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 $\text{ N}$)

If potential (in volts) in a region is expressed as $V(x, y, z) = 6xy - y + 2yz$,the electric field (in $N/C$) at point $(1, 1, 0)$ is:

The electric potential $V$ at any point $(x, y, z)$ (all in metres) in space is given by $V = 4x^2 \text{ volt}$. The electric field at the point $(1 \text{ m}, 0, 2 \text{ m})$ in $\text{volt/metre}$ is

The electric potential at a place is varying as $V = \frac{1}{2}(y^2 - 4x) \text{ V}$. Then the electric field at $x = 1 \text{ m}$ and $y = 1 \text{ m}$ is

The electric potential at any point as a function of distance $(x)$ in meters is given by $V = 5x^2 + 10x - 9 \, (V)$. The value of the electric field at $x = 1 \, m$ is ...... $V m^{-1}$.

The electric field in a region surrounding the origin is uniform and along the $x$-axis. $A$ small circle is drawn with the centre at the origin,cutting the axes at points $A, B, C, D$ having coordinates $(a, 0), (0, a), (-a, 0), (0, -a)$ respectively,as shown in the figure. Then,the potential is minimum at the point: