# Conservation of Energy and Electric Potential

## Homework Statement

A very long, thin straight line of charge has a constant charge density of 2.0pC/cm. An electron is initially 1.0cm from the line and moving away with a speed of 1000km/s. How far does the electron go before it comes back?

## Homework Equations

ΔU = ΔV*q
ΔKE + ΔU = 0
ΔV = -2Kλln(rb/ra)
W = 2Kλqln(rb/ra)
W = -ΔU

## The Attempt at a Solution

first off:
2.0pC/cm = 200pC/m

1000km/s = 1x10^6m/s

KEa + Ua = KEb + Ub
1/2mv^2 + Ua = 0 + Ub
1/2mv^2 = Ub-Ua
1/2 * (9.1x10^-31)*(1x10^6)^2 = -ΔU
4.55x10^-19 = -ΔU
W = -ΔU
4.55x10^-19 = 2Kλqln(rb/ra)
4.55x10^-19 = 2K(200x10^-4)(1.6x10^-19)ln(rb/.01)
7.899x10^-9 = ln(rb) - ln.01
ln(rb) = 4.605
rb = 100m

Well, that is the process that I followed, and I feel like I did ok, except that the majority of the rest of my class got different answers. They all got .66m, which seems more realistic, 100m just seems too far. I have another question that again, I seem to get a different value than everyone else. I think I may be screwing up the conservation part in the beginning, but I have no idea where I'm going wrong. I asked my teacher, and he said to use ΔU = ΔV*q, but other than considering ΔV = -2Kλln(rb/ra) & W = 2Kλqln(rb/ra), I don't know where I would use it. Any help would be greatly appreciated.

ehild
Homework Helper
You might made a mistake by converting pC to coulombs. 1pC =10-12 C.

ehild

Simon Bridge
Homework Helper
I find it helps to do the whole algebra before putting numbers into the equations.
I get: ##\frac{1}{2}mv^2=2ke\lambda\ln(r_0/r)## ... find r.
Hmm ... at first glance - I think you have rb and ra swapped over in the log.

Well, using the observation that you made ehild, it did turn out that was a big chunk of my error. Replacing all of the 10^-4 with 10^-12 did give me a lower result, but not the one I was hoping for, I got .022m. I'm not sure what to make of it, it seems more reasonable than 100m, but is now relatively short. I know that when working with particles with such small mass it can be extremely hard to predict your results, and this is why I am unsure of it. I know that I shouldn't base my expectations for my answer of off my classmates work, but when the majority get the same answer it's hard to ignore.

As far as switching the ra & rb, I did give it a go, and I ended up with 220.3m. Also going off the intergration 2Kλq∫ a->b 1/r dr = 2Kλq : b-a : ∫1/r dr = lnr evaluated at b-a lnb - lna = ln(b/a). I do appreciate the help though, and if I goofed this somehow, let me know. The detailss always get me, and it's the details that count :/

I'll post my other problem, and see if you guys see any mistakes there, if you do, maybe it will reveal the flaw in my thinking. Perhaps I can get some insight into where my problem is.

I get: ##\frac{1}{2}mv^2=2ke\lambda\ln(r_0/r)## ... find r.

If you find the work = q ∫ E dr, the integration is from ro to r which leads to ln(r/ro).

The steps taken look right to me.

Simon Bridge
Homework Helper
Like I said: lets take it formally and do all the algebra first.
Double-check the conversions and list all the values used.

Starting with: $$mv^2+4ke\lambda\ln\left( \frac{r_i}{r_0} \right )=4ke\lambda\ln\left( \frac{r_f}{r_0} \right )$$... where ##r_0## is the radius of some reference potential - I get: $$mv^2=4ke\lambda\ln\left( \frac{r_f}{r_0} \right )-4ke\lambda\ln\left( \frac{r_i}{r_0} \right )=4ke\lambda\ln\left( \frac{r_f}{r_0}\frac{r_0}{r_i} \right ) = 4ke\lambda\ln\left ( \frac{r_f}{r_i} \right )$$... (ahh right!) and solve for ##r_f## :$$r_f=r_i\exp\left [ \frac{mv^2}{4ke\lambda} \right ]$$... using the following values:

##\lambda = 2.0\text{pC/cm} = 2.0\times 10^{-12}\text{C/cm}=2.0\times 10^{-10}\text{C/m}##
##r_i=1.0\text{cm}##
##v=1000\text{km/s} = 1000000\text{m/s}##
##e=1.60\times 10^{-19}\text{C}=1.60\times 10^{-7}\text{pC}##
##m=9.11\times 10^{-31}kg##
##k=8.99\times 10^9\text{Nm$^2$/C$^2$}##

I am getting: ##r_f=2.2\text{cm}## (too!) ...
If the rest of the class is correct, then the both of us have missed out a factor of 30 someplace. We need some other way to evaluate this to check.

Last edited:
I have to say I like how neat your method is, without even putting numbers in you have a nice neat formula to follow. As far as that factor of 30, I suppose there is a possiblity that they are wrong? Our professor never really confirmed it was correct, so there is potential for that situation. This is like a take home quiz, so we like to at least compare answers, so there is also the possibility people ar just copying each other. I think I'll keep mulling it over, maybe ask my professor about it, at least I'll probably get partial credit if it is wrong.

I really appreciate your help with this! It's nice to know I have somewhere to go when I really get stumped or confused.

Simon Bridge