The electric potential in a region along the x-axis varies with x according to the relation V(x)=4+5x2. Then

V(x)=4+5x2V(x=1)=9VAnd V(X=−2)=24VHence, ΔV=15VE=−(ΔV/Δx)=−10x∴      E( at x=−1m)=10NC−1So, F=qE=+10N (along positive x-axis).