How the distance from centre decides type of charge motion?

[jbriggs444]

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?

 

[gracy]

what about my post #34?

 

[jbriggs444]

what about my post #34?

What about it?

 

[gracy]

correct or not?What's wrong n it?where am I missing?

 

[jbriggs444]

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?

 

[gracy]

where am I missing?

What's wrong in it?Clear my doubt.

 

[jbriggs444]

What's wrong in it?Clear my doubt.

Asked and answered. It is neither right nor wrong because it makes no statement.

 

[gracy]

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.

 

[jbriggs444]

We increased the size of the shell and we moved the test charge.

Edit: Hence the words "had been" -- to emphasize that something changed.

 

[gracy]

By what factor has the attraction of the test charge to the enclosed charge increased or decreased?

increased ,2 times..

 

[jbriggs444]

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?

 

[gracy]

Now, generalize this. Is the attraction of the test charge toward the origin proportional to its distance, r, from the origin?

yes.

 

[jbriggs444]

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?

 

[gracy]

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??

 

[jbriggs444]

Why should anything happen? There's nothing at the origin but more of the same uniform charge cloud that surrounds the origin.

 

[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).