Variation in electric potential is maximum if one goes

Electric potential in a region is varying according to the relation $V = \frac{3x^2}{2} - \frac{y^2}{4}$,where $x$ and $y$ are in $m$ and $V$ is in $V$. Electric field intensity (in $N/C$) at a point $(1 \, m, 2 \, m)$ is:

The potential is varying with distance $(x, y)$ as $V = \frac{1}{2} (y^2 - 4x) \text{ V}$. The electric field at $x = 1 \text{ m}$ and $y = 1 \text{ m}$ is:

$A$ uniform electric field of $400 \, V/m$ is directed $45^\circ$ above the $x$-axis. The potential difference $V_A - V_B$ is: (Coordinates are in cm) (in $, V$)

In a region,if the electric field is defined as $\vec{E} = (\hat{i} + 2\hat{j} + \hat{k}) \text{ V/m}$,then the potential difference between two points $A(0, 0, 0)$ and $B(2, 3, 4)$ in that region is ...... $V$.

If the electric potential in a region is given by $V = 4x^2$ volts,then the electric field at the point $(1, 0, 2) \ m$ is: