$A$ uniform electric field of $400 \, V/m$ is directed $45^\circ$ above the $x$-axis. The potential difference $V_A - V_B$ is: (Coordinates are in cm) (in $, V$)

The electrostatic potential inside a charged spherical ball is given by $V = b - ar^2$,where $r$ is the distance from the centre; $a$ and $b$ are constants. Then,the charge density inside the ball is:

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:

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 electric potential at a point $(x, y, z)$ is given by $V = -x^2y - xz^3 + 4$. The electric field at that point is:

The electric potential in some region is expressed by $V = 6x - 8xy^2 - 8y + 6yz - 4z^2 \text{ volt}$. The magnitude of the electric force acting on a charge of $2 \text{ C}$ situated at the origin will be $-$ (in $\text{ N}$)