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

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 potential $V(x, y, z)$ for a planar charge distribution is given by:
$V(x, y, z) = \begin{cases} 0 & \text{for } x < -d \\ -V_0(1 + \frac{x}{d})^2 & \text{for } -d \le x < 0 \\ -V_0(1 + 2\frac{x}{d}) & \text{for } 0 \le x < d \\ -3V_0 & \text{for } x \ge d \end{cases}$
where $-V_0$ is the potential at the origin and $d$ is a distance. The graph of the electric field as a function of position is:

Two plates are at potentials $-10\, V$ and $+30\, V$. If the separation between the plates is $2\, cm$,the electric field between them is.......$V/m$.

Determine the electric field strength vector if the potential of this field depends on $x, y$ coordinates as $V = 10axy$.

Write the relation between the electric field of an electric charge and electrostatic potential at any point.