The variation of potential $V$ with distance $x$ from a fixed point is as shown in the figure. The electric field at $x = 13\,m$ is......$V/m$.

When a charge of $3 \, C$ is placed in a uniform electric field, it experiences a force of $3000 \, N$. Within this field, the potential difference between two points separated by a distance of $1 \, cm$ is: (in $V$)

The electrostatic potential inside a charged spherical ball is given by $V = ar^2 + b$,where $r$ is the distance from its centre and $a$ and $b$ are constants. The volume charge density of the ball is [$\varepsilon_0$ = permittivity of free space].

The figure shows the electric potential $V$ as a function of distance through four regions on the $x$-axis. Which of the following is true for the magnitude of the electric field $E$ in these regions?

Potential in the $x-y$ plane is given as $V = 5(x^2 + xy) \, V$. The electric field at the point $(1, -2)$ will be

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: