One particle, of charge q1, is held at rest at the origin. Another particle, of charge q2, approaches along the x axis, in hyperbolic motion: x(t) = √b^2 + (ct)^2 ; it reaches the closest point, b, at time t = 0, and then returns out to infinity. (a) What is the force F2 on q2 (due to q1) at time

Question

One particle, of charge [latex]q_{1}[/latex], is held at rest at the origin. Another particle, of charge [latex]q_{2}[/latex], approaches along the x axis, in hyperbolic motion:

[latex]x(t)=\sqrt{b^{2}+(c t)^{2}}[/latex];

it reaches the closest point, b, at time t = 0, and then returns out to infinity.

(a) What is the force [latex]F_{2} ext { on } q_{2}[/latex] (due to q1) at time t?

(b) What total impulse [latex]\left(I_{2}=\int_{-\infty}^{\infty} F_{2} d t ight)[/latex] is delivered to [latex]q_{2} ext { by } q_{1}[/latex]?

(c) What is the force [latex]F_{1} ext { on } q_{1}\left( ext { due to } q_{2} ight)[/latex] at time t?

(d) What total impulse [latex]\left(I_{1}=\int_{-\infty}^{\infty} F_{1} d t ight) ext { is delivered to } q_{1} ext { by } q_{2}[/latex] ? [Hint: It might help to review Prob. 10.17 before doing this integral. Answer: [latex]I_{2}=-I_{1}=\left.q_{1} q_{2} / 4 \epsilon_{0} b c ight][/latex]

Accepted Answer

[latex] ext { (a) } F _{2}=\frac{q_{1} q_{2}}{4 \pi \epsilon_{0}} \frac{1}{\left(b^{2}+c^{2} t^{2} ight)} \hat{ x }[/latex].

(This is just Coulomb’s law, since [latex]q_{1}[/latex] is at rest.)

[latex] ext { (b) } I_{2}=\frac{q_{1} q_{2}}{4 \pi \epsilon_{0}} \int_{-\infty}^{\infty} \frac{1}{\left(b^{2}+c^{2} t^{2} ight)} d t=\left.\frac{q_{1} q_{2}}{4 \pi \epsilon_{0}}\left[\frac{1}{b c} an ^{-1}(c t / b) ight] ight|_{-\infty} ^{\infty}=\frac{q_{1} q_{2}}{4 \pi \epsilon_{0} b c}\left[ an ^{-1}(\infty)- an ^{-1}(-\infty) ight][/latex]

[latex]=\frac{q_{1} q_{2}}{4 \pi \epsilon_{0} b c}\left[\frac{\pi}{2}-\left(-\frac{\pi}{2} ight) ight]=\frac{q_{1} q_{2}}{4 \pi \epsilon_{0}} \frac{\pi}{b c}[/latex].

(c) From Prob. 10.20, [latex]E =-\frac{q_{2}}{4 \pi \epsilon_{0}} \frac{1}{x^{2}}\left(\frac{c-v}{c+v} ight) \hat{ x }[/latex] . Here x
and υ are to be evaluated at the retarded time tr, which is given by [latex] c\left(t-t_{r} ight)=x\left(t_{r} ight)=\sqrt{b^{2}+c^{2} t_{r}^{2}} \Rightarrow c^{2} t^{2}-2 c t t_{r}+c^{2} t_{r}^{2}=b^{2}+c^{2} t_{r}^{2} \Rightarrow t_{r}=\frac{c^{2} t^{2}-b^{2}}{2 c^{2} t}[/latex] . Note: As we found in Prob. 10.17, [latex]q_{2}[/latex] first “comes into view” [latex] ext { (for } q_{1} ext { ) }[/latex] at time t = 0. Before that it can exert no force on [latex]q_{1}[/latex] , and there is no retarded time. From the graph of [latex]t_{r}[/latex] versus t we see that [latex]t_{r}[/latex] ranges all the way from [latex]-\infty ext { to } \infty ext { while } t>0[/latex].

[latex]x\left(t_{r} ight)=c\left(t-t_{r} ight)=\frac{2 c^{2} t^{2}-c^{2} t^{2}+b^{2}}{2 c t}=\frac{b^{2}+c^{2} t^{2}}{2 c t}( ext { for } t>0) . \quad v(t)=\frac{1}{2} \frac{2 c^{2} t}{\sqrt{b^{2}+c^{2} t^{2}}}=\frac{c^{2} t}{x}[/latex] , so

[latex]v\left(t_{r} ight)=\left(\frac{c^{2} t^{2}-b^{2}}{2 t} ight)\left(\frac{2 c t}{b^{2}+c^{2} t^{2}} ight)=c\left(\frac{c^{2} t^{2}-b^{2}}{c^{2} t^{2}+b^{2}} ight)( ext { for } t>0)[/latex] . Therefore

[latex]\frac{c-v}{c+v}=\frac{\left(c^{2} t^{2}+b^{2} ight)-\left(c^{2} t^{2}-b^{2} ight)}{\left(c^{2} t^{2}+b^{2} ight)+\left(c^{2} t^{2}-b^{2} ight)}=\frac{2 b^{2}}{2 c^{2} t^{2}}=\frac{b^{2}}{c^{2} t^{2}}( ext { for } t>0) . \quad E =-\frac{q_{2}}{4 \pi \epsilon_{0}} \frac{4 c^{2} t^{2}}{\left(b^{2}+c^{2} t^{2} ight)^{2}} \frac{b^{2}}{c^{2} t^{2}} \hat{ x } \Rightarrow[/latex]

[latex]F _{1}= \begin{cases}0, & t<0 \ -\frac{q_{1} q_{2}}{4 \pi \epsilon_{0}} \frac{4 b^{2}}{\left(b^{2}+c^{2} t^{2} ight)^{2}} \hat{ x }, & t>0\end{cases}[/latex]

[latex] ext { (d) } I_{1}=-\frac{q_{1} q_{2}}{4 \pi \epsilon_{0}} 4 b^{2} \int_{0}^{\infty} \frac{1}{\left(b^{2}+c^{2} t^{2} ight)^{2}} d[/latex] . The integral is

[latex]\frac{1}{c^{4}} \int_{0}^{\infty} \frac{1}{\left[(b / c)^{2}+t^{2} ight]^{2}} d t=\frac{1}{c^{4}}\left(\frac{c^{2}}{2 b^{2}} ight)\left[\left.\frac{t}{(b / c)^{2}+t^{2}} ight|_{0} ^{\infty}+\int_{0}^{\infty} \frac{1}{\left.\left[(b / c)^{2}+t^{2} ight) ight]} d t ight]=\frac{1}{2 c^{2} b^{2}}\left(\frac{\pi c}{2 b} ight)=\frac{\pi}{4 c b^{3}}[/latex]

[latex] ext { So } I_{1}=-\frac{q_{1} q_{2}}{4 \pi \epsilon_{0}} \frac{\pi}{b c} ext {. }[/latex]

(e) [latex]F _{1} eq- F _{2}[/latex] , so Newton’s third law is not obeyed. On the other hand, [latex]I_{1}=-I_{2}[/latex] in this instance, which suggests that the net momentum delivered from (1) to (2) is equal and opposite to the net momentum delivered from (2) to (1), and hence that the total mechanical momentum is conserved. (In general the fields might carry off some momentum, leaving the mechanical momentum altered; but that doesn’t happen in the present case.)

Question 10.28: One particle, of charge q1, is held at rest at the origin. A...

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