Answer:

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