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

(a) field at p due to charge + 10 `muc`
`=(10^(-5) c)/(4pi(8.854 xx10^(-12)C^(2)N^(-1)m^(-2)))xx(1)/((15+0.25)^(2)xx10^(-4)m^(2))`
`=3.86 xx10^(6) NC^(-1)` along PA
in this example the ratios OP/OB is quite large (=60) thuis we can expect to get approximately the saem result as above by directly using field at a distance r form the centre on the axis of the dipole has a magnitude field at Q due to charge =-10 `muc` at a
clearly the components of these two forces with equal magnitudes cancel along the direction OQ but add up along the direction parallel to BA therefore the resultant electric field at Q due to the two charges at A and B is
`=1.33xx10^(5) NC^(-1)` along BA
as in (a) we can expect to get approximately the same result by directly using the formula for dipole field at a point on the normal to the axis of the dipole
`E=(P)/(4pi epsilon_(0)r^(3))`
`=1.33 xx10^(5) NC^(-1)`
The direction of electric field in the this case is opposite to the direction of the dipole maoment vector again the result agrees with that obtained before