Obtain an expression for the electric field intenstiy at a point on the equatorial line of an electric dipole.

Explanation

The magnitudes of the electric field due to the two charges `+qand-q` are given by,
`E_(+q)=(q)/(4piepsilon_(0))(1)/((r^(2)+a^(2)))` … (i)
`E_(-q)=(q)/(4piepsilon_(0))(1)/((r^(2)+a^(2)))` …(ii)
`thereforeE_(+q)=E_(-q)`
The directions of `E_(+q)andE_(-q)` are as shown in the figure . The components normal to the dipole axis cancel away . The components along the dipole axis add up.
`therefore` Total electric field
`E=-(E_(+q)+E_(-q)cos thetahatp`
[Negative sign shows that field is opposite to `hatp`]
`E=(-2qa)/(4piepsilon_(0)(r^(2)+a^(2))^(3//2))hatp` ...(iii)
At large diatances `(rgtgta)`, this reduces to
`E=(2qa)/(4piepsilon_(0)r^(3))hatp` ... (iv)
`because=2qahatp=vecp`
`thereforeE=(vecP)/(4piepsilon_(0)r^(3))(rgtgta)`