Two charges `+- 10 mu C` are placed `5*0 mm` apart. Determine the electric field at (a) point P on the axis of dipole `15 cm` away from its center on the side of the positive charge. As shown in Figure and at (b) a point Q. `15 cm` away form O on a line passing through O and a line passing through O and

normal to the axis of the dipole as shown in Fig.

Explanation

Here, ` q = +- 10 muC = +- 10^(-5)C`
`2a = 5*0 mm = 5xx10^(3) m`
`r = OP = 15 cm = 15xx10^(-2) m`
`|p| = qxx 2a = 10^(-5)xx5xx10^(-3)`
`5xx10^(-8) C-m`
(a) As P lies on axial line of diople.
`:. = E_(1) = (2| vec(p) | r)/(4pi in_(0) (r^(2) - a^(2))^(2))` , along BP
`As a ltlt r, :.`
`(2| vec(p) | r)/(4pi in_(0) r^(3)) = (2xx5xx10^(-8)xx9xx10^(9))/((15xx10^(-2))^(3))`
`2*67xx10^(5) N//C`, along BP
(b) As Q lies on equatorail line of dipole,
`:. E_(2) = (| vec(p) |)/(4pi in_(0) (r^(2) + a^(2))^(3//2)) = (| vec(p) |)/(4pi in_(0) r^(3))`
`(because a lt lt r)`
`(1)/(2) E_(1) = (1)/(2)xx2*67xx10^(5) = 1*33xx10^(5) N//C`
`vec(E_(2))` is along a line parallet to BA as shown in Fig.