Electric potential at any point is $V = -5x + 3y + \sqrt{15} z$,then the magnitude of the electric field is

The electric potential $V$ at any point $(x, y, z)$ in space is given by the equation $V = 4x^2 \, V$,where $x, y$,and $z$ are all in meters. The electric field at the point $(1 \, m, 0, 2 \, m)$ in $V/m$ is:

The potential gradient is a

The electric potential at a point in free space due to charge $Q$ coulomb is $V = Q \times 10^{11} \, V$. The electric field at that point is:

The electric potential $(V)$ as a function of distance $(x)$ [in meters] is given by $V = (5x^2 + 10x - 9) \, V$. The value of the electric field at $x = 1 \, m$ would be ...... $V/m$.

If the electric potential at any point $(x, y, z) \, m$ in space is given by $V = 3x^2$ volt,the electric field at the point $(1, 0, 3) \, m$ will be ............