E}=-\frac{q_2}{4{\mathrm{蟺蔚}}_0}\frac{1}{{\left({\displaystyle \frac{c^2{t}^2+b^2}{2ct}}\right)}^2}\left(\frac{b^2}{c^2{t}^2}\right)\hat{x} \\ \stackrel{鈬赌}{E}=-\frac{q_2}{4{\mathrm{蟺蔚}}_0}\frac{4c^2{t}^2}{{\left(c^2{t}^2+b^2\right)}^2}\left(\frac{b^2}{c^2{t}^2}\right)\hat{x} \\ \stackrel{鈬赌}{E}=-\frac{q_2}{4{\mathrm{蟺蔚}}_0}\frac{4b^2}{{\left(c^2{t}^2+b^2\right)}^2}\hat{x} \end{gathered}$$

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

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

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

(d)

Write the expression for the total impulse$I_1$ .

${\mathit{I}}_{\mathbf{1}}\mathbf{=}{\mathbf{鈭�}}_{\mathbf{-}\mathbf{鈭�}}^{\mathbf{鈭�}}{\mathit{F}}_{\mathbf{1}}\mathit{d}\mathit{t}$

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

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

On further solving, the above equation becomes,

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

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