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More linear charge density troubles...

  • #1
778
36

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}}##
XFmLOon.png


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?
 

Answers and Replies

  • #2
cnh1995
Homework Helper
Gold Member
3,295
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An infinitely long line of charge has linear charge density λ=4.0010−12Cmλ=4.0010−12Cmλ=4.00_{10^{−12}} \frac{C}{m}. A proton (mass mp=1.6710−27kgmp=1.6710−27kgm_p=1.67_{10^{-27}}kg, charge e=1.60210−19Ce=1.60210−19Ce=1.602_{10^{-19}}C) is ra=0.18mra=0.18mr_a=0.18m from the line and moving directly toward the line at v=1000msv=1000msv=1000\frac{m}{s}.
So what is the question?
 
  • #3
778
36
Oops. Forgot to copy-paste that. Basically, it's: "How close does the proton get to the line of charge before bouncing back?"
 
  • #4
778
36
So I tested it out, and it turns out I'm right. I won't be needing any more input as a result.
 
  • #5
alejandromeira
I have reviewed your calculations and see the process all correct. It is a very clear problem of application of energy theorems.Remenber $$W_{conservative}=-\Delta U$$ The minus sign is very important. Maybe you've applied a recipe formula but you've done well.

On the other hand, because there are only conservative forces (electrostatical force is conservative), you can apply the principle of conservation of energy between the initial and final points (this is another point of view). $$K_{0}+U_{0}=K_{1}+U_{1}$$ where the final kinetic energy K1 is zero, since it stops.
*************************************************************************
I remember you the two principal energetic theorems in mechanics because, they are very very importants. If you meditate and solve problems with them you will learn a lot. The minus sign is very important. $$W_{total}=\Delta K$$ $$W_{conservative}=-\Delta U$$
 

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