In a certain region,the electric potential is given by $V = 6x - 8xy^2 - 8y + 6yz - 4z^2 \, V$. The magnitude of the force on a charge of $2 \, C$ at the origin is ........ $N$.

The electric potential at any point $(x, y, z)$ (all in meters) in space is given by $V = 5x^2$ volt. The electric field at the point $(1, 2, 3) \text{ m}$ is $\overrightarrow{E} = $ . . . . . . $\text{N/C}$.

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 following figure shows the electric potential $V$ as a function of position $x$ across $4$ regions on the $x$-axis. Which of the following is correct for the electric field $E$ in these regions?

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:

Let $V$ and $E$ be the potential and electric field intensity at a point,respectively. Then: