Two charges `10 muC` are placed 5.0 mm apart. Determine the electric field at a point Q, 15 cm away from O on a line passing through O and normal to the axis of the dipole, as shown in Fig.

Explanation

Field at Q due to charge `+10 muC` at B
`=(10^(-5) C)/(4pi(8.854xx10^(-12) C^(2) N^(-1) m^(-2)))xx(1)/((15^(2)+(0.25)^(2)]xx10^(-4) m^(2))x`
`=3.99xx10^(6) NC^(-1)` alpng BQ
Field at Q due to charge `-10 muC` at A
`=(10^(-5)C)/(4pi(8.854xx10^(-12) C^(2) N^(-1) m^(-2)))xx(1)/([15^(2)+(0.25)^(2)]xx10^(-4) m^(2))`
`=3.99xx10^(6)xxNC^(-1)` along QA
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
`=2xx0.25/sqrt(15^(2)+(0.25)^(2))xx3.99xx10^(6) NC^(-1)` along BA
`=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 epsi_(0) r^(3)) (r//a gt gt 1)`
`=(5xx10^(-8)Cm)/(4pi(8.854xx10^(-12) C^(2) N^(-1) m^(-2)))xx1/((15)^(3)xx10^(-6) m^(3))`
`=1.33xx10^(5) NC^(-1)`
The direction of electric field in this case is opposite to the direction of the dipole moment vector. Again the result agrees with that obtained before.