Finding the displacement of an electron between 2 charged rods

In summary: 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}##In summary, the equations shown in the conversation involve finding the velocity and acceleration of an electron in an electric field, using the formula v^2 = u^2 + 2as. The value of the electric field is found using the formula a = qe/m, where q is the charge of the electron and m is its mass. The conversation
  • #1
jisbon
476
30
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?
 
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  • #2
jisbon said:
Are the equations correct?
What are the equations, what do the symbols represent ?
Did you check the dimensions ?
 
  • #3
BvU said:
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:

1569897962449.png


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?
 
  • #4
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.
 
  • #5
Delta2 said:
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?
 
  • #6
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.
 
  • #7
Delta2 said:
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)
##
 
  • #8
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.
 
  • #9
Delta2 said:
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?
 
  • #10
jisbon said:
##\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.
 
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  • #11
Delta2 said:
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
 
  • #12
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}##
 

FAQ: Finding the displacement of an electron between 2 charged rods

1. What is the definition of displacement in terms of electrons between charged rods?

The displacement of an electron between two charged rods refers to the distance and direction an electron moves when subjected to the electric field created by the charged rods.

2. How do you calculate the displacement of an electron between two charged rods?

The displacement of an electron can be calculated by using the formula: displacement = (electric field strength) x (distance between the rods).

3. What factors affect the displacement of an electron between two charged rods?

The displacement of an electron can be affected by the strength of the electric field, the distance between the rods, and the charge on the rods.

4. What units are used to measure the displacement of an electron?

The displacement of an electron is typically measured in meters (m) or centimeters (cm).

5. How does the displacement of an electron between two charged rods relate to electric potential?

The displacement of an electron is directly proportional to the electric potential difference between the two charged rods. This means that as the potential difference increases, the displacement of an electron also increases.

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