For the given figure,the value of the electric field is ......... $V/m$.

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

$A$ uniform electric field having a magnitude $E_0$ and direction along the positive $X$-axis exists. If the potential $V$ is zero at $x = 0$,then its value at $x = +x$ will be:

Electric potential in a region is varying according to the relation $V = \frac{3x^2}{2} - \frac{y^2}{4}$,where $x$ and $y$ are in $m$ and $V$ is in $V$. Electric field intensity (in $N/C$) at a point $(1 \, m, 2 \, m)$ is:

The electric field vector in a region is given by $E = (3 \hat{i} + 4y \hat{j}) \ V \ m^{-1}$. The potential at the origin is zero. Then,the potential at a point $(2, 1) \ m$ is: (in $V$)

In a region,the electric field is $(30 \hat{i} + 40 \hat{j}) \text{ NC}^{-1}$. If the electric potential at the origin is zero,the electric potential at the point $(1 \text{ m}, 2 \text{ m})$ is