The electric potential $V$ is given as a function of distance $x$ (metre) by $V = (4x^2 + 8x - 3) \ V$. The value of the electric field at $x = 0.5 \ m$,in $V/m$,is:

The variation of electric potential $V$ with distance $x$ from a fixed point is shown in the figure. What is the value of the electric field at $x = 2 \ m$?

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

In a certain region of space with volume $0.2 \ m^3$,the electric potential is found to be $5 \ V$ throughout. The magnitude of the electric field in this region is . . . . . . $N/C$.

$A$ 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: