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## Homework Statement

An infinitely long line of charge has linear charge density ##λ=4.00_{10^{−12}} \frac{C}{m}##. A proton (mass ##m_p=1.67_{10^{-27}}kg##, charge ##e=1.602_{10^{-19}}C##) is ##r_a=0.18m## from the line and moving directly toward the line at ##v=1000\frac{m}{s}##.

## Homework Equations

##dQ=λdl=λdy##

##\hat{r}=\frac{r}{\sqrt{r^2+y^2}}##

## The Attempt at a Solution

Okay, I start by equating initial potential and kinetic energy to final mechanical energy, which is just final potential energy, since the proton has stopped moving. So... ##K_0+U_0=U_f##. What I had trouble doing was expressing ##U## in terms of the linear charge density. This is what I did.

##F=\frac{1}{4\pi ε_0}(q)⋅∫\frac{dQ}{r^2+y^2}\hat {r}=\frac{1}{4\pi ε_0}(λq)⋅∫\frac{dy}{r^2+y^2}\hat {r}##

##F=\frac{1}{4\pi ε_0}(λq)⋅∫\frac{rdy}{\sqrt{(r^2+y^2)^3}}=\frac{λqr}{4\pi ε_0}∫\frac{dy}{\sqrt{(r^2+y^2)^3}}##

Let ##y=rtan\theta##. Then ##dy=rsec^2\theta d\theta## and ##\sqrt{(r^2+y^2)^3}=r^3sec^3\theta##. So...

##F=\frac{λqr}{4\pi ε_0}∫\frac{rsec^2\theta d\theta}{r^3sec^3\theta}=\frac{λq}{4\pi ε_0⋅r}∫cos\theta d\theta=\frac{λq}{4\pi ε_0⋅r}sin\theta##

Or, switching back to regular coordinates...

##F=\frac{λq}{4\pi ε_0⋅r}(\frac{y}{\sqrt{y^2+r^2}})## on ##-\infty < y < \infty##

So I get ##F=\frac{2λq}{4\pi ε_0⋅r}##

Then I want to integrate over ##dr##.

##∫F⋅dr=W=\frac{2λq}{4\pi ε_0}⋅\int_{r_a}^{r_b} \frac{dr}{r}=\frac{2λq}{4\pi ε_0}⋅ln|r|## on ##r_b<r<r_a##

Then I plug this into my formula for ##K+U_0=U_f##, given that ##W=-ΔU=-(U_f-U_0)##. I solve for ##r_b##, and that's my process. Can anyone tell me if I messed up anywhere?