Potential in the $x-y$ plane is given as $V = 5(x^2 + xy) \, V$. The electric field at the point $(1, -2)$ will be

Due to some charges on the $x$-axis,the potential at a point on the $x$-axis is given by $V(x) = 20/(x^2 - 4) \ V$. The electric field at $x = 4 \ \mu m$ is given by:

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

$A$ 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 and electric potential at a point due to a point charge are $500 \ NC^{-1}$ and $30 \ V$ respectively. The magnitude of the charge 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 $\text{ N}$)