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

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?

$A$ uniform electric field of $20\, N/C$ exists along the $x$-axis in a space. The potential difference $(V_B - V_A)$ for the points $A(4\,m, 2\,m)$ and $B(6\,m, 5\,m)$ is.....$V$.

As shown in the figure, if the values of the electric potential at three points $A, B$ and $C$ in a uniform electric field $(\vec{E})$ are $V_A, V_B$, and $V_C$ respectively, then

The figure represents two equipotential lines in the $x-y$ plane for an electric field. The $x$-component $E_{x}$ of the electric field in the space between these equipotential lines is, (in $V/m$)

In a certain region,the electric potential is given by $V = 6x - 8xy^2 - 8y + 6yz - 4z^2 \, V$. The magnitude of the force on a charge of $2 \, C$ at the origin is ........ $N$.