The potential difference between two points $A(2, 1, 0) \ m$ and $B(0, 2, 4) \ m$ in an electric field $\vec{E} = (x \hat{i} - 2y \hat{j} + z \hat{k}) \ Vm^{-1}$ is: (in $V$)

The electric field in a region surrounding the origin is uniform and along the $x$-axis. $A$ small circle is drawn with the centre at the origin,cutting the axes at points $A, B, C, D$ having coordinates $(a, 0), (0, a), (-a, 0), (0, -a)$ respectively,as shown in the figure. Then,the potential is minimum at the point:

Two large plane parallel conducting plates are kept $10 \ cm$ apart as shown in the figure. The potential difference between them is $V$. The potential difference between the points $A$ and $B$ (shown in the figure) 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 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.....$V$.

If $V = -5x + 3y + \sqrt{15}z$,then find the magnitude of the electric field $E(x, y, z)$ in units.