Two charges `10 muC` are placed 5.0 mm apart. Determine the electric field at a point P on the axis of the dipole 15 cm away from its centre O on the side of the positive charge, as shown in Fig.

Explanation

Field at P due to charge `+10 mu C`
`= (10^(-5)C)/(4pi (8.854 xx 10^(-12) C^(2)N^(-1)m^(-2))) xx (1)/((15-0.25)^(2) xx 10^(-4)m^(2))`
`= 4.13 xx 10^(6)NC^(-1)` along BP
Field at P due to charge `-10 mu C`
`= (10^(-5)C)/(4pi (8.854 xx 10^(-12) C^(2)N^(-1)m^(-2))) xx (1)/((15-0.25)^(2) xx 10^(-4)m^(2))`
`= 3.86 xx 10^(6) NC^(-1)` along PA
The resultant electric field at P due to the teo charges at A and B is `2.7 xx 10^(5) NC^(-1)` along BP.
In this example, the ratio OP/OB is quite large (= 60). Thus, we can expect to get approximately the same result as above by directly using the formula for electric field at a far-away point on the axis of a dipole. For a dipole consisting of charges `+- 1, 2a` distance apart, the electric field at a distance r from the centre on the axis of the dipole has a magnitude.
`E = (2p)/(4pi epsilon_(0) r^(3)) (r/a gt gt 1)`
Where p = 2aq is the magnitude of the dipole moment.
The direction of electric field on the dipole axis is always along the direction of the dipole moment vector (i.e., from -q to q). Here`, p = 10^(-5) xx C, 5 xx 10^(-3) m = 5 xx 10^(-8)C m`
Therefore,
`E = (2 xx 5 xx 10^(-8) Cm)/(4pi(8.854 xx 10^(-12) C^(2)N^(-1)m^(-2)))xx (1)/((15)^(3)xx 10^(-6)m^(3)) = 2.6 xx 10^(5)N C^(-1)`.
Along the dipolemoment direction AB, which is close to the result obtained earlier.