Assume that an electric field $\overrightarrow{E} = 30x^2 \hat{i}$ exists in space. Then the potential difference $V_A - V_O$,where $V_O$ is the potential at the origin and $V_A$ is the potential at $x = 2 \, m$,is: (in $, V$)

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

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

$A$ uniform electric field of $400 \, V/m$ is directed at an angle of $45^\circ$ above the positive $x$-axis. Calculate the potential difference $V_A - V_B$ in $V$. The coordinates of points $A$ and $B$ are $(0, 2) \, cm$ and $(3, 0) \, cm$ respectively.

In a space having electric field $\vec{E}=A(x \hat{i}+y \hat{j})$,the potential at a point $(10 \ m, 20 \ m)$ is zero. Then the potential at the origin is $\left[A=10 \ Vm^{-2}\right]$. (in $V$)

As shown in the figure, if the values of the electric potential at three points $A, B$ and $C$ in a uniform electric field $(\vec{E})$ are $V_A, V_B$, and $V_C$ respectively, then