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

If the electric potential is given by $V = (5x^2 + 10x - 9) \ V$,what is the electric field at $x = 1 \ m$ in $V/m$?

An infinite non-conducting sheet has a surface charge density of $7 \times 10^{-7} \text{ C m}^{-2}$ on one side. The distance between equipotential surfaces whose potentials differ by $19.8 \text{ V}$ will be (Take $\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^9 \text{ SI units}$) (in $\text{ mm}$)

Let $V$ and $E$ be the potential and electric field intensity at a point,respectively. Then:

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 $V$ is given as a function of distance $x$ (metre) by $V = (5x^2 + 10x - 9) \text{ V}$. The value of the electric field at $x = 1 \text{ m}$ is ...... $V/m$.