Finding the displacement of an electron between 2 charged rods

[jisbon]

Homework Statement

An electron 1m from the left rod (2 micro C/m) released with initial velocity $1.04*10^8 m/s$ towards the right rod (-2 micro C/m). The length between 2 rods in 4m.
Find maximum distance it can go to the right.

Relevant Equations

-

Since electron will stop eventually due to efield,
equation is : $v^2 = u^2 +2as$
Where v = 0 , u = $1.04*10^8 m/s$

$a = \frac{qe}{m} =\frac{(1.6*10^{-19})(e)}{9.11*10^{-31}}$
$e = \frac{Q}{4\pi\epsilon(1)}+\frac{Q}{4\pi\epsilon(4-1)}$

Are the equations correct? Or is my concepts wrong?

 

[BvU]

Are the equations correct?

What are the equations, what do the symbols represent ?
Did you check the dimensions ?

 

[jisbon]

What are the equations, what do the symbols represent ?
Did you check the dimensions ?

Well,

to find maximum distance it can go, it means v=0, where final speed equals to zero.
u is the initial speed and s is the value I'm trying to find.

As for $a = \frac{qe}{m} =\frac{(1.6*10^{-19})(e)}{9.11*10^{-31}}$ , it's to find out the acceleration of the electron in an e-field using the formula $F=qe$, where $F=ma$
Since being able to find out the above formula, I need to find out the formula of the efield, which can be explained by:

To find efield in that position, I will have to find where

$E_{sum} = E_{left rod}+E_{right rod} = \frac{2}{4\pi \epsilon_{0}(1)}+\frac{-2}{4\pi \epsilon_{0}(4-1)}$

Am I on the right track so far?

 

[Delta2]

The E-field will not be uniform in the line in-between the rods, so the acceleration will not be uniform (constant) either so the equation $v^2=u^2+2as$ cannot be applied because that equation is for constant acceleration through out the motion.
I think you have to work with potentials here instead of electric field and acceleration. What is the total (combined due to both rods) potential at the initial position of 1m? What is the potential at the final position (let it be x). The work done will be $(V_{1m}-V_x)q_e$. Use work - energy theorem too.

 

[jisbon]

The E-field will not be uniform in the line in-between the rods, so the acceleration will not be uniform (constant) either so the equation $v^2=u^2+2as$ cannot be applied because that equation is for constant acceleration through out the motion.
I think you have to work with potentials here instead of electric field and acceleration. What is the total (combined due to both rods) potential at the initial position of 1m? What is the potential at the final position (let it be x). The work done will be $(V_{1m}-V_x)q_e$. Use work - energy theorem too.

To get potential, is it correct to say I can integrate $\frac{2}{4\pi \epsilon_{0}(r)}$
or should I derive the whole thing again?

 

[Delta2]

Yes that's how you get the potential , you integrate the e-field from the position r to infinite but I thought you already knew the potential for the case of a linear rod. Don't forget to get the combined potential $V_{rod1,r}+V_{rod2,r}$ at a position $r$ due to both rods.

 

[jisbon]

Yes that's how you get the potential , you integrate the e-field from the position r to infinite but I thought you already knew the potential for the case of a linear rod. Don't forget to get the combined potential $V_{rod1,r}+V_{rod2,r}$ at a position $r$ due to both rods.

Just would like to double-check, but is the potential due to the first and second rod the respective equations below?
Due to left rod:
$\Delta V = -\int \vec E \cdot d\vec r = -\int_{1}^{\infty} \vec E \cdot \frac{d\vec r}{dt} dt = -\frac{\lambda}{2\pi\epsilon_0}\int_{1}^{\infty} \frac{dt}{t} dt = -\frac{\lambda}{2\pi\epsilon_0}\ln\left(\frac{1}{\infty}\right)$

 

[Delta2]

I wonder why you enter dt into the equation but anyway I see your point, we going to mess with infinities.
Forget about potentials, just integrate the total E-field from position $r=1m$ to position $r=x$ and use work energy theorem.

 

[jisbon]

I wonder why you enter dt into the equation but anyway I see your point, we going to mess with infinities.
Forget about potentials, just integrate the total E-field from position $r=1m$ to position $r=x$ and use work energy theorem.

$\int_{1}^{x} \frac{2}{4\pi \epsilon_{0}(r)} +\int_{1}^{x} \frac{2}{4\pi \epsilon_{0}(3)}$ which means something like this?

 

[Delta2]

$\int_{1}^{x} \frac{2}{4\pi \epsilon_{0}(r)} +\int_{1}^{x} \frac{2}{4\pi \epsilon_{0}(3)}$ which means something like this?

yes almost you need to have 4-r instead of just 3 there.

 

[jisbon]

yes almost you need to have 3-r instead of just 3 there.

Where r=1. So 4-1=3 no? Or am I doing something wrong

 

[Delta2]

No it isn't r=1, r is the position of the electron as it moves from r=1 to r=x. We need a variable of integration there to be able to integrate something ! We going to calculate work done by the E-field as $\int_1^x q_e\vec{E_{total}}(r)\cdot d\vec{r}$