V(x)=5−4x2
V(1)=5−4=1V,V(2)=5−4(22)=−11V
Potential difference between x=1m and x=2m is V12=V1−V2=1−(−11)=12V
here, Ex=−dxdV=8x,Ey=−dydV=0 and Ez=−dzdV=0
The electric force on 1 coulomb charge at x=1 is F=qEx=1(8)=8N