Potential in the $x-y$ plane is given as $V = 5(x^2 + xy) \, V$. The electric field at the point $(1, -2)$ will be

Potential gradient is defined as

Assume that an electric field $E = 30x^2 \hat{i}$ exists in space. If $V_0$ is the potential at the origin and $V_A$ is the potential at $x = 2 \ m$,then the potential difference $(V_A - V_0)$ is: (in $V$)

Two points $A$ and $B$ on the $Y$-axis are at distances of $12.3 \ cm$ and $12.5 \ cm$ from the origin,respectively. The potentials at these points are $56 \ V$ and $54.8 \ V$,respectively. What is the component of the force on a charge of $4 \ \mu C$ placed at point $A$ on the $Y$-axis?

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 . . . . . . $V$.

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: