How the distance from centre decides type of charge motion?

[gracy]

I mean will the charge go to the other side of the origin?

 

[jbriggs444]

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?

 

[gracy]

What is the force on the test charge at the origin? What does Newton's first law have to say?

mass into acceleration

 

[jbriggs444]

No. That is a mangling of Newton's second law. Please try again.

 

[gracy]

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.

 

[jbriggs444]

And the first part of my question asked about the force on the test charge at the origin. Can you answer that?

 

[gracy]

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?

 

[jbriggs444]

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?

 

[gracy]

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.

 

[gracy]

As origin is equilibrium position total force has to be zero

I mean will the charge go to the other side of the origin?

Is the answer no then?

 

[jbriggs444]

Is the answer no then?

Suppose the object approaches the origin with some non-zero speed. What does Newton's first law say again?

 

[gracy]

What does Newton's first law say again?

Yes.It will travel with the same speed.

 

[gracy]

And the test charge should be of opposite nature(whether positive or negative) to that of charge enclosed in sphere,right?

 

[jbriggs444]

Right. My post 58 objected to your reasoning, not to your conclusion.

Now can you agree that the motion of the test charge in the hypothetical spherical cloud of uniform charge will be simple harmonic motion?

With that agreement can you agree that my guess in post #2 is a physical scenario compatible with the behavior you described in post#1?

 

[jbriggs444]

And the test charge should be of opposite nature(whether positive or negative) to that of charge enclosed in sphere,right?

Right. You had originally described the test charge as "-Q" so I assumed you wanted it to be negatively charged. I never mentioned a sign on the uniform charge and assumed that you would assume that it was positive.

 

[gracy]

Now can you agree that the motion of the test charge in the hypothetical spherical cloud of uniform charge will be simple harmonic motion?

only when r is small,right?

 

[jbriggs444]

only when r is small,right?

Yes.

Which explains why one might have simple harmonic motion for charges near the origin and periodic oscillation different from simple harmonic motion for charges further away.

 

[gracy]

There is still a bit of confusion in my mind.I want to clear that to thoroughly understand this concept so that your hardwork behind this does not go in vain and I have crystal clear understanding.

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.

these two are part of spherical shell theorem which kind of makes symmetrical situation of sphere,I want to ask is this applicable for both cases whether r is small or big
r is distance of test charge from the center of charged sphere i.e origin?

 

[gracy]

I think the answer to my question in previous post is no.Because if spherical shell theorem were to be applicable in both cases then for bigger /larger r also the motion will be simple harmonic motion which is not the fact as depicted in my OP.Right?

 

[jbriggs444]

these two are part of spherical shell theorem which kind of makes symmetrical situation of sphere,I want to ask is this applicable for both cases whether r is small or big
r is distance of test charge from the center of charged sphere i.e origin?

The passage you quoted is in the context of an explanation about how simple harmonic motion is the expected result as long as the test charge stays within the area of uniform charge. So it is within a context where r is always "small".

When r is large compared to the size of the charged cloud, the situation is different.

Edit: In particular, for large r, the amount of charge enclosed by a sphere of radius r no longer scales with r. It is fixed (it cannot be more than the total charge on the cloud).

 

[gracy]

Sir,please say "yes" to my post #69.

 

[jbriggs444]

I think the answer to my question in previous post is no.Because if spherical shell theorem were to be applicable in both cases then for bigger /larger r also the motion will be simple harmonic motion which is not the fact as depicted in my OP.Right?

The spherical shell theorem is a theorem. It always applies as long as its premises hold. They do hold in the case of large r.

In particular:

1. The net force from the portion of the charged cloud outside the shell of radius r is zero.
2. The net force from the portion of the charged cloud inside the shell of radius r is equivalent to the force that would exist if the charge were concentrated at the origin.

 

[gracy]

Then why not in this case?

 

[gracy]

Oh! I thought I understood the whole thing.

 

[jbriggs444]

Then why not in this case?

Re-read post #60 and, in particular, the edit there.

Edit: 70, darnit.

 

[gracy]

post 60?it's mine.

 

[jbriggs444]

post 60?it's mine.

Yeah, yeah, I corrected myself but you were just too quick.

 

[gracy]

Oh! I thought I understood the whole thing.

I was correct in thinking so!

 

[gracy]

Ok.With all credits to you I now understand why in case of big r it will not be simple hatmonic motion but why it would be periodic anyway?

 

[jbriggs444]

Ok.With all credits to you I now understand why in case of big r it will not be simple hatmonic motion but why it would be periodic anyway?

It seems intuitively obvious. But let me see if I can come up with a solid and understandable argument...

We are working in one dimension. The only forces present are along the x axis. So there are no side to side complications to worry about.

We are dealing with a central force that depends only on distance from the origin. That means that we are working in a conservative field. However, let's try a more elementary argument than that. How about time reversal symmetry...

We know we have an attractive force. If the test charge is ever motionless or if it is ever moving toward the origin, it will pass through the origin. Let us hand-wave away the case where the test charge is moving at a speed greater than or equal to escape velocity and never hits the origin.

When the test mass passes through the origin, it will go outward for some distance, turn around and fall back. [Remember that we hand-waved away the possibility of escape]

On its path back to the origin, its acceleration at each r value on the way in will be equal to its acceleration at that r value on the way out. It will have the same inward velocity at each r value on the way in as it had on its way out. It will return to the origin at the same speed it left. The time taken for that trip is purely a function of its speed at the origin. And its speed each time it passes through the origin is identical.

It follows that its trajectory is periodic.

 

[gracy]

I understood it.Thanks Thanks Thanks.You just made my day I was worried throughout the day for this problem.Now I can be relax.Just because of you.You were so patient throughout this thread coping with my silly questions.Thanks again.All the helpers here on physics forum are doing selfless help which is remarkable.Love physics forum!

 

[jbriggs444]

The thank you's make it all worth while.

 

[gracy]

Your avatar/profile pic suggests it's your pleasure

 

[jbriggs444]

Who, me? I know nothing. Nothing. The commandant says that there is to be no fraternizing with the prisoners.

 

[gracy]

Who, me? I know nothing. Nothing. The commandant says that there is to be no fraternizing with the prisoners.

Oooooooooooooo!THAT'S WHAT YOU KEEP SAYING!