If the electric potential is given by $V = 4x + 3y$,what is the electric field at the point $(2, 1)$?

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 field $\vec E$ between two points is constant in both magnitude and direction. Consider a path of length $d$ at an angle $\theta = 60^o$ with respect to the field lines as shown in the figure. The potential difference between points $1$ and $2$ is

If $V = -5x + 3y + \sqrt{15}z$,then find the magnitude of the electric field $E(x, y, z)$ in units.

Two plates are separated by a distance of $20 \, cm$. The potential difference between them is $10 \, V$. The electric field between the two plates is ...... $V m^{-1}$.

Some equipotential surfaces are shown in the figure below. The magnitude of the electric field will be $:-$ (in $V/m$)