The electric field intensity just sufficient to balance the earth's gravitational attraction on an electron will be: (given mass and charge of an electron respectively are $9.1 \times 10^{-31}\,kg$ and $1.6 \times$ $10^{-19}\,C$.)

Two point charges $A$ and $B$ of magnitude $+8 \times 10^{-6}\,C$ and $-8 \times 10^{-6}\,C$ respectively are placed at a distance $d$ apart. The electric field at the middle point $O$ between the charges is $6.4 \times 10^{4}\,NC ^{-1}$. The distance ' $d$ ' between the point charges $A$ and $B$ is..............$m$

Find the electric field at point $P$ (as shown in figure) on the perpendicular bisector of a uniformly charged thin wire of length $L$ carrying a charge $Q.$ The distance of the point $P$ from the centre of the rod is $a=\frac{\sqrt{3}}{2} L$.

The number of electrons to be put on a spherical conductor of radius $0.1\,m$ to produce an electric field of $0.036\, N/C$ just above its surface is

What is called electric field ?

Two point charges $q_1\,(\sqrt {10}\,\,\mu C)$ and $q_2\,(-25\,\,\mu C)$ are placed on the $x-$ axis at $x = 1\,m$ and $x = 4\,m$ respectively. The electric field (in $V/m$ ) at a point $y = 3\,m$ on $y-$ axis is, [ take ${\mkern 1mu} {\mkern 1mu} \frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}{\mkern 1mu} {\mkern 1mu} N{m^2}{C^{ - 2}}{\rm{ }}$ ]