$A$ 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 region,if the electric field is defined as $\vec{E} = (\hat{i} + 2\hat{j} + \hat{k}) \text{ V/m}$,then the potential difference between two points $A(0, 0, 0)$ and $B(2, 3, 4)$ in that region is ...... $V$.

The electric potential $V$ at any point $(x, y, z)$ in space is given by the equation $V = 4x^2 \, V$,where $x, y$,and $z$ are all in meters. The electric field at the point $(1 \, m, 0, 2 \, m)$ in $V/m$ is:

$A$ charge of $5\,C$ experiences a force of $5000\,N$ when it is kept in a uniform electric field. What is the potential difference $V$ (in volts) between two points separated by a distance of $1\,cm$?

What is the angle between the maximum value of the potential gradient and the equipotential surface?

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: