00:01
Solution of this question is now part one here to find electric field to find electric field intensity due to dipole at axial line so for this consider electrical electrical dipl of length to l.
01:01
This is charge minus q and this is charge plus q.
01:06
Here take a point p at a distance are from the midpoint o of the dipole.
01:22
At point p we need to calculate the electric feel intensity.
01:27
So for this first electric field intensity, intensity due to electric field intensity at point p due to charge charge plus q so this is given by e1 equal to 1 divide by 4 pi multiply q divided by r.
02:22
Here r is the distance between charge q plus q charge and point p.
02:28
So this is given by this distance is l and this is also l.
02:42
So the distance between charge plus q and p is r minus l.
02:53
So this is r minus l square.
02:58
Now the direction of this electric field is represented by it is e1 is in this direction.
03:12
Now electric field, electric field intensity at p due to minus q charge.
03:37
So it is given by e2 equal to 1 divided by 4 pi epsilon not q divided by here the distance between point p and minus q charge is given as r plus l so elective fuel intensity will be e2 equal to 1 divided by 4 pi epsilon not qr plus l square now the direction of this field is this is e2.
04:23
Now we need to calculate electric field at point p.
04:27
So electric field intensity at point p is e equal to e1 minus e2.
05:02
Here e1 is 1 divided by 4 pi epsilon 0 q r minus l square minus 1 divide by 4 pi epsilon not q r plus l square so now e is equal to 1 divide by 4 pi epsilon not multiply q we can write this as r plus l square minus r minus l square divide by r square minus l square whole square so on solving this e is equal to 1 divided by 4 pi e multiply q divided by r square minus l square whole square.
06:22
Here, p is equal to electric dipole movement is equal to 2l multiply by q.
06:32
So we can write above equation as e is equal to 1 divided by 4 pi epsilon not multiplied to multiply to r.
06:47
This is 2l.
06:51
X x x multiply r divide by let here r is much much greater than else so we can write this as r raised to par four so on solving this we get lactic field intensity at axial line due to dipole is 1 divided by 4 pi epsom 0 to p b divide by r now we need to calculate electric field intensity due to dipole at equatorial line...