Tow charges `pm10 mu` C are placed 5.0 mm apart determine the electric field at (a) a point p on the axis of the dipole 15 cm away from its centre o ont the side of the positive charge as (a) and (b) a point q 15 cm away from o on a line passing through o and normal to the axis of the dipole as

Explanation

Here electric dipole moment of given electric dipole,
`p = (2a)q`
`=(5 xx 10^(-3)) (10 xx 10^(-6))`
`therefore p = 5 xx 10^(-8)` cm
Here, r = 15 cm = 0.15 m
2a = 0.005 m
`rArr r gt gt gt gt 2a`
Given dipole can be taken as point dipole.
For a point dipole, electric field at far axial point is,
`E_(a) =(2kp)/r^(3)`
`=(2)(9 xx 10^(9))(5 xx 10^(-8))/(0.15)^(3)`
`therefore E_(a) = 2.667 xx 10^(5) N//C` (in the direction of `vecp`)
For a point dipole, electric field at far equatorial point is,
`E_(e) = (kp)/r^(3)`
`=(9 xx 10^(9))(5 xx 10^(-8))/(0.15)^(3)`
`therefore E_(e) = 1.335 xx 10^(5) N//C`