Electric field in a region is given by $\vec{E} = Ax\hat{i} + By\hat{j}$, where $A = 10 \ V/m^2$ and $B = 5 \ V/m^2$. If the electric potential at a point $(10, 20)$ is $500 \ V$, then the electric potential at the origin is . . . . . . $V$.

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

The electrostatic potential in a charged spherical region of radius $r$ varies as $V = ar^3 + b$, where $a$ and $b$ are constants. The total charge in the sphere of unit radius is $\alpha \times \pi a \epsilon_0$. The value of $\alpha$ is . . . . . . .

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:

In which region is the magnitude of the $x$-component of the electric field maximum,if the potential $(V)$ versus distance $(X)$ graph is as shown?

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 $V$)