00:01
So in order to find the electric field at the middle of the square surrounded by these charges at each corner of the square, all we need to do is add up each of the electric fields produced by the individual charges.
00:17
So in order to do that, i'll just, i labeled the charges here, one, two, three, and four, and i'll just construct my equation here for each one.
00:30
So e1 k won't change q is equal to plus q over r squared and r here is given by this we can just use pythagorean theorem to solve that r will be equal to l over the square root of 2 so i'll plug then at the end to keep this r simple but anyway onto e2 our q here is now minus 2 q over r squared e3 minus q e4 plus 2 q now each of these goes to be acting 90 degrees apart from one another but as a result of that these two two two and three will be acting along the same line and one and four will also be acting along the same line but each in opposite directions so we can simplify this a little bit more by making e plus and e minus, where e plus is the field line along this diagonal of 1 to 4, and e minus is the field line on the diagonal between 2 and 3.
02:09
So these forces will be working in opposite directions, since they are the same sign but on opposite sides of our point.
02:20
This will be the bigger one minus the smaller one.
02:23
So this will be 2 kq over r squared minus kq over r squared.
02:33
2kq over r squared minus 1 kq over r squared will just be equal to kq over r squared.
02:41
E minus will follow a very similar way of doing this.
02:46
So minus 2kq over r squared minus...