For a given electric field $\vec{E} = 2x\hat{i} + 3y\hat{j}$,find the potential at $(X, Y)$ if the potential at the origin is $5\, V$.

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 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:

$A$ charge of $5\,C$ experiences a force of $5000\,N$ when it is kept in a uniform electric field. What is the potential difference $V$ (in volts) between two points separated by a distance of $1\,cm$?

The electric field in a region is given by $\vec{E} = (Ax + B)\hat{i}$ where $E$ is in $N\,C^{-1}$ and $x$ is in meters. The values of constants are $A = 20\, SI\, \text{unit}$ and $B = 10\, SI\, \text{unit}$. If the potential at $x = 1$ is $V_1$ and that at $x = -5$ is $V_2$, then $V_1 - V_2$ is.....$V$.

$A$ uniform electric field of $20\, N/C$ exists along the $x$-axis in a space. The potential difference $(V_B - V_A)$ for the points $A(4\,m, 2\,m)$ and $B(6\,m, 5\,m)$ is.....$V$.