
Use the concept of potential gradient, then the relation, {E}_{r}=-\frac{dV}{dr}\Rightarrow {E}_{\theta }=-\frac{1}{r}\times \frac{dV}{d\theta }&V=\frac{KP}{{r}^{2}}\mathrm{cos}\theta, for general point,
⇒Eθ=−r1×dθd(r2KPcosθ)⇒E=−r3KP1+3cos2θ, for the equatorial position, θ=90∘
⇒E=−r3kP=−4πϵ0r3P.