The electric potential in a region is represented as $V = 2x + 3y -z$ ; then the expression of electric field strength is

A particle $A$ has charge $+q$ and particle $B$ has charge $+ 4q$ with each of them having the same mass $m$. When allowed to fall from rest through same electrical  potential difference, the ratio of their speed $V_A : V_B$ will be :-

Figure shows three points $A$, $B$ and $C$ in a region of uniform electric field $\overrightarrow E $. The line $AB$ is perpendicular and $BC$ is parallel to the field lines. Then which of the following holds good. Where ${V_A} > {V_B}$ and ${V_C}$ represent the electric potential at points $A$, $B$ and $C$ respectively

$A B C$ is a right angled triangle situated in a uniform electric field $\vec{E}$ which is in the plane of the triangle. The points $A$ and $B$ are at the same potential of $15 \,V$ while the point $C$ is at a potential of $20 \,V . A B=3 \,cm$ and $B C=4 \,cm$. The magnitude of electric field is (in $S.I.$ Units)

The potential at a point $x$ (measured in $μ\ m$) due to some charges situated on the $ x$-axis is given by $V(x)$ =$\frac{{20}}{{{x^2} - 4}}$ $volt$ The electric field $E$ at $x = 4\ μ m$ is given by

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