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Question Number 104572 by ajfour last updated on 22/Jul/20

Commented by ajfour last updated on 22/Jul/20

$$Apointchargeqwithmassm$$$$isreleasedatadistance\mathit{a}from$$$$afixeduniformlychargedrod$$$$oflengthL,chargeQ.Findthe$$$$speedacquiredbypointcharge$$$$qasitreachesaninfinite$$$$distancefromtherod.$$

Answered by OlafThorendsen last updated on 22/Jul/20

$$\mathrm{E}=\frac{\mathrm{Q}}{2\pi {ϵ}_0}.\frac{1}{x}.\frac{1}{x+\mathrm{L}}$$$$\mathrm{F}=q\mathrm{E}=m\stackrel{\bullet \bullet }{x}$$$$\frac{q\mathrm{Q}}{2\pi {ϵ}_0}.\frac{1}{x(x+\mathrm{L})}=m\stackrel{\bullet \bullet }{x}$$$$\frac{q\mathrm{Q}}{2\pi {ϵ}_0\mathrm{L}}.\frac{\stackrel{\bullet }{x}}{x(x+\mathrm{L})}=m\stackrel{\bullet \bullet }{x}\stackrel{\bullet }{x}(\mathrm{with}v=\stackrel{\bullet }{x})$$$$\frac{q\mathrm{Q}}{2\pi {ϵ}_0{\mathrm{L}}^2}\stackrel{\bullet }{x}[\frac{1}{x}-\frac{1}{x+\mathrm{L}}]=m\stackrel{\bullet \bullet }{x}\stackrel{\bullet }{x}$$$$\frac{q\mathrm{Q}}{2\pi {ϵ}_0{\mathrm{L}}^2}.\ln \frac{x}{x+\mathrm{L}}=\frac{1}{2}{mv}^2+\mathrm{C}$$$$t=0,v=0\Rightarrow \mathrm{C}=-\frac{q\mathrm{Q}}{2\pi {ϵ}_0{\mathrm{L}}^2}.\ln \frac{a}{a+\mathrm{L}}$$$$\frac{1}{2}{mv}^2=\frac{q\mathrm{Q}}{2\pi {ϵ}_0{\mathrm{L}}^2}(\ln \frac{x}{x+\mathrm{L}}-\ln \frac{a}{a+\mathrm{L}})$$$$v_{\mathrm{\infty }}^2=\frac{q\mathrm{Q}}{\pi {ϵ}_{0}m{\mathrm{L}}^2}\ln \frac{a+\mathrm{L}}{a}$$$$...maybe$$

Commented by ajfour last updated on 22/Jul/20

$$DontyouthinkSir,$$$$E=\frac{Q}{4\pi {ϵ}_0}\frac{1}{x(x+L)}?$$

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