The variation of electrostatic potential $V$ along the $x$-direction is shown in the graph. Which of the following statements about the electric field is correct?

The electric potential $V(x, y, z)$ for a planar charge distribution is given by:
$V(x, y, z) = \begin{cases} 0 & \text{for } x < -d \\ -V_0(1 + \frac{x}{d})^2 & \text{for } -d \le x < 0 \\ -V_0(1 + 2\frac{x}{d}) & \text{for } 0 \le x < d \\ -3V_0 & \text{for } x \ge d \end{cases}$
where $-V_0$ is the potential at the origin and $d$ is a distance. The graph of the electric field as a function of position 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:

The electric potential at a point $(x, y, z)$ is given by $V = -x^2y - xz^3 + 4$. The electric field at that point is:

In a space having electric field $\vec{E}=A(x \hat{i}+y \hat{j})$,the potential at a point $(10 \ m, 20 \ m)$ is zero. Then the potential at the origin is $\left[A=10 \ Vm^{-2}\right]$. (in $V$)

If the electric potential at any point $(x, y, z) \, m$ in space is given by $V = 3x^2$ volt,the electric field at the point $(1, 0, 3) \, m$ will be ............