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 $V$)

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 field in a region of space is given as $E = (5x) \hat{i} \text{ N/C}$. Consider point $A$ on the $Y$-axis at $y = 5 \text{ m}$ and point $B$ on the $X$-axis at $x = 2 \text{ m}$. If the potentials at points $A$ and $B$ are $V_A$ and $V_B$ respectively, then $(V_B - V_A)$ is (in $\text{ V}$)

If the electric potential at a point $P(x, y)$ is given by $V = axy$,then the electric field at a distance $r$ from the origin is proportional to:

$A$ uniform electric field of $20\, N/C$ exists along the $x$-axis in a space. The potential difference $(V_B - V_A)$ for the points $A(4\,m, 2\,m)$ and $B(6\,m, 5\,m)$ is.....$V$.

The electric field in a region is given by $\vec{E} = (Ax + B)\hat{i}$ where $E$ is in $N\,C^{-1}$ and $x$ is in meters. The values of constants are $A = 20\, SI\, \text{unit}$ and $B = 10\, SI\, \text{unit}$. If the potential at $x = 1$ is $V_1$ and that at $x = -5$ is $V_2$, then $V_1 - V_2$ is.....$V$.