The figure shows the variation of electric field intensity $E$ versus distance $x$. What is the potential difference between the points at $x = 2 \, m$ and $x = 6 \, m$ from $O$ (in $V$)?

The figure shows a set of equipotential surfaces. The magnitude and direction of the electric field that exists in the region is .........

The potential $\phi(x, y)$ of an electrostatic field $\vec{E} = a(y \hat{i} + x \hat{j})$ is [where $a$ is a constant and $\hat{i}$ and $\hat{j}$ are unit vectors along $X$ and $Y$ axes].

The electric potential at a point in free space due to charge $Q$ coulomb is $V = Q \times 10^{11} \, V$. The electric field at that point is:

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 ...... $V$.

If potential (in volts) in a region is expressed as $V(x, y, z) = 6xy - y + 2yz$,the electric field (in $N/C$) at point $(1, 1, 0)$ is: