As shown in the figure, if the values of the electric potential at three points $A, B$ and $C$ in a uniform electric field $(\vec{E})$ are $V_A, V_B$, and $V_C$ respectively, then

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 figure shows two equipotential lines in the $x, y$ plane for an electric field. The scales are marked in $\text{cm}$. The $x$-component $E_x$ and $y$-component $E_y$ of the electric field in the space between these equipotential lines are respectively:

The electric field vector in a region is given by $E = (3 \hat{i} + 4y \hat{j}) \ V \ m^{-1}$. The potential at the origin is zero. Then,the potential at a point $(2, 1) \ m$ is: (in $V$)

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:

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 ...... $V$.