Equipotential surfaces are shown in figure. Then the electric field strength will be

Using △V=−E→⋅r→⇒    ΔV=E.Δrsin⁡30∘⇒     E=ΔVΔrsin⁡30∘⇒     E=(20−10)10×10−2sin⁡30∘⇒     E=1010×10−2sin⁡30∘=1021/2=200V/mDirection of E→ be perpendicular to the equipotential surface and towards the decreasing value of potential i.e. at 120° (90+30)with x-axis.