express the result in vector form of equiltorial line
Answers
Explanation:
Now, suppose that the point P is situated on the right-bisector of the dipole AB at a distance r metre from its mid-point 0 (Fig. (a)]
Again Let E
1
and E
2
be the magnitudes of the intensities of the electric field at P due to the charges +q and −q of the dipole respectively. The distance of P from each charge is
r
2
+l
2
. Therefore,
and E
1
=
4πε
0
1
(r
2
+l
2
)
q
away from +q
The magnitudes of E
1
and E
2
are equal (but directions are different). On resolving E
1
and E
2
into two components parallel and perpendicular to AB, the components perpendicular to AB (E
1
sinθandE
2
sinθ) cancel each other (because they are equal and opposite), while the components parallel to AB (E
1
cosθandE
2
cosθ ), being in the same direction, add up [Fig. (b)]. Hence the resultant intensity of electric field at the point P is
E=E
1
cosθ+E
2
cosθ
=
4πε
0
1
(r
2
+l
2
)
q
cosθ+
4πε
0
1
(r
2
+l
2
)
q
cosθ
=
4πε
0
1
(r
2
+l
2
)
q
2cosθ
But 2ql=p (moment of electric dipole)
∴ =
4πε
0
1
(r
2
+l
2
)
32
p
The direction of electric field E is 'antiparallel' to the dipole axis.
If r is very large compared to 2l (r>>2l), then l
2
may be neglected in comparison to r
2
.
E=
4πε
0
1
r
3
p
Newton/Coulomb
solution