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 place is varying as $V = \frac{1}{2}(y^2 - 4x) \text{ V}$. Then the electric field at $x = 1 \text{ m}$ and $y = 1 \text{ m}$ is

$ABC$ is a right-angled triangle situated in a uniform electric field $\vec{E}$ which is in the plane of the triangle. The points $A$ and $B$ are at the same potential of $15 \, V$,while the point $C$ is at a potential of $20 \, V$. Given $AB = 3 \, cm$ and $BC = 4 \, cm$. The magnitude of the electric field is (in $S.I.$ units):

The potential $V$ varies with $x$ and $y$ as $V = \frac{1}{2}(y^2 - 4x) \text{ volts}$. The electric field at $(1 \text{ m}, 1 \text{ m})$ is:

Potential gradient is defined as

The following figure shows the electric potential $V$ as a function of position $x$ across $4$ regions on the $x$-axis. Which of the following is correct for the electric field $E$ in these regions?