The electric potential as a function of $x, y$ is given by $V = 5(x^2 - y^2) \text{ V}$. The electric field at a point $(2, 3) \text{ m}$ is . . . . . . $\text{V/m}$.

The electric potential at a point $(x, y)$ in the $x-y$ plane is given by $V = -kxy$. The field intensity at a distance $r$ from the origin varies as

$A$ uniform electric field having a magnitude $E_0$ and direction along the positive $X$-axis exists. If the potential $V$ is zero at $x = 0$,then its value at $x = +x$ 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:

Obtain the relation between electric field and electric potential.

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 ...... $V$.