Conservation of Energy and Electric Potential

1. Nov 7, 2012

fornax

1. The problem statement, all variables and given/known data
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?

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

3. 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.

2. Nov 7, 2012

ehild

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

ehild

3. Nov 7, 2012

Simon Bridge

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.

4. Nov 8, 2012

fornax

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.

5. Nov 8, 2012

aralbrec

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.

6. Nov 8, 2012

Simon Bridge

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: Nov 8, 2012
7. Nov 8, 2012

fornax

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.

8. Nov 8, 2012

Simon Bridge

This is why I keep banging on about doing the algebra first.
See how easy it is to follow what I'm doing? Of course that means it is easier for people to see I've made a mistake ... but that's why we are here right?

But nobody ever listens :(

BTW: notice how aralbrec challenged my earlier result?
He used a quick reality check to show what I got was inconsistent with the physical situation.