00:01
So in this problem, we have an object that has a negative charge of q, negative 6 times 10 to negative 9culems.
00:10
And it starts at rest at point a and is released.
00:14
And at point b, 0 .5 meters to the right of point a, it's found to have a kinetic energy of 3 times 10 to the negative 7 joules.
00:22
So in this question, we're asked to find a vb.
00:26
So the question is, what do we need to know, what formulas do we need to use in order to solve this problem? so what we need to remember is that energy is conserved.
00:34
So the total energy at point a, so kinetic and potential, is going to be equal to that of point b.
00:42
So what does this look like when we write it out? so we'll have the kinetic energy at point a plus the potential energy at point a.
00:54
It's going to be equal to the kinetic energy at point b plus the potential energy at point b.
01:04
So the second thing that we need to do is we don't directly have the potential energy, but we can find it using the charge and the potential that we're given.
01:14
So you should remember that the potential energy is going to be equal to potential times the charge, the object that's moving in this field.
01:28
So now that we have this substitution that we can make, we can go ahead and rewrite this.
01:37
So we have ka plus q times va, and that's equal to kb plus q times vb...