00:01
For this problem on the topic of electric fields, we have protons which are being projected with an initial speed of 9 .55 times 10 to 3 meters per second.
00:09
They move into a region where uniform electric field of minus 720 j.
00:15
Newton's per kulom is present, as we can see in the figure.
00:18
The protons are to hit a target that lies a horizontal distance of 1 .27 millimeters away from the point at which they are launched.
00:26
We want to find the two projection angles theta that will result in a hit, as well as the total time of flight for each trajectory.
00:35
Now we know the initial speed is 9 .55 times 10 to the 3 meters per second, and we'll calculate the y component of the acceleration, which is the electrostatic force applied on the protons little e, the chart of the proton, times the electric field strength, capital e, over the mass.
00:59
Of the proton.
01:00
This is by newton second law.
01:02
And this is 1 .6 times 10 to the minus 19 cooloms.
01:08
We'll suppress the units here times an electric field strength of 720 newtons per coulom over the mass of the proton 1 .67 times 10 to the minus 27 kg.
01:24
This gives us the acceleration, the vertical acceleration of the protons to be 6 .7...