E = -grad(V) = -(dV/dx $x_{hat}$ + dV/dy $y_{hat}$ + dV/dz $z_{hat}$). In 1D: E = -dV/dr. The negative sign means E points from high V to low V (toward decreasing potential). If V is given as a function of coordinates, E components are found by partial differentiation. Conversely, V = -integral(E.dr) from reference to point. For a uniform field E along x: V = -Ex + constant. For a non-uniform field, the integral must be evaluated along a specific path (but the result is path-independent since E is conservative). In JEE, common problems give V = $ax^2$ + by + c and ask for E components: $E_x$ = -2ax, $E_y$ = -b, $E_z$ = 0. The field magnitude and direction at any point follow from these components.