Two charges, each of `5 mu C` but opposite in sign, are placed 4 cm apart. Calculate the electric field intensity of a point that is a distance 4 cm from the mid point on the axial line of the dipole.

Answer

`10^(8) NC^(-1)`

Explanation

Here, `q = 5 muC = 5xx10^(-6)C`
`2a = 4cm = 4xx10^(-2)m, E = ?`
`r = 4 cm = 4xx10^(-2) m`.
`E_("axial") = (2 pr)/(4 pi in_(0) (r^(2) - a^(2))^(2)) = (2 (q) (2a) r)/(4pi in_(0) (r^(2) - a^(2))^(2))`
`= (9xx10^(9)xx2(5xx10^(-6))(4xx10^(-2)) xx4xx10^(-2))/([0.04^(2) - 0.02^(2)]^(2))`
`E_("axial") = 10^(8) NC^(-1)`