potential due to the quadrupole is;
V=qd²[3cos²θ-1]/4╥εx³
Now,
Radial component of electric field, Ex=-∂V/∂x=-∂/∂x[qd²(3cos²θ-1)/4╥εx³]
and
Transverse component of field, Eθ= - (1/x)[∂V/∂θ]
=-(1/r){∂/∂θ[qd²(3cos²θ-1)/4╥εx³]}
therefore,resultant field is, E=√(Ex²+Eθ²) which will be...