00:01
Hi, in the given problem there is an electric dipole kept vertically like this.
00:11
The two charges positive plus q and negative minus q.
00:25
These two charges are kept as shown in the figure and observation point is at the equatorial line.
00:37
Here, this observation point is at a distance, this is named as p, and this is at a distance from the center of this dipole r, and this distance r is quite large as compared with the gap between the two charges of this dipole, means the dipole length.
01:02
So this r is very much larger than this d.
01:06
Now we have to find net elective field at this observation point p, for which, first of all, we join the charges with the observation point and find their distances.
01:23
So as the observation point p is at equator of the dipole.
01:47
So we can say distance of observation point from both of the charges should be same.
01:54
So if we consider this plus q to be put at a point a minus q to be put at a point b, then we can say this ap is equal to b .p.
02:05
And this will be d by 2 as square root of r square plus d by 2 whole square means d square plus d square by 4.
02:22
Now as the distances are same, so these two angles will also be same, theta h.
02:28
And now the two electric fields at this observation point, as we know, electric field always goes away from the positive charge.
02:35
So due to plus q put at a, it will be going away like this.
02:41
And due to minus q put at b, it will be approaching the negative charge like this.
02:46
So these are the two electric fields.
02:51
Ea and e b.
02:54
Now as the magnitude of charges is same and the distances ap and bp are also same...