00:01
For this problem on the topic of electrostatics, we want to find the magnitude and direction of the electric field that the nucleus of the ion atom produces, firstly, just outside the surface of the nucleus, and then at the distance of the outermost electron.
00:17
And we then want to find the magnitude and direction of the acceleration of the outermost electron that is due only to the nucleus, neglecting any force due to the other electrons.
00:26
Now, the electric field produced e is equal to the electric constant k times the magnitude of the charge q over the distance r squared.
00:39
So in this case, this is k times 26 times the charge e divided by the radius of the nucleus squared.
00:54
Putting in our values, we get this to be the electric constant 8 .99 times 10 to the power 9 newton meter squared per coulomb squared times 26 times the magnitude of 1 .602 times 10 to the minus 19 couloms divided by 4 .6 times 10 to the minus 15 meters.
01:32
Squared.
01:34
And so at the, just outside the surface of the nucleus, the electric field strength is 1 .8 times 10 to the power 21 newtons per coulom.
01:51
Now we want to calculate the electric field and this electric field points radially outward.
02:05
And now we want to calculate the electric field at the outermost electrons distance...