00:03
In this problem, we have a charge queue that is placed inside a uniform electric field initially at point a, and then it travels 0 .5 meters to the right and lands at point b with the kinetic energy of 3 into 10 to the negative 7 joules.
00:22
We need to, and we need to find the potential at b.
00:27
So first, we need to find the amount of work done in transporting the particle from a to b.
00:33
So the formula for the work done is equal to the change in potential times the charge.
00:53
So the change in potential is what we're solving for.
00:59
And the charge is three, sorry, rather the magnitude of the charge is.
01:11
So we're only concerned with the magnitude of the charge.
01:15
So that would be six into 10 to the negative nine coulom.
01:21
And note that the kinetic energy gained will be equal to the work done so we can substitute that in equals to delta v to 6 into 10 to the negative 9 and for what we're left with is the change in potential being equal to 50 volts now note that the particle is negatively charged and negatively charged particles tend to go from a lower potential to a higher potential.
02:01
They behave the exact opposite to a positive charge.
02:10
So the potential at b will be 50 plus 30 volts, which gives us the potential at b at 80 volts.
02:26
Now we need to find the magnitude and direction of the electric field...