Electric potential is given by $V = 6x - 8xy^2 - 8y + 6yz - 4z^2$. Then,the electric force acting on a $2 \, C$ point charge placed at the origin will be......$N$.

Find the potential $V$ of an electrostatic field $\vec{E} = a(y\hat{i} + x\hat{j})$,where $a$ is a constant.

Two plates are $20 \ cm$ apart and a potential difference of $10 \ V$ is applied between them. The electric field between the plates is . . . . . . . (in $Vm^{-1}$)

If on the $x$-axis electric potential decreases uniformly from $60 \,V$ to $20 \,V$ between $x=-2 \,m$ to $x=+2 \,m$,then the magnitude of electric field at the origin

The figure shows two equipotential lines in the $x, y$ plane for an electric field. The scales are marked in $\text{cm}$. The $x$-component $E_x$ and $y$-component $E_y$ of the electric field in the space between these equipotential lines are respectively:

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 $V$)