00:01
Hi, here in this given problem, in the first part of the problem, we have to find net electric field at point p due to these three charges.
00:10
So we know electric field goes away from the positive charge.
00:15
So due to this plus 3 .00 nanoculum kept at the top of the arm, electric field will be going like this.
00:24
Let it be even.
00:26
Then due to this plus 3 .00 nanoculum put at the bottom of the r, the electric field will be going.
00:30
Are electric field going away again like this and this is e2 and due to this minus 2 .00 nanoculum elective field is this towards the negative charge name it e now we'll take components we will resolve even in e2 because these angles are 30 degrees so these angles here these will also be 30 degree so horizontal components e1 x e to x.
01:06
Vertical components e to y, e1y.
01:15
Okay.
01:16
Now we start solving for this.
01:20
Radius of the arc, this is given as 4 .40 centimeter.
01:26
So these electric fields, e1 and e2 will be having same magnitude given by k into q by r2.
01:38
Because the magnitude of charges same means the charges are alike identical and radius that is also same so these two electric fields will be given by for k this is 9 into 10 dash to the part 9 q 3 nanokulam or 3 into 10th per minus 9 kulum divided by this square of distance the square of radius 4 .40 centimeter or 4 .40 into 10 dash per minus 2 square so these electric fields, e1 and e2, calculated to be equal to 1 .39 into 10 .000 % percent of 4 newton per gula.
02:21
So as electric fields are same in magnitude, so what can we say about vertical components? vertical component of e1, that will be e1, sine theta...