How the distance from centre decides type of charge motion?

[jbriggs444]

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?

 

[gracy]

If the test charge is on the surface of this shell

That's where it was in the first place,wasn't it?

 

[gracy]

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.

That's where it was in the first place,wasn't it?

 

[gracy]

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.

If the test charge is on the surface of this shell

That's where it was in the first place,wasn't it?

 

[gracy]

t is now twice as far from the origin as it had been.

if an inverse square force law is in effect, doubling distance reduces attraction by what factor?

4 times.

 

[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?