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

Given that V˙(x)=4+5x2When x=1, V(1)=9 volt When x=−2, V(−2)=24 volt ∴ ΔV=V(−2)−V(1)=15 volt              E=−ΔVΔx=−10x         E at x=−1m will be −10×(−1)=+10N/C∴ F=qE=+(1)(10)=+10N along positive x-axis.