Two plates are $2\,cm$ apart, a potential difference of $10\;volt$ is applied between them, the electric field between the plates is.........$N/C$

Electric potential in a region is varying according to the relation $V=\frac{3 x^2}{2}-\frac{y^2}{4}$, where $x$ and $y$ are in metre and $V$ is in volt. Electric field intensity (in $N/C$) at a point $(1 \,m , 2 \,m$ ) is ......

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

$A B C$ 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)

Two metal pieces having a potential difference of $800 \;V$ are $0.02\; m$ apart horizontally. A particle of mass $1.96 \times 10^{-15} \;kg$ is suspended in equilibrium between the plates. If $e$ is the elementary charge, then charge on the particle is

Two plates are at potentials $-10\, V$ and $+30\, V$. If the separation between the plates be $2\, cm$. The electric field between them is.......$V/m$