How the distance from centre decides type of charge motion?

AI Thread Summary
The discussion centers on how the distance from the center of a charge distribution affects the type of motion exhibited by a test charge. A charge far from the center experiences oscillatory motion due to an inverse square force law, while a charge close to the center undergoes simple harmonic motion (SHM) because the force becomes directly proportional to the distance from the origin. The spherical shell theorem is referenced to explain that only the charge within a certain radius contributes to the net electric field experienced by the test charge. The conversation also clarifies that the acceleration in SHM is proportional to displacement, distinguishing it from general oscillatory motion. Ultimately, the relationship between force and distance is crucial in determining the type of motion.
gracy
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If charge -Q of mass m is kept on y-axis at large distance from the center,then it will execute oscillatory and periodic motion but if it is released very close to the origin then it will execute SHM .I want to know why?I mean how the distance from centre decides type of motion oscillatory or SHM?
 
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gracy said:
If charge -Q of mass m is kept on y-axis at large distance from the center,then it will execute oscillatory and periodic motion but if it is released very close to the origin then it will execute SHM .I want to know why?I mean how the distance from centre decides type of motion oscillatory or SHM?
What is at the origin to cause this? What, in your opinion distinguishes "oscillatory and periodic" motion from simple harmonic motion?

Edit: at a guess, we have a diffuse, spherically symmetric cloud with a uniform positive charge density centered on the origin. A charge far away sees an inverse square force law (-1/r^2) and a charge located within the cloud sees a force law that is directly proportional to distance (-r).
 
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jbriggs444 said:
oscillatory and periodic" motion from simple harmonic motion?
Both are periodic. The basic difference between the two is that in shm acceleration is directly proportional to displacement which is not there in case of oscillation.
 
jbriggs444 said:
A charge far away sees an inverse square force law (-1/r^2)
This is coulomb's force,I know.
jbriggs444 said:
and a charge located within the cloud sees a force law that is directly proportional to distance (-r)
But I don't know which force are you talking about here?
 
gracy said:
This is coulomb's force,I know.
But I don't know which force are you talking about here?
The Coulomb force.

According to the spherical shell theorem, the portion of the charge cloud farther from the origin than the -Q test charge cancels out to a net of zero. So only the portion of the cloud nearer the origin than the test charge matters. That portion of the cloud has volume proportional to r3 and inverse square attraction proportional to 1/r2. That means that the net attraction is proportional to r.
 
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jbriggs444 said:
spherical shell theorem
but it is for gravitation.
 
gracy said:
but it is for gravitation.
That is for any force that follows an inverse square law.
 
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Ok.Please don't go anywhere.I have to some queries.
 
jbriggs444 said:
That portion of the cloud has volume proportional to r3 and inverse square attraction proportional to 1/r2. That means that the net attraction is proportional to r.
could you please elaborate this part?
 
  • #10
jbriggs444 said:
spherical shell theorem
That means overall(total) charge can be treated as/assumed to be concentrated at a point at its centre.
 
  • #11
jbriggs444 said:
on the origin
You meant centre of the sphere.
 
  • #12
gracy said:
That means overall(total) charge can be treated as/assumed to be concentrated at a point at its centre.
Right.

Uniform charge density. Sphere with volume that goes as r3. So total enclosed charge proportional to r3. Field equivalent to that charge at origin. Inverse square force law. So net field strength at radius r scales as r3/r2.
 
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  • #13
gracy said:
You meant centre of the sphere.
Yes, as in post #2, I am guessing that you are discussing a situation with a cloud of charge of uniform density centered at the origin. The center of a sphere of radius r centered at the origin is at the origin (by definition).
 
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  • #14
jbriggs444 said:
the portion of the charge cloud farther from the origin than the -Q test charge
If we take distance of -Q from centre of sphere i.e origin as "a" & the portion of the charge cloud to be q and radius of sphere to be r
let b be any value less than r .If r>b or r>a
that's what it means?
 
  • #15
gracy said:
If we take distance of -Q from centre of sphere i.e origin as "a" & the portion of the charge cloud to be q and radius of sphere to be r let b be any value less than r .If r>b or r>a
that's what it means?
That is nonsense. You stated that -Q is a quantity of charge. It cannot also be a distance.
 
