00:01
So here we are given four charges that are placed on the edges of a square.
00:08
Square is the side a, and the four charges are, as given, 4q, 3, q, 2, and 1 q.
00:17
And first, for the part of this problem, we need to answer what is the resultant, or the net, electric field which is exerted on, which is at the position of the charge q, or this position.
00:31
So here we see that we have three electric fields that are contributing to this net electric field.
00:41
We have the electric field from the charge 2q.
00:46
Here we have from the charge 3 q that is the iono and we have for the electric field which is coming from the charge 4 q.
00:55
And we should express every the intensity of every field the standard formulas so e1 is coming from the 2q charge and it will be constant k times 2q over the distance which is a squared e2 is coming from this diagonal here and of course this is 45 degrees and of course this is going to be it is going to be from the from the textbook formula k times and in this case what is the what's the distance this is not a but the diagonal of of square with the side a so that is going to be squared 2 times a squared so this is going to be 3 q over square 2 times a squared.
02:14
And for the e3 that is corresponding on this axis here.
02:19
It is corresponding to the 4q charge and it is 4q over a squared.
02:31
So this resultant electric field will be a vector field.
02:39
So we cannot say just a field, it is a vector field.
02:42
So to find, first to express the final electric vector field, we need to add the point, actually the value of the vector field at the point where the charge q is, we need to find the components of the vector field of electric vector field at the point where the charge q is.
03:04
So the horizontal component of this electric vector vector field is the sum of, well, we can see from the diagram that horizontally only 2q charge and 3q charge.
03:27
So the electric field coming from the charged 3q is only having a component that is vertical.
03:37
It only has only one of its components.
03:41
And from the charge 2q, the electric field, which arises from it, is completely vertical at the point is completely horizontal at the point where the charge q is...