# Velocity of charge at the center thrown from far away point

1. Feb 3, 2016

### gracy

here from time 0:53 to 0:58 velocity at center would be more than or approximately equal to zero .I want to ask why approximately zero why not zero?

2. Feb 3, 2016

### BvU

Hi !

If the particle slows down to zero speed it doesn't cross any more.
So the use of the $\ge$ is a bit debatable indeed -- as you sort of detected.
Matter of wording. Somewhat unfortunate.

3. Feb 3, 2016

### gracy

So to cross it should have velocity more than zero .That's it.He Should only use > symbol i.e to cross the center ,velocity >0 why he goes on and uses equality sign and says approximately zero?
Do you consider it unfortunate?"That's nice!"

Last edited: Feb 3, 2016
4. Feb 3, 2016

### BvU

The limit is for the = sign. For any $\epsilon > 0$ with which the speed exceeds this limit the thing will cross. That's more or less the physical definition of a limit. Doesn't get much better unless you go into the realm of mathematical analysis.
Let's not draw this into a 60+ posts thread and give the presenter a bit of latitude....

5. Feb 3, 2016

### gracy

Relax ! It was just 4th post.
I did not understand.

Last edited: Feb 3, 2016
6. Feb 3, 2016

### cnh1995

If the speed at the center is 0, the charge won't cross the center. You got it right. You are supposed to find the minimum velocity of projection "above which" the particle will cross the center.

7. Feb 3, 2016

Which limit?

8. Feb 3, 2016

### cnh1995

Limit of the velocity of projection. For a particular Vprojection, Vc will be 0. Above this limiting Vprojection, any value will make the charge cross the center.

9. Feb 5, 2016

### gracy

In spite of lots of good answers on this ,I'am still having the same question/doubt.

10. Feb 5, 2016

### BvU

11. Feb 5, 2016

### gracy

I have quoted what's my question

12. Feb 5, 2016

### BvU

It doesn't matter what he should or should not. And it doesn't matter why he goes on the way he does.
What matters is to find out what it is that prevents you from saying: I understand the exercise. I know what a limit is and how I can calculate it.
So: what is it that prevents you from saying: I understand the exercise. I know what a limit is and how I can calculate it.

I'm going to need some time to drive to work, then to do work and then for sports. So you have plenty of time to think this over and formulate a response ...

13. Feb 5, 2016

### cnh1995

Well, indeed Vc>0 for the charge to "cross" the center. But for the charge to "reach" the center, there is some limiting velocity of projection. So, concerned path of the charge is- from the center and ahead. Hence, mathematically, he wrote Vc≥0. When you "just exceed" the limiting value of velocity of projection, Vc will be "slightly" greater than 0. Hence, he said it will be approximately 0. Vc≥0 is the right equation to begin with.

14. Feb 5, 2016

### gracy

I understood it now but not by applying "limit method" rather with the help of @cnh1995 answer in above post.I would like to understand limit method as well.Is it the same limit which is used in mathematics?

15. Feb 5, 2016

### cnh1995

I believe here "limit" means a limiting value which separates the two possibilities,
a)The charge will cross the center
b) It won't.
At a particular value, Vprojection will be such that it is the boundary condition for b. Above that, a will be true forever. So, for b to be true, maximum limit of Vprojection results in Vc=0(no math, only intuition!). Hence, from mathematical point of view, you are calculating the limit of Vprojection as Vc→0.

16. Feb 5, 2016

### cnh1995

If you want to do the exact mathematical analysis by limit method, you'll need to find Vprojection as a function of Vc, which will involve some calculus.

17. Feb 5, 2016

### gracy

And any value below that would result in b?

18. Feb 5, 2016

### BvU

What makes you ask that ?

19. Feb 5, 2016

### gracy

My understanding.It could be obviously wrong!

20. Feb 5, 2016

### BvU

Have some faith. You did the calculations ! That's what came out. No doubt whatsoever possible. Obviously.