00:01
Talk about this question it says that all electrons are located at the corner of a square so if this is a square and the side length is also given as as 10 nanometers so for the moment let's represent it represented by d so we are given the distances d and there are four electrons on each of the corner and at the center we have an alpha particle so this is the center we have an alpha particle with charge 2e and here we have all the electrons uh we need to find the work to move the alpha particle to the midpoint of one of the sides of the square so let's say the alpha particle is moved to here from a to b so what is the work done so work done is going to be the change uh the charge times the work done is going to be its charge times the difference of the potential between the two points uh now uh let's let's let's find the potential at all the point.
00:59
So at va, the potential is due to all the four electrons, and that's going to be identical.
01:04
So we just need to find this distance, and this is nothing but the diagonal of the square, which is d -r -2.
01:14
So if d -r -2 is the total diagonal, then half of the diagonal will be the distance between the electron on the electron and the alpha particle.
01:22
So the potential is going to be minus of k -e -e -e.
01:28
Minus of k e over uh minus of k e over r where r is d root 2 over 2 and this is just and this this is this should be multiplied with 4 because there are four such electrons so that's going to be negative 8 k e over d root 2 so that's going to be negative 4 root 2 k e over d over d so that's going to be negative uh 4 root 2 k e over d so that's the potential at a we are going to place the value of k and t at last then we have to find a potential at b so the potential at b is going to be uh uh the due to these two electrons it will be same so that's going to be two times minus k e over r and r is nothing but half the length of the square which is d over two plus we have these two electrons with a different distance as compared to the other.
02:30
So if this is x and this is d and this is d over two, so using the pythagoras theorem, x square is equal to d square plus d square over four, which means that the value of x is root 5 over 2t.
02:44
So that's going to be two times minus ke over root 5 over 2d.
02:51
So if you're going to simplify this, that's going to be negative.
02:55
That's going to be negative 4ke over d minus 4ke over root 5d.
03:05
So this is the potential at b.
03:08
So we need to find a work done.
03:10
So here is the formula for the work done.
03:12
So we're going to substitute everything over here...