00:01
For this problem on the topic of electric fields, we are told that a proton is moving in the horizontal direction at a speed of 4 .5 times 10 to the 5 meters per second.
00:10
It then enters a uniform vertical electric field, which has a magnitude of 9 .6 times 10 to the 3 newtons per coulom.
00:17
We want to find the time that it takes the proton to travel 5 centimeters horizontally.
00:23
We want to then find the vertical displacement of the proton after it has traveled these 5 centimeters horizontally.
00:29
And then we want to find the horizontal and vertical components of its velocity after it has traveled the 5 centimeters.
00:38
So firstly, we know that time taken is the horizontal distance covered, we'll call it x, divided by its horizontal component of velocity, vx.
00:49
And this is a distance of 0 .05 meters, 5 centimeters, divided by its horizontal speed, 4 .05 meters, 5 centimeters, divided by its horizontal speed, 4 .4.
01:00
0 .5 times 10 to the 5 meters per second.
01:05
This gives us a time of 1 .11 times 10 to the power 7, 10 to the power minus 7 seconds, which is 111 nanoseconds.
01:22
So that's the time it takes the proton to move a horizontal distance of 5 centimeters...