00:01
All right, hello, in this question, we have this charge configuration.
00:02
So we have two fixed charges here on the x -axis with the same charge and the same distance away from the origin.
00:08
And then we have a third charge here.
00:10
We're also given its mass.
00:12
And we're told that it's released from rest at y1.
00:15
And then we want to find the speed when it hits this point y2 up here.
00:19
We're asked to consider only electrical forces.
00:22
So in order to do this, we could do some of the forces.
00:27
But if we do that, as this particle moves away, it's going to get less and less force on it.
00:32
So we'd have to do some integral thing.
00:33
And while it's possible, it would be very messy.
00:36
Instead, i'm going to opt to use a conservation of energy approach.
00:40
So e1 and e2 are equal.
00:43
Now, what is our source of energy? well, we have electric potential here.
00:48
So initially, we have some electric potential at 1.
00:52
At the end, we're going to have some electric potential, not gravitational potential.
00:57
So at the end, we have some electric potential as well.
00:59
But then we're also going to have some kinetic energy.
01:02
So what is our electric potential? well, it's going to be our charge times the potential at 0 .1.
01:09
That is equal to charge at our potential at 0 .2 plus 1 half mv squared.
01:15
This v is speed.
01:16
It's not potential.
01:19
So i'll call it v final just to differentiate.
01:21
So we need to figure out what our potential is at 1 and 2...