- #1
gracy
- 2,486
- 83
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?
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?BvU said:If the particle slows down to zero speed it doesn't cross any more.
Do you consider it unfortunate?"That's nice!"BvU said:Matter of wording. Somewhat unfortunate.
Relax ! It was just 4th post.BvU said:Let's not draw this into a 60+ posts thread
I did not understand.BvU said:For any ϵ> 0 with which the speed exceeds this limit the thing will cross
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.gracy said:Relax ! It was just 4th post.
I did not understand.
Which limit?BvU said:with which the speed exceeds this limit the thing will cross.
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.gracy said:Which limit?
In spite of lots of good answers on this ,I'am still having the same question/doubt.gracy said: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?
I have quoted what's my questionBvU said:Please explain.
gracy said: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?
And any value below that would result in b?cnh1995 said:Above that, a will be true forever.
My understanding.It could be obviously wrong!BvU said:What makes you ask that ?
Yes. And I just did the problem using the limit method and got the same answer, butgracy said:And any value below that would result in b?
It's "a lot of" calculus..I believe it is best to go by the conservation of energy method.cnh1995 said:which will involve some calculus.
The formula for calculating the velocity of charge at the center thrown from a far away point is v = (kQ)/r, where v is the velocity, k is the Coulomb's constant, Q is the charge, and r is the distance between the charge and the center point.
The velocity of charge at the center thrown from a far away point is inversely proportional to the distance between the charge and the center point. This means that as the distance increases, the velocity decreases, and vice versa.
Coulomb's constant, denoted by k, is a proportionality constant that relates the electrostatic force between two charged particles to the product of their charges and the distance between them. It affects the velocity of charge by determining the strength of the electrostatic force, which in turn affects the velocity of the charge.
No, the velocity of charge at the center thrown from a far away point can never be zero. This is because the charge will always experience an electrostatic force from the source of the charge, causing it to move with a non-zero velocity.
The charge of the particle directly affects its velocity at the center thrown from a far away point. A higher charge will result in a higher velocity, while a lower charge will result in a lower velocity, assuming all other variables are constant.