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 electric potentials of two plates are $-10\, V$ and $+30\, V$ respectively. If the distance between the two plates is $2\, cm$,then the electric field between them is ... $V/m$.

The figure shows the variation of electric field intensity $E$ versus distance $x$. What is the potential difference between the points at $x = 2 \, m$ and $x = 6 \, m$ from $O$ (in $V$)?

The electric potential $V$ at any point $(x, y, z)$ in space is given by the equation $V = 4x^2 \, V$,where $x, y$,and $z$ are all in meters. The electric field at the point $(1 \, m, 0, 2 \, m)$ in $V/m$ is:

Distinguish between electric field and electrostatic potential.