How the distance from centre decides type of charge motion?

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Discussion Overview

The discussion revolves around the motion of a charge -Q in relation to its distance from the center of a uniformly charged sphere. Participants explore the differences between oscillatory motion and simple harmonic motion (SHM), particularly how the distance from the center influences the type of motion exhibited by the charge. The conversation includes theoretical considerations and interpretations of force laws.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants propose that a charge -Q at a large distance from the center will exhibit oscillatory and periodic motion, while a charge released close to the origin will execute SHM.
  • One participant questions what is at the origin that causes this difference in motion and seeks to distinguish between "oscillatory and periodic" motion and SHM.
  • It is suggested that a diffuse, spherically symmetric cloud with uniform positive charge density centered on the origin influences the motion of the charge -Q.
  • Participants discuss the implications of the spherical shell theorem, noting that only the portion of the charge cloud closer to the origin affects the net force on the test charge.
  • There is a clarification that the net attraction experienced by the charge is proportional to its distance from the origin, due to the volume of the charge cloud being proportional to r³ and the force law being inversely proportional to r².
  • Some participants express confusion over the definitions and relationships between the distances and charges involved in the discussion.
  • There is an agreement that the portion of the charge cloud farther from the origin than the test charge produces no net electrical field on the test charge.

Areas of Agreement / Disagreement

Participants generally agree on the application of the spherical shell theorem and its implications for the net electrical field. However, there remains some confusion and lack of consensus regarding the definitions and implications of oscillatory motion versus SHM, as well as the specific relationships between the distances and charges discussed.

Contextual Notes

Some participants express uncertainty about the definitions of variables and the implications of the spherical shell theorem in this context. The discussion includes unresolved mathematical relationships and assumptions regarding the charge distribution and its effects on motion.

  • #31
gracy said:
8 times?
Correct.

Now, with eight times the charge you have an electric field that is eight times as strong. But remember that we doubled the radius of the shell. If the test charge is on the surface of this shell, it is now twice as far from the origin as it had been.

Ignoring everything else, if an inverse square force law is in effect, doubling distance reduces attraction by what factor?
 
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  • #32
jbriggs444 said:
If the test charge is on the surface of this shell
That's where it was in the first place,wasn't it?
 
  • #33
jbriggs444 said:
Somewhere else within this cloud of charge we have a test charge.
We measure the distance from the origin to the test charge.
We call this distance "r".
We draw an imaginary spherical shell that is centered on the origin and has radius r.
gracy said:
That's where it was in the first place,wasn't it?
 
  • #34
jbriggs444 said:
Somewhere else within this cloud of charge we have a test charge.
We measure the distance from the origin to the test charge.
We call this distance "r".
We draw an imaginary spherical shell that is centered on the origin and has radius r.
jbriggs444 said:
If the test charge is on the surface of this shell
gracy said:
That's where it was in the first place,wasn't it?
wasn't it.png
 
  • #35
jbriggs444 said:
t is now twice as far from the origin as it had been.
jbriggs444 said:
if an inverse square force law is in effect, doubling distance reduces attraction by what factor?
4 times.
 
  • #36
gracy said:
4 times.
So if you put that together, moving the test charge to twice its original distance, the charge enclosed by a spherical shell centered on the origin and just touching the test charge has increased by a factor of 8. Due to inverse square, the attraction induced by a given charge has decreased by a factor of 4.

By what factor has the attraction of the test charge to the enclosed charge increased or decreased? Has it increased? Has it decreased?
 
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  • #37
what about my post #34?
 
  • #38
gracy said:
what about my post #34?
What about it?
 
  • #39
correct or not?What's wrong n it?where am I missing?
 
  • #40
gracy said:
correct or not?What's wrong n it?where am I missing?
Your post 34 does not make a statement. How can it be right or wrong?
 
  • #41
gracy said:
where am I missing?
What's wrong in it?Clear my doubt.
 
  • #42
gracy said:
What's wrong in it?Clear my doubt.
Asked and answered. It is neither right nor wrong because it makes no statement.
 
  • #43
jbriggs444 said:
If the test charge is on the surface of this shell, it is now twice as far from the origin as it had been.
How?As per my understanding ,it is at the same distance.r is not doubled.
 
  • #44
We increased the size of the shell and we moved the test charge.

Edit: Hence the words "had been" -- to emphasize that something changed.
 
  • #45
jbriggs444 said:
By what factor has the attraction of the test charge to the enclosed charge increased or decreased?
increased ,2 times..
 
  • #46
gracy said:
increased ,2 times..
Right.

Now, generalize this. Is the attraction of the test charge toward the origin proportional to its distance, r, from the origin?
 
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  • #47
jbriggs444 said:
Now, generalize this. Is the attraction of the test charge toward the origin proportional to its distance, r, from the origin?
yes.
 
  • #48
gracy said:
yes.
Is this the same relationship that holds for simple harmonic motion -- that attraction toward the equilibrium point is proportional to displacement from the equilibrium point?
 
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  • #49
jbriggs444 said:
Is this the same relationship that holds for simple harmonic motion -- that attraction toward the equilibrium point is proportional to displacement from the equilibrium point?
yes.but what happens when it reaches at origin??
 
  • #50
Why should anything happen? There's nothing at the origin but more of the same uniform charge cloud that surrounds the origin.
 
  • #51
I mean will the charge go to the other side of the origin?
 
  • #52
gracy said:
I mean will the charge go to the other side of the origin?
I have been implicitly assuming that the test charge is on an object with some non-zero mass. (A reasonable assumption because all charged objects have non-zero mass).

What is the force on the test charge at the origin? What does Newton's first law have to say?
 
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  • #53
jbriggs444 said:
What is the force on the test charge at the origin? What does Newton's first law have to say?
mass into acceleration
 
  • #54
No. That is a mangling of Newton's second law. Please try again.
 
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  • #55
oh,sorry.First law,ok.It says An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.
 
  • #56
And the first part of my question asked about the force on the test charge at the origin. Can you answer that?
 
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  • #57
As origin is equilibrium position total force has to be zero,but that's what I want to ask why this center of charge i..e origin is equilibrium position?
 
  • #58
You are reasoning backwards.

You are apparently not yet convinced that the origin is an equilibrium position. You cannot argue that force is zero there based on something that you do not even believe to be true. Instead, you need to convince yourself that force is zero there.

So... What is the total electrostatic force on the test charge when it is in the center of a uniform spherical cloud of charge?
 
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  • #59
jbriggs444 said:
So... What is the total electrostatic force on the test charge when it is in the center of a uniform spherical cloud of charge?
I now understand force from both side will cancel each other.Resultant force will be zero.
 
  • #60
gracy said:
As origin is equilibrium position total force has to be zero
gracy said:
I mean will the charge go to the other side of the origin?
Is the answer no then?
 

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