Question

34)The expression for the intensity of electric field on equatorial position of the dipole is: *​

Answer 1

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 E1​ and E2​ 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 r2+l2

​. Therefore,

and E1​=4πε0​1​(r2+l2)q​ away from +q

The magnitudes of E1​ and E2​ are equal (but directions are different). On resolving E1​ and E2​ into two components parallel and perpendicular to AB, the components perpendicular to AB (E1​sinθandE2​sinθ) cancel each other (because they are equal and opposite), while the components parallel to AB (E1​cosθandE2​cosθ ), being in the same direction, add up [Fig. (b)]. Hence the resultant intensity of electric field at the point P is

E=E1​cosθ+E2​cosθ

=4πε0​1​(r2+l2)q​cosθ+4πε0​1​(r2+l2)q​cosθ

=4πε0​1​(r2+l2)q​2cosθ

But 2ql=p (moment of electric dipole)

∴    =4πε0​1​(r2+l2)32p​

The direction of electric field E is 'antiparallel' to the dipole axis.

If r is very large compared to 2l (r>>2l), then l2 may be neglected in comparison to r2.

E=4πε0​1​r3p​Newton/Coulomb

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