  • #16
jbriggs444 said:
It cannot also be a distance.
Where i said -Q is distance?
 
  • #17
jbriggs444 said:
That is nonsense. You stated that -Q is a quantity of charge. It cannot also be a distance.
Edit: I apologize. You mean the distance from the origin to the charge whose value is -Q. And you want to denote that as a. I am not sure what you want q and r and b to denote.
 
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  • #18
jbriggs444 said:
You mean the distance from the origin to the charge whose value is -Q. And you want to denote that as a.
yes
 
  • #19
please make it more clear.It's my humble request
jbriggs444 said:
Uniform charge density. Sphere with volume that goes as r3. So total enclosed charge proportional to r3. Field equivalent to that charge at origin. Inverse square force law. So net field strength at radius r scales as r3/r2.
 
  • #20
gracy said:
please make it more clear.It's my humble request
What is it that you would like clarified?
 
  • #21
Uniform charge density. Sphere with volume that goes as r3. So total enclosed charge proportional to r3
Till here I understood.
 
  • #22
Field equivalent to that charge at origin. Inverse square force law. So net field strength at radius r scales as r3/r2.
I am unable to link
 
  • #23
So you agree that if we have a uniform cloud of charge centered on the origin and a test charge at distance r from the origin that the total charge closer to the origin than the test charge is proportional to r3

Do you agree that if this cloud of charge is spherical then the portion which is farther than r from the origin produces no net electrical field on the test charge? The spherical shell theorem should tell you this.

Do you agree that the field from the portion of the could which is nearer than r from the origin produces an electrical field on the test charge identical to that which would be produced if its entire charge were concentrated at the origin? The spherical shell theorem should also tell you this.
 
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  • #24
jbriggs444 said:
Do you agree that if this cloud of charge is spherical then the portion which is farther than r from the origin produces no net electrical field on the test charge? The spherical shell theorem should tell you this?
yes.
jbriggs444 said:
Do you agree that the field from the portion of the could which is nearer than r from the origin produces an electrical field on the test charge identical to that which would be produced if its entire charge were concentrated at the origin? The spherical shell theorem should also tell you this.
yes
jbriggs444 said:
So you agree that if we have a uniform cloud of charge centered on the origin and a test charge at distance r from the origin that the total charge closer to the origin than the test charge is proportional to r3
I am so sorry ;did not get you.
 
  • #25
jbriggs444 said:
So you agree that if we have a uniform cloud of charge centered on the origin and a test charge at distance r from the origin that the total charge closer to the origin than the test charge is proportional to r3
Should I think of coordinate system.
 
  • #26
jbriggs444 said:
So you agree that if we have a uniform cloud of charge centered on the origin and a test charge at distance r from the origin that the total charge closer to the origin than the test charge is proportional to r3
Yes.I understand it now .
 
  • #27
jbriggs444 said:
So you agree that if we have a uniform cloud of charge centered on the origin and a test charge at distance r from the origin that the total charge closer to the origin than the test charge is proportional to r3

Do you agree that if this cloud of charge is spherical then the portion which is farther than r from the origin produces no net electrical field on the test charge? The spherical shell theorem should tell you this.

Do you agree that the field from the portion of the could which is nearer than r from the origin produces an electrical field on the test charge identical to that which would be produced if its entire charge were concentrated at the origin? The spherical shell theorem should also tell you this.
How does it answer my OP?
 
  • #28
gracy said:
Should I think of coordinate system.
No. The coordinate system is irrelevant.

We have a uniform cloud of charge.
Somewhere within this cloud of charge is a point we call the origin.
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.
This spherical shell encloses a certain volume -- given by 4/3 pi r3
It encloses a certain amount of charge.
The amount of charge is proportional to the volume.
The volume is proportional to r3
Can you see that the amount of charge enclosed by this shell is proportional to r3

If you keep everything else constant and double r, by what factor will the amount of enclosed charge have increased?
 
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  • #29
gracy said:
How does it answer my OP?
Your OP was unclear. I had to guess what you were talking about. 27 posts later and you still have not clarified it.
 
  • #30
jbriggs444 said:
If you keep everything else constant and double r, by what factor will the amount of enclosed charge have increased?
8 times?
 
  • #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.
 

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