Equipotential surfaces are shown in figure. Then the electric field strength will be
100Vm−1 along X -axis
100Vm−1 along Y -axis
200Vm−1 at an angle 120∘ with X -axis
50Vm−1 at an angle 120∘ with X -axis
Using △V=−E→⋅r→
⇒ ΔV=E.Δrsin30∘⇒ E=ΔVΔrsin30∘⇒ E=(20−10)10×10−2sin30∘⇒ E=1010×10−2sin30∘=1021/2=200V/m
Direction of E→ be perpendicular to the equipotential surface and towards the decreasing value of potential i.e. at 120° (90+30)with x-axis.