Two plates are $2\,cm$ apart,a potential difference of $10\,V$ is applied between them. The electric field between the plates is ......... $N/C$.

Electric potential at any point is $V = -5x + 3y + \sqrt{15} z$,then the magnitude of the electric field is

The potential at a point $x$ (measured in $\mu m$) due to some charges situated on the $x$-axis is given by $V(x) = \frac{20}{x^2 - 4} \text{ volt}$. The electric field $E$ at $x = 4 \mu m$ is given by:

The potential $V$ varies with $x$ and $y$ as $V = \frac{1}{2}(y^2 - 4x) \text{ volts}$. The electric field at $(1 \text{ m}, 1 \text{ m})$ is:

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

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