If on the $x$-axis electric potential decreases uniformly from $60 \,V$ to $20 \,V$ between $x=-2 \,m$ to $x=+2 \,m$,then the magnitude of electric field at the origin

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 $\phi(x, y)$ of an electrostatic field $\vec{E} = a(y \hat{i} + x \hat{j})$ is [where $a$ is a constant and $\hat{i}$ and $\hat{j}$ are unit vectors along $X$ and $Y$ axes].

If $V = -5x + 3y + \sqrt{15}z$,then find the magnitude of the electric field $E(x, y, z)$ in units.

In space,the electric potential varies as $V = 20|\vec{r}|$ volt,where $\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$ is the position vector. Then,the electric field in $N C^{-1}$ at the point $(4 \ m, 3 \ m, -5 \ m)$ is:

The electric potential $V$ at any point $(x, y, z),$ all in metres in space is given by $V = 4x^2$ volt. The electric field at the point $(1, 0, 2)$ in volt/meter,